Use the Wronskian to show whether the give set of functions is linearly dependent or linearly independent.
a. y 1 = x y 2 = x + 1 y_1=x \qquad y_2=x+1 y 1 = x y 2 = x + 1
y 1 = x , y 1 ′ = 1 y 2 = x + 1 , y 2 ′ = 1 \begin{aligned}
y_1&=x, \qquad &y'_1&=1\\
y_2&=x+1, \qquad &y'_2&=1
\end{aligned} y 1 y 2 = x , = x + 1 , y 1 ′ y 2 ′ = 1 = 1 W ( y 1 , y 2 ) = ∣ x x + 1 1 1 ∣ = x − ( x − 1 ) = 1 ≠ 0 \begin{aligned}
W(y_1,y_2)&=\left| \begin{array}{cc}x&x+1\\1&1\end{array} \right|\\
&=x-(x-1)\\
&=1 \neq 0
\end{aligned} W ( y 1 , y 2 ) = x 1 x + 1 1 = x − ( x − 1 ) = 1 = 0 Since W ( x , x + 1 ) ≠ 0 W(x,x+1)\neq 0 W ( x , x + 1 ) = 0 { x , x + 1 } \{ x, x+1 \} { x , x + 1 }
b. y 1 = e α x sin β x y 2 = e α x cos β x y_1=e^{\alpha x}\sin\beta x \qquad y_2=e^{\alpha x}\cos\beta x y 1 = e αx sin β x y 2 = e αx cos β x
y 1 = e α x sin β x , y 1 ′ = β e α x cos β x + α e α x sin β x y 2 = e α x cos β , y 2 ′ = − β e α x sin β x + α e α x cos β x \begin{aligned}
y_1&=e^{\alpha x}\sin\beta x, \qquad &y'_1&=\beta e^{\alpha x}\cos\beta x+ \alpha e^{\alpha x}\sin \beta x\\
y_2&=e^{\alpha x}\cos\beta, \qquad &y'_2&=-\beta e^{\alpha x}\sin\beta x+ \alpha e^{\alpha x}\cos \beta x
\end{aligned} y 1 y 2 = e αx sin β x , = e αx cos β , y 1 ′ y 2 ′ = β e αx cos β x + α e αx sin β x = − β e αx sin β x + α e αx cos β x W ( y 1 , y 2 ) = ∣ e α x sin β x e α x cos β x β e α x cos β x + α e α x sin β x − β e α x sin β x + α e α x cos β x ∣ = − β e 2 α x sin 2 β x + α e 2 α x sin β x cos β x − β e 2 α x cos 2 β x − α e 2 α x sin β x cos β x = − β e 2 α x ( s i n 2 β x + cos 2 β x ) = − β e 2 α x \begin{aligned}
W(y_1,y_2)&=\left| \begin{array}{cc}e^{\alpha x}\sin\beta x&e^{\alpha x}\cos\beta x\\\beta e^{\alpha x}\cos\beta x+ \alpha e^{\alpha x}\sin \beta x&-\beta e^{\alpha x}\sin\beta x+ \alpha e^{\alpha x}\cos \beta x\end{array} \right|\\
&=-\beta e^{2\alpha x}\sin^2 \beta x+\alpha e^{2\alpha x}\sin\beta x\cos\beta x - \beta e^{2\alpha x}\cos^2\beta x - \alpha e^{2\alpha x}\sin\beta x\cos\beta x\\
&=-\beta e^{2\alpha x}(sin^2\beta x+\cos^2\beta x)\\
&=-\beta e^{2\alpha x}
\end{aligned} W ( y 1 , y 2 ) = e αx sin β x β e αx cos β x + α e αx sin β x e αx cos β x − β e αx sin β x + α e αx cos β x = − β e 2 αx sin 2 β x + α e 2 αx sin β x cos β x − β e 2 αx cos 2 β x − α e 2 αx sin β x cos β x = − β e 2 αx ( s i n 2 β x + cos 2 β x ) = − β e 2 αx Since β ≠ 0 \beta\neq0 β = 0 e 2 α x ≠ 0 e^{2\alpha x}\neq 0 e 2 αx = 0 W ( y 1 , y 2 ) ≠ 0 W(y_1,y_2)\neq 0 W ( y 1 , y 2 ) = 0
∴ \therefore ∴
Verify that each of the given functions is a solution of the differential equation, and use their Wronskian to show that these solutions are linearly independent.
Verify the linear combination of the solutions is also a solution.
a. y 1 = e x y 2 = e − 2 x y ′ ′ + y ′ − 2 y = 0 y_1=e^x\qquad y_2=e^{-2x}\qquad y''+y'-2y=0 y 1 = e x y 2 = e − 2 x y ′′ + y ′ − 2 y = 0
y 1 = e x , y 1 ′ = e x y 1 ′ ′ = e x y 2 = e − 2 x y 2 ′ = − 2 e − 2 x y 2 ′ ′ = 4 e − 2 x \begin{aligned}
y_1&=e^x,\qquad &y_1'&=e^x\qquad &y_1''&=e^x\\
y_2&=e^{-2x}\qquad &y_2'&=-2e^{-2x}\qquad &y_2''&=4e^{-2x}
\end{aligned} y 1 y 2 = e x , = e − 2 x y 1 ′ y 2 ′ = e x = − 2 e − 2 x y 1 ′′ y 2 ′′ = e x = 4 e − 2 x For
y 1 → e x + e x − 2 e x = 0 y 2 → 4 e − 2 x + ( − 2 e − 2 x ) − 2 ( e − 2 x ) = 0 \begin{aligned}
y_1&\rightarrow e^x+e^x-2e^x=0\\
y_2&\rightarrow 4e^{-2x}+(-2e^{-2x})-2(e^{-2x})=0
\end{aligned} y 1 y 2 → e x + e x − 2 e x = 0 → 4 e − 2 x + ( − 2 e − 2 x ) − 2 ( e − 2 x ) = 0 Both are solution.
W ( y 1 , y 2 ) = ∣ e x e − 2 x e x − 2 e − 2 x ∣ = − 2 e − x − e − x = − 3 e − x ≠ 0 \begin{aligned}
W(y_1,y_2)&=\left| \begin{array}{cc}e^x&e^{-2x}\\e^x&-2e^{-2x}\end{array} \right|\\
&=-2e^{-x}-e^{-x}\\
&=-3e^{-x}\neq 0
\end{aligned} W ( y 1 , y 2 ) = e x e x e − 2 x − 2 e − 2 x = − 2 e − x − e − x = − 3 e − x = 0 Since e − x ≠ 0 e^{-x}\neq0 e − x = 0 W ( y 1 , y 2 ) ≠ 0 W(y_1,y_2)\neq 0 W ( y 1 , y 2 ) = 0
∴ \therefore ∴
b. y 1 = x 2 y 2 = x − 1 x 2 y ′ ′ − 2 y = 0 y_1=x^2\qquad y_2=x^{-1}\qquad x^2y''-2y=0 y 1 = x 2 y 2 = x − 1 x 2 y ′′ − 2 y = 0
y 1 = x 2 , y 1 ′ = 2 x , y 1 ′ ′ = 2 y 2 = x − 1 , y 2 ′ = − x − 2 , y 2 ′ ′ = − 2 x − 3 \begin{aligned}
y_1&=x^2,\qquad &y_1'&=2x,\qquad &y_1''=2\\
y_2&=x^{-1},\qquad &y_2'&=-x^{-2},\qquad &y_2''=-2x^{-3}
\end{aligned} y 1 y 2 = x 2 , = x − 1 , y 1 ′ y 2 ′ = 2 x , = − x − 2 , y 1 ′′ = 2 y 2 ′′ = − 2 x − 3 For
y 1 → x 2 y ′ ′ − 2 y = x 2 ( 2 ) − 2 ( x 2 ) = 0 y 2 → x 2 ( 2 x − 3 ) − 2 ( x − 1 ) = 2 x − 1 − 2 x − 1 = 0 \begin{aligned}
y_1&\rightarrow x^2y''-2y=x^2(2)-2(x^2)=0\\
y_2&\rightarrow x^2(2x^{-3})-2(x^{-1})=2x^{-1}-2x^{-1}=0
\end{aligned} y 1 y 2 → x 2 y ′′ − 2 y = x 2 ( 2 ) − 2 ( x 2 ) = 0 → x 2 ( 2 x − 3 ) − 2 ( x − 1 ) = 2 x − 1 − 2 x − 1 = 0 Both are solution.
W ( y 1 , y 2 ) = ∣ x 2 x − 1 2 x − x − 2 ∣ = − x 0 − 2 x 0 = − 3 ≠ 0 \begin{aligned}
W(y_1,y_2)&=\left| \begin{array}{cc}x^2&x^{-1}\\2x&-x^{-2}\end{array} \right|\\
&=-x^0-2x^0\\
&=-3\neq 0
\end{aligned} W ( y 1 , y 2 ) = x 2 2 x x − 1 − x − 2 = − x 0 − 2 x 0 = − 3 = 0 ∴ \therefore ∴
c. y 1 = e − x y 2 = x e − x y ′ ′ + 2 y ′ + y = 0 y_1=e^{-x}\qquad y_2=xe^{-x}\qquad y''+2y'+y=0 y 1 = e − x y 2 = x e − x y ′′ + 2 y ′ + y = 0
y 1 = e − x y 1 ′ = − e − x y 1 ′ ′ = e − x y 2 = x e − x y 2 ′ = − x e − x + e − x y 2 ′ ′ = x e − x − e − x − e − x = x e − x − 2 e − x \begin{aligned}
y_1&=e^{-x} \qquad &y_1'&=-e^{-x} \qquad &y_1''&=e^{-x}\\
y_2&=xe^{-x} \qquad &y_2'&=-xe^{-x}+e^{-x} \qquad &y_2''&=xe^{-x}-e^{-x}-e^{-x}=xe^{-x}-2e^{-x}
\end{aligned} y 1 y 2 = e − x = x e − x y 1 ′ y 2 ′ = − e − x = − x e − x + e − x y 1 ′′ y 2 ′′ = e − x = x e − x − e − x − e − x = x e − x − 2 e − x For
y 1 → y ′ ′ + 2 y ′ + y = e − x + 2 ( − e − x ) + e − x = 0 y 2 → ( x e − x − 2 e − x ) + 2 ( − x e − x + e − x ) + 2 e − x = 0 \begin{aligned}
y_1&\rightarrow y''+2y'+y=e^{-x}+2(-e^{-x})+e^{-x}=0\\
y_2&\rightarrow (xe^{-x}-2e^{-x})+2(-xe^{-x}+e^{-x})+2e^{-x}=0
\end{aligned} y 1 y 2 → y ′′ + 2 y ′ + y = e − x + 2 ( − e − x ) + e − x = 0 → ( x e − x − 2 e − x ) + 2 ( − x e − x + e − x ) + 2 e − x = 0 Both are solution.
W ( y 1 , y 2 ) = ∣ e − x x e − x − e − x x e − x + e − x ∣ = ( − x e − 2 x + e − 2 x ) + ( x e − 2 x ) = e − 2 x ≠ 0 \begin{aligned}
W(y_1,y_2)&=\left| \begin{array}{cc}e^{-x}&xe^{-x}\\-e^{-x}&xe^{-x}+e^{-x}\end{array} \right|\\
&=(-xe^{-2x}+e^{-2x})+(xe^{-2x})\\
&=e^{-2x}\neq 0
\end{aligned} W ( y 1 , y 2 ) = e − x − e − x x e − x x e − x + e − x = ( − x e − 2 x + e − 2 x ) + ( x e − 2 x ) = e − 2 x = 0 ∴ \therefore ∴
Solve the following second order ODEs
a. y ′ ′ − 4 y = 0 y''-4y=0 y ′′ − 4 y = 0
y ′ ′ − 4 y = 0 m 2 − 4 = 0 m = ± 2 ⇒ y = C 1 e 2 x + C 2 e − 2 x \begin{aligned}
y''-4y&=0\\
m^2-4&=0\\
m&=\pm 2\\
\Rightarrow y&=C_1e^{2x}+C_2 e^{-2x}
\end{aligned} y ′′ − 4 y m 2 − 4 m ⇒ y = 0 = 0 = ± 2 = C 1 e 2 x + C 2 e − 2 x b. y ′ ′ − y ′ − 6 y = 0 y''-y'-6y=0 y ′′ − y ′ − 6 y = 0
y ′ ′ − y ′ + 7 y = 0 m 2 − m + 6 = 0 ( m − 3 ) ( m + 2 ) = 0 m = 3 , − 2 ⇒ y = C 1 e 3 x + C 2 e − 2 x \begin{aligned}
y''-y'+7y&=0\\
m^2-m+6&=0\\
(m-3)(m+2)&=0\\
m&=3,-2\\
\Rightarrow y&=C_1e^{3x}+C_2e^{-2x}
\end{aligned} y ′′ − y ′ + 7 y m 2 − m + 6 ( m − 3 ) ( m + 2 ) m ⇒ y = 0 = 0 = 0 = 3 , − 2 = C 1 e 3 x + C 2 e − 2 x c. 6 y ′ ′ + y ′ − y = 0 6y''+y'-y=0 6 y ′′ + y ′ − y = 0
6 y ′ ′ + y ′ − y = 0 6 m 2 + m − 1 = 0 ( 3 m − 1 ) ( 2 m + 1 ) = 0 m = 1 3 , − 1 2 ⇒ y = C 1 e 1 3 x + C 2 e − 1 2 x \begin{aligned}
6y''+y'-y&=0\\
6m^2+m-1&=0\\
(3m-1)(2m+1)&=0\\
m&=\frac{1}{3}, -\frac{1}{2}\\
\Rightarrow y&=C_1e^{\frac{1}{3}x}+C_2e^{-\frac{1}{2}x}
\end{aligned} 6 y ′′ + y ′ − y 6 m 2 + m − 1 ( 3 m − 1 ) ( 2 m + 1 ) m ⇒ y = 0 = 0 = 0 = 3 1 , − 2 1 = C 1 e 3 1 x + C 2 e − 2 1 x d. y ′ ′ + y = 0 y''+y=0 y ′′ + y = 0
y ′ ′ + y = 0 m 2 + 1 = 0 m = − 1 = ± i y 1 = C 1 e i x , y 2 = C 1 e − i x ⇒ y = e i x ( C 1 cos x + C 2 sin x ) = C 1 cos x + C 2 sin x \begin{aligned}
y''+y&=0\\
m^2+1&=0\\
m&=\sqrt{-1}=\pm i \\
y_1=C_1e^{ix} &, y_2=C_1e^{-ix}\\
\Rightarrow y&=e^{ix}(C_1\cos x+C_2\sin x)\\
&=C_1\cos x+C_2\sin x
\end{aligned} y ′′ + y m 2 + 1 m y 1 = C 1 e i x ⇒ y = 0 = 0 = − 1 = ± i , y 2 = C 1 e − i x = e i x ( C 1 cos x + C 2 sin x ) = C 1 cos x + C 2 sin x e. y ′ ′ + 4 y ′ + 8 y = 0 y''+4y'+8y=0 y ′′ + 4 y ′ + 8 y = 0
y ′ ′ + 4 y ′ + 8 y = 0 m 2 + 4 m + 8 = 0 m = − 4 ± 4 2 − 4 ( 1 ) ( 8 ) 2 ( 1 ) = 1 2 ( − 4 ± 16 − 32 ) = 1 2 ( − 4 ± 4 i ) = − 2 ± 2 i y 1 = C 1 e ( − 2 + 2 i ) x , y 2 = C 2 e ( − 2 − 2 i ) x y = e − 2 x ( C 1 cos 2 x + C 2 sin 2 x ) \begin{aligned}
y''+4y'+8y&=0\\
m&2+4m+8&=0\\
m&=\frac{-4\pm\sqrt{4^2-4(1)(8)}}{2(1)}\\
&=\frac{1}{2}\left(-4\pm\sqrt{16-32}\right)\\
&=\frac{1}{2}(-4\pm 4i)\\&=-2\pm 2i\\
y_1=C_1e^{(-2+2i)x} &, y_2=C_2e^{(-2-2i)x} \\
y&=e^{-2x}(C_1\cos 2x + C_2\sin 2x)
\end{aligned} y ′′ + 4 y ′ + 8 y m m y 1 = C 1 e ( − 2 + 2 i ) x y = 0 2 + 4 m + 8 = 2 ( 1 ) − 4 ± 4 2 − 4 ( 1 ) ( 8 ) = 2 1 ( − 4 ± 16 − 32 ) = 2 1 ( − 4 ± 4 i ) = − 2 ± 2 i , y 2 = C 2 e ( − 2 − 2 i ) x = e − 2 x ( C 1 cos 2 x + C 2 sin 2 x ) = 0
Solve the following second order ODEs with initial-value conditions
a. y ′ ′ − 4 y = 0 y ( 0 ) = 4 y ′ ( 0 ) = 12 y''-4y=0\qquad\qquad y(0)=4\qquad y'(0)=12 y ′′ − 4 y = 0 y ( 0 ) = 4 y ′ ( 0 ) = 12
y ′ ′ − 4 y = 0 m 2 = 4 m = ± 2 \begin{aligned}
y''-4y&=0\\m^2&=4\\m&=\pm2\\
\end{aligned} y ′′ − 4 y m 2 m = 0 = 4 = ± 2 General Solution, y = C 1 e 2 x + C 2 e − 2 x y=C_1e^{2x}+C_2e^{-2x} y = C 1 e 2 x + C 2 e − 2 x y ′ = 2 C 1 e 2 x − 2 C 2 e − 2 x y'=2C_1e^{2x}-2C_2e^{-2x} y ′ = 2 C 1 e 2 x − 2 C 2 e − 2 x
When
x = 0 , y = 4 ⇒ 4 = C 1 + C 2 (1) x=0, y=4 \qquad \Rightarrow \qquad 4=C_1+C_2 \tag{1} x = 0 , y = 4 ⇒ 4 = C 1 + C 2 ( 1 ) x = 0 , y ′ = 12 ⇒ 12 = 2 C 1 − 2 C 2 (2) x=0, y'=12 \qquad \Rightarrow \qquad 12=2C_1-2C_2 \tag{2} x = 0 , y ′ = 12 ⇒ 12 = 2 C 1 − 2 C 2 ( 2 ) Take ( 2 ) + 2 × ( 1 ) (2)+2\times(1) ( 2 ) + 2 × ( 1 )
( 12 + 8 ) = ( 2 C 1 + 2 C 2 ) + ( 2 C 1 − 2 C 2 ) 20 = 4 C 1 C 1 = 5 \begin{aligned}
(12+8)&=(2C_1+2C_2)+(2C_1-2C_2)\\
20&=4C_1\\
C_1&=5
\end{aligned} ( 12 + 8 ) 20 C 1 = ( 2 C 1 + 2 C 2 ) + ( 2 C 1 − 2 C 2 ) = 4 C 1 = 5 Into ( 1 ) (1) ( 1 )
4 = 5 + C 2 ⇒ C 2 = − 1 4=5+C_2 \qquad \Rightarrow \qquad C_2=-1 4 = 5 + C 2 ⇒ C 2 = − 1 Thus,
y = 5 e 2 x − e − 2 x y=5e^{2x}-e^{-2x} y = 5 e 2 x − e − 2 x b. y ′ ′ − 6 y + 9 y = 0 y ( 0 ) = − 1.4 y ′ ( 0 ) = 4.6 y''-6y+9y=0 \qquad \qquad y(0)=-1.4 \qquad y'(0)=4.6 y ′′ − 6 y + 9 y = 0 y ( 0 ) = − 1.4 y ′ ( 0 ) = 4.6
y ′ ′ − 6 y ′ + 9 y = 0 m 2 − 6 m + 0 = 0 ( m − 3 ) 2 = 0 m = 3 \begin{aligned}
y''-6y'+9y&=0\\m^2-6m+0&=0\\(m-3)^2&=0\\m&=3
\end{aligned} y ′′ − 6 y ′ + 9 y m 2 − 6 m + 0 ( m − 3 ) 2 m = 0 = 0 = 0 = 3 General Solution, y = C 1 e 3 x + C 2 x e 3 x y=C_1e^{3x}+C_2xe^{3x} y = C 1 e 3 x + C 2 x e 3 x y ′ = 3 C 1 e 3 x + C 2 e 3 x + 3 x C 2 e 3 x y'=3C_1e^{3x}+C_2e^{3x}+3xC_2e^{3x} y ′ = 3 C 1 e 3 x + C 2 e 3 x + 3 x C 2 e 3 x
When
x = 0 , y = − 1.4 ⇒ − 1.4 = C 1 + 0 C 1 = − 1.4 \begin{aligned}
x=0, y=-1.4 \qquad \Rightarrow \qquad -1.4&=C_1+0 \\
C_1&=-1.4
\end{aligned} x = 0 , y = − 1.4 ⇒ − 1.4 C 1 = C 1 + 0 = − 1.4 x = 0 , y ′ = 4.6 ⇒ 4.6 = 3 C 1 + C 2 + 0 4.6 = 3 ( − 1.4 ) + C 2 C 2 = 8.8 \begin{aligned}
x=0,y'=4.6 \qquad \Rightarrow \qquad 4.6&=3C_1+C_2+0 \\
4.6 &=3(-1.4)+C_2 \\
C_2&=8.8
\end{aligned} x = 0 , y ′ = 4.6 ⇒ 4.6 4.6 C 2 = 3 C 1 + C 2 + 0 = 3 ( − 1.4 ) + C 2 = 8.8 Thus,
y = − 1.4 e − 3 x + 8.8 x e − 3 x y=-1.4e^{-3x}+8.8xe^{-3x} y = − 1.4 e − 3 x + 8.8 x e − 3 x c. 4 y ′ ′ − 8 y + 3 y = 0 y ( 1 ) = 4 5 e 1 2 y ′ ( 1 ) = 2 5 e 1 2 4y''-8y+3y=0 \qquad\qquad y(1)=\frac{4}{5}e^{\frac{1}{2}} \qquad y'(1)=\frac{2}{5}e^{\frac{1}{2}} 4 y ′′ − 8 y + 3 y = 0 y ( 1 ) = 5 4 e 2 1 y ′ ( 1 ) = 5 2 e 2 1
4 y ′ ′ − 8 y ′ + 3 y = 0 4 m 2 − 8 m + 3 = 0 ( 2 m − 3 ) ( 2 m − 1 ) = 0 m = 1 2 , 3 2 \begin{aligned}
4y''-8y'+3y&=0\\
4m^2-8m+3&=0\\
(2m-3)(2m-1)=0\\
m&=\frac{1}{2}, \frac{3}{2}
\end{aligned} 4 y ′′ − 8 y ′ + 3 y 4 m 2 − 8 m + 3 ( 2 m − 3 ) ( 2 m − 1 ) = 0 m = 0 = 0 = 2 1 , 2 3 General solution. y = C 1 e 1 2 x + C 2 e 3 2 x y=C_1e^{\frac{1}{2}x}+C_2e^{\frac{3}{2}x} y = C 1 e 2 1 x + C 2 e 2 3 x y ′ = 1 2 C 1 e 1 2 x + 3 2 C 2 e 3 2 x y'=\frac{1}{2}C_1e^{\frac{1}{2}x}+\frac{3}{2}C_2e^{\frac{3}{2}x} y ′ = 2 1 C 1 e 2 1 x + 2 3 C 2 e 2 3 x
When
x = 1 , y = 4 5 e 1 2 ⇒ 4 5 e 1 2 = C 1 e 1 2 + C 2 e 3 2 ( 4 5 − C 1 ) e 1 2 = C 2 e 3 2 (1) \begin{aligned}
x=1, y=\frac{4}{5}e^{\frac{1}{2}} \qquad \Rightarrow \qquad \frac{4}{5}e^{\frac{1}{2}}&=C_1e^{\frac{1}{2}}+C_2e^{\frac{3}{2}} \\
\left(\frac{4}{5}-C_1\right)e^{\frac{1}{2}}&=C_2e^{\frac{3}{2}} \tag{1}
\end{aligned} x = 1 , y = 5 4 e 2 1 ⇒ 5 4 e 2 1 ( 5 4 − C 1 ) e 2 1 = C 1 e 2 1 + C 2 e 2 3 = C 2 e 2 3 ( 1 ) x = 1 , y ′ = 2 5 e 1 2 ⇒ 2 5 e 1 2 = 1 2 C 1 e 1 2 + 3 2 C 2 e 3 2 ( 2 5 − 1 2 C 1 ) e 1 2 = 3 2 C 2 e 3 2 2 3 ( 2 5 − 1 2 C 1 ) e 1 2 = C 2 e 3 2 (2) \begin{aligned}
x=1, y'=\frac{2}{5}e^{\frac{1}{2}} \qquad \Rightarrow \qquad \frac{2}{5}e^{\frac{1}{2}} &= \frac{1}{2}C_1e^{\frac{1}{2}}+\frac{3}{2}C_2e^{\frac{3}{2}} \\
\left(\frac{2}{5}-\frac{1}{2}C_1\right) e^{\frac{1}{2}} &=\frac{3}{2}C_2e^{\frac{3}{2}} \\
\frac{2}{3}\left(\frac{2}{5}-\frac{1}{2}C_1\right)e^{\frac{1}{2}} &= C_2e^{\frac{3}{2}} \tag{2}
\end{aligned} x = 1 , y ′ = 5 2 e 2 1 ⇒ 5 2 e 2 1 ( 5 2 − 2 1 C 1 ) e 2 1 3 2 ( 5 2 − 2 1 C 1 ) e 2 1 = 2 1 C 1 e 2 1 + 2 3 C 2 e 2 3 = 2 3 C 2 e 2 3 = C 2 e 2 3 ( 2 ) Let ( 1 ) = ( 2 ) (1) = (2) ( 1 ) = ( 2 )
( 4 5 − C 1 ) e 1 2 = ( 4 15 − 1 3 C 1 ) e 1 2 ( 4 5 − 4 15 ) = ( 1 − 1 3 ) C 1 2 3 C 1 = 8 15 C 1 = 4 5 \begin{aligned}
\left(\frac{4}{5}-C_1\right)e^{\frac{1}{2}}&=\left(\frac{4}{15}-\frac{1}{3}C_1\right)e^{\frac{1}{2}}\\
\left(\frac{4}{5}-\frac{4}{15}\right)&=\left(1-\frac{1}{3}\right)C_1\\
\frac{2}{3}C_1&=\frac{8}{15} \\
C_1&=\frac{4}{5}
\end{aligned} ( 5 4 − C 1 ) e 2 1 ( 5 4 − 15 4 ) 3 2 C 1 C 1 = ( 15 4 − 3 1 C 1 ) e 2 1 = ( 1 − 3 1 ) C 1 = 15 8 = 5 4 Into ( 1 ) (1) ( 1 )
( 4 5 − 4 5 ) e 1 2 = C 2 e 3 2 C 2 e 3 2 = 0 e 3 2 ≠ 0 C 2 = 0 \begin{aligned}
\left(\frac{4}{5}-\frac{4}{5}\right)e^{\frac{1}{2}}&=C_2e^{\frac{3}{2}} \\
C_2e^{\frac{3}{2}} &=0 \qquad e^{\frac{3}{2}} \neq 0 \\
C_2&=0
\end{aligned} ( 5 4 − 5 4 ) e 2 1 C 2 e 2 3 C 2 = C 2 e 2 3 = 0 e 2 3 = 0 = 0 Thus,
y = 4 5 e 1 2 x y=\frac{4}{5}e^{\frac{1}{2}x} y = 5 4 e 2 1 x d. y ′ ′ + y = 0 y ( 0 ) = 3 y ′ ( 0 ) = − 0.5 y''+y=0 \qquad\qquad y(0)=3 \qquad y'(0)=-0.5 y ′′ + y = 0 y ( 0 ) = 3 y ′ ( 0 ) = − 0.5
y ′ ′ + y = 0 m 2 + 1 = 0 m = ± i \begin{aligned}
y''+y&=0\\m^2+1&=0\\m&=\pm i
\end{aligned} y ′′ + y m 2 + 1 m = 0 = 0 = ± i General Solution. y = C 1 sin x + C 2 cos x y=C_1\sin x + C_2\cos x y = C 1 sin x + C 2 cos x y ′ = C 1 cos x − C 2 sin x y'=C_1\cos x - C_2\sin x y ′ = C 1 cos x − C 2 sin x
When
x = 0 , y = 3 ⇒ 3 = C 1 ( 0 ) + C 2 ( 1 ) C 2 = 3 \begin{aligned}
x=0, y=3 \qquad \Rightarrow \qquad 3&=C_1(0)+C_2(1) \\ C_2&=3
\end{aligned} x = 0 , y = 3 ⇒ 3 C 2 = C 1 ( 0 ) + C 2 ( 1 ) = 3 x = 0 , y ′ = − 0.5 ⇒ − 0.5 = C 1 ( 1 ) + C 2 ( 0 ) C 1 = − 0.5 \begin{aligned}
x=0, y'=-0.5 \qquad \Rightarrow \qquad -0.5&=C_1(1)+C_2(0) \\ C_1&=-0.5
\end{aligned} x = 0 , y ′ = − 0.5 ⇒ − 0.5 C 1 = C 1 ( 1 ) + C 2 ( 0 ) = − 0.5 Thus,
y = − 1 2 sin x + 3 cos x y=-\frac{1}{2}\sin x+3\cos x y = − 2 1 sin x + 3 cos x e. y ′ ′ + 2 y ′ + 2 y = 0 y ( 0 ) = 1 y ′ ( 0 ) = − 1 y''+2y'+2y=0 \qquad\qquad y(0)=1 \qquad y'(0)=-1 y ′′ + 2 y ′ + 2 y = 0 y ( 0 ) = 1 y ′ ( 0 ) = − 1
y ′ ′ + 2 y ′ + 2 y = 0 m 2 + 2 m + 2 = 0 m = 1 2 ( 1 ) [ − 2 ± 2 2 − 4 ( 1 ) ( 2 ) ] = 1 2 [ − 2 ± 4 ] = − 1 ± i \begin{aligned}
y''+2y'+2y&=0\\
m^2+2m+2&=0\\
m=\frac{1}{2(1)}\left[-2\pm\sqrt{2^2-4(1)(2)}\right]\\
&=\frac{1}{2}\left[-2\pm\sqrt{4}\right]\\
&=-1\pm i
\end{aligned} y ′′ + 2 y ′ + 2 y m 2 + 2 m + 2 m = 2 ( 1 ) 1 [ − 2 ± 2 2 − 4 ( 1 ) ( 2 ) ] = 0 = 0 = 2 1 [ − 2 ± 4 ] = − 1 ± i General solution.
y = e − x ( C 1 cos x + C 2 sin x ) y ′ = − e x ( C 1 cos x + C 2 sin x ) + e − x ( − C 1 sin x + C 2 cos x ) \begin{aligned}
y&=e^{-x}(C_1\cos x+C_2\sin x) \\
y'&=-e^x(C_1\cos x+C_2 \sin x)+e^{-x}(-C_1\sin x+C_2\cos x)
\end{aligned} y y ′ = e − x ( C 1 cos x + C 2 sin x ) = − e x ( C 1 cos x + C 2 sin x ) + e − x ( − C 1 sin x + C 2 cos x ) When
x = 0 , y = 1 ⇒ 1 = C 1 ( 1 ) + C 2 ( 0 ) C 1 = 1 \begin{aligned}
x=0, y=1 \qquad \Rightarrow \qquad 1&=C_1(1)+C_2(0) \\ C_1&=1
\end{aligned} x = 0 , y = 1 ⇒ 1 C 1 = C 1 ( 1 ) + C 2 ( 0 ) = 1 x = 0 , y ′ = − 1 ⇒ − 1 = − [ C 1 ( 1 ) + C 2 ( 0 ) ] + [ − C 1 ( 0 ) + C 2 ( 1 ) ] − 1 = − C 1 + C 2 C 2 = C 1 − 1 C 2 = 0 \begin{aligned}
x=0, y'=-1 \qquad \Rightarrow \qquad -1&=-[C_1(1)+C_2(0)]+[-C_1(0)+C_2(1)]\\-1&=-C_1+C_2\\C_2&=C_1-1\\C_2&=0
\end{aligned} x = 0 , y ′ = − 1 ⇒ − 1 − 1 C 2 C 2 = − [ C 1 ( 1 ) + C 2 ( 0 )] + [ − C 1 ( 0 ) + C 2 ( 1 )] = − C 1 + C 2 = C 1 − 1 = 0 Thus,
y = e − x cos x y=e^{-x}\cos x y = e − x cos x
Obtain a general solution for nonhomogeneous problems using the method of undetermined coefficients.
a. y ′ ′ + 3 y ′ + 2 y = 2 x 2 y''+3y'+2y=2x^2 y ′′ + 3 y ′ + 2 y = 2 x 2
m 2 + 3 m + 2 = 0 ( m + 1 ) ( m + 2 ) = 0 m = − 1 , − 2 y c = C 1 e − x + C 2 e − 2 x \begin{aligned}
m^2+3m+2&=0\\
(m+1)(m+2)&=0\\
m&=-1,-2\\ \\
y_c=C_1e^{-x}+C_2e^{-2x}
\end{aligned} m 2 + 3 m + 2 ( m + 1 ) ( m + 2 ) m y c = C 1 e − x + C 2 e − 2 x = 0 = 0 = − 1 , − 2 y p = A x 2 + B x + c y ′ p = 2 A x + B y ′ ′ p = 2 A ⇒ 2 ( A x 2 + B x + C ) + 3 ( 2 A x + B ) + 2 ( 2 A ) = 2 x 2 ( 2 A ) x 2 + ( 2 B + 6 A ) x + ( 2 C + 3 B + 2 A ) = 2 x 3 \begin{aligned}
y_p&=Ax^2+Bx+c\\
y'p&=2Ax+B\\
y''p&=2A\\ \\
\Rightarrow 2(Ax^2+Bx+C)+3(2Ax+B)+2(2A)&=2x^2\\
(2A)x^2+(2B+6A)x+(2C+3B+2A)=2x^3
\end{aligned} y p y ′ p y ′′ p ⇒ 2 ( A x 2 + B x + C ) + 3 ( 2 A x + B ) + 2 ( 2 A ) ( 2 A ) x 2 + ( 2 B + 6 A ) x + ( 2 C + 3 B + 2 A ) = 2 x 3 = A x 2 + B x + c = 2 A x + B = 2 A = 2 x 2 Thus, A = 1 A=1 A = 1 B = − 3 B=-3 B = − 3 C = 7 2 C=\frac{7}{2} C = 2 7
y p = x 2 − 3 x + 7 2 y_p=x^2-3x+\frac{7}{2} y p = x 2 − 3 x + 2 7 ∴ y = C 1 e − x + C 2 e − 2 x + x 2 − 3 x + 7 2 \therefore y=C_1e^{-x}+C_2e^{-2x}+x^2-3x+\frac{7}{2} ∴ y = C 1 e − x + C 2 e − 2 x + x 2 − 3 x + 2 7 b. y ′ ′ + 4 y = 4 x 2 + 6 y''+4y=4x^2+6 y ′′ + 4 y = 4 x 2 + 6
m 2 + 4 = 0 m = ± 2 i y c = C 1 e − 2 i x + C 2 e 2 i x y c = C 1 cos 2 x + C 2 sin 2 x \begin{aligned}
m^2+4&=0\\
m&=\pm 2i\\ \\
y_c&=C_1e^{-2ix}+C_2e^{2ix}\\
y_c&=C_1\cos 2x+C_2\sin 2x
\end{aligned} m 2 + 4 m y c y c = 0 = ± 2 i = C 1 e − 2 i x + C 2 e 2 i x = C 1 cos 2 x + C 2 sin 2 x y p = A x 2 + B x + c y ′ p = 2 A x + B y ′ ′ p = 2 A ⇒ ( 2 A ) + 4 ( A x 2 + B x + C ) = 4 x 2 + 6 ( 4 A ) x 2 + ( 4 B ) x + ( 2 A + 4 C ) = 4 x 2 + 6 \begin{aligned}
y_p&=Ax^2+Bx+c\\
y'p&=2Ax+B\\
y''p&=2A\\ \\
\Rightarrow (2A)+4(Ax^2+Bx+C)&=4x^2+6 \\
(4A)x^2 + (4B)x+(2A+4C)&=4x^2+6
\end{aligned} y p y ′ p y ′′ p ⇒ ( 2 A ) + 4 ( A x 2 + B x + C ) ( 4 A ) x 2 + ( 4 B ) x + ( 2 A + 4 C ) = A x 2 + B x + c = 2 A x + B = 2 A = 4 x 2 + 6 = 4 x 2 + 6 Thus,
4 A = 4 4 B = 0 2 A + 4 C = 6 A = 1 B = 0 C = 1 \begin{aligned}
4A &=4 \qquad &4B&=0 \qquad &2A+4C&=6 \\
A&=1 \qquad &B&=0 \qquad &C&=1
\end{aligned} 4 A A = 4 = 1 4 B B = 0 = 0 2 A + 4 C C = 6 = 1 y p = x 2 + 1 y_p=x^2+1 y p = x 2 + 1 ∴ y = C 1 cos 2 x + C 2 sin 2 x + x 2 + 1 \therefore y=C_1\cos 2x+C_2\sin 2x+x^2+1 ∴ y = C 1 cos 2 x + C 2 sin 2 x + x 2 + 1 c. y ′ ′ − y = e x y''-y=e^x y ′′ − y = e x
m 2 − 1 = 0 m = ± 1 y c = C 1 e − x + C 2 e x α = 1 ⇒ m 1 ≠ m 2 \begin{aligned}
m^2-1&=0\\
m&=\pm 1 \\ \\
y_c&=C_1e^{-x}+C_2e^x \\
\alpha&=1 \Rightarrow m_1\neq m_2
\end{aligned} m 2 − 1 m y c α = 0 = ± 1 = C 1 e − x + C 2 e x = 1 ⇒ m 1 = m 2 y p = A x e x y p ′ = A e x + A x e x y p ′ ′ = 2 A e x + A x e x \begin{aligned}
y_p&=Axe^x \\
y'_p&=Ae^x+Axe^x \\
y''_p&=2Ae^x+Axe^x
\end{aligned} y p y p ′ y p ′′ = A x e x = A e x + A x e x = 2 A e x + A x e x Thus
( 2 A e x + A x e x ) − ( A x e x ) = e x 2 A = 1 A = 1 2 \begin{aligned}
(2Ae^x+Axe^x)-(Axe^x)=e^x \\
2A&=1 \\
A&=\frac{1}{2}
\end{aligned} ( 2 A e x + A x e x ) − ( A x e x ) = e x 2 A A = 1 = 2 1 y p = 1 2 x e x y_p=\frac{1}{2}xe^x y p = 2 1 x e x ∴ y = C 1 e − x + C 2 e x + 1 2 x e x \therefore y=C_1e^{-x}+C_2e^x+\frac{1}{2}xe^x ∴ y = C 1 e − x + C 2 e x + 2 1 x e x d. y ′ ′ + 3 y ′ − 4 y = 10 e x y''+3y'-4y=10e^x y ′′ + 3 y ′ − 4 y = 10 e x
m 2 + 3 m − 4 = 0 ( m − 1 ) ( m + 4 ) = 0 m = 1 , − 4 y c = C 1 e x + X 2 e − 4 x α = 1 ⇒ m 1 ≠ m 2 \begin{aligned}
m^2+3m-4&=0\\
(m-1)(m+4)&=0\\
m&=1,-4 \\ \\
y_c&=C_1e^x+X_2e^{-4x} \\
\alpha&=1 \Rightarrow m_1\neq m_2
\end{aligned} m 2 + 3 m − 4 ( m − 1 ) ( m + 4 ) m y c α = 0 = 0 = 1 , − 4 = C 1 e x + X 2 e − 4 x = 1 ⇒ m 1 = m 2 y p = A x e x y p ′ = A x e x + A e x y p ′ ′ = 2 A e x + A x e x \begin{aligned}
y_p&=Axe^x \\
y'_p&=Axe^x+Ae^x \\
y''_p &=2Ae^x+Axe^x
\end{aligned} y p y p ′ y p ′′ = A x e x = A x e x + A e x = 2 A e x + A x e x Thus
( 2 A e x + A x e x ) + 3 ( A x e x + A e x ) − 4 ( A x e x ) = 10 e x ( A + 3 A − 4 A ) x e x + ( 2 A + 3 A ) e x = 10 e x 5 A = 10 A = 2 \begin{aligned}
(2Ae^x+Axe^x)+3(Axe^x+Ae^x)-4(Axe^x)&=10e^x \\
(A+3A-4A)xe^x+(2A+3A)e^x &=10e^x \\
5A&=10\\
A&=2
\end{aligned} ( 2 A e x + A x e x ) + 3 ( A x e x + A e x ) − 4 ( A x e x ) ( A + 3 A − 4 A ) x e x + ( 2 A + 3 A ) e x 5 A A = 10 e x = 10 e x = 10 = 2 y p = 2 x e x y_p=2xe^x y p = 2 x e x ∴ y = C 1 e x + C 2 e − 4 x + 2 x e x \therefore y=C_1e^x+C_2e^{-4x}+2xe^x ∴ y = C 1 e x + C 2 e − 4 x + 2 x e x e. y ′ ′ − 4 y ′ + 3 y = 2 e 3 x y''-4y'+3y=2e^{3x} y ′′ − 4 y ′ + 3 y = 2 e 3 x
m 2 − 4 m + 3 = 0 ( m − 3 ) ( m − 1 ) = 0 m = 1 , 3 α = 3 ⇒ m 2 ≠ m 1 \begin{aligned}
m^2-4m+3&=0\\
(m-3)(m-1)&=0\\
m=1,3 \\
\alpha&=3 \Rightarrow m_2\neq m_1
\end{aligned} m 2 − 4 m + 3 ( m − 3 ) ( m − 1 ) m = 1 , 3 α = 0 = 0 = 3 ⇒ m 2 = m 1 y p = A x e 3 x y p ′ = 3 A x e 3 x + A e 3 x y p ′ ′ = 9 A x e 3 x + 6 A e 3 x \begin{aligned}
y_p&=Axe^{3x} \\
y'_p=3Axe^{3x}+Ae^{3x} \\
y''_p=9Axe^{3x}+6Ae^{3x} \\
\end{aligned} y p y p ′ = 3 A x e 3 x + A e 3 x y p ′′ = 9 A x e 3 x + 6 A e 3 x = A x e 3 x Thus
( 9 A x e 3 x + 6 A e 3 x ) − 4 ( 3 A x e 3 x + A e 3 x ) + 3 ( A x e 3 x ) = 2 e 3 x ( 9 A − 12 A + 3 A ) x e 3 x + ( 6 A − 4 A ) = 2 e 3 x 2 A = 2 A = 1 \begin{aligned}
(9Axe^{3x}+6Ae^{3x})-4(3Axe^{3x}+Ae^{3x})+3(Axe^{3x})&=2e^{3x} \\
(9A-12A+3A)xe^{3x}+(6A-4A)&=2e^{3x} \\
2A&=2\\
A&=1
\end{aligned} ( 9 A x e 3 x + 6 A e 3 x ) − 4 ( 3 A x e 3 x + A e 3 x ) + 3 ( A x e 3 x ) ( 9 A − 12 A + 3 A ) x e 3 x + ( 6 A − 4 A ) 2 A A = 2 e 3 x = 2 e 3 x = 2 = 1 y p = x e 3 x y_p=xe^{3x} y p = x e 3 x ∴ y = C 1 e x + C 2 e 3 x + x e 3 x \therefore y=C_1e^x+C_2e^{3x}+xe^{3x} ∴ y = C 1 e x + C 2 e 3 x + x e 3 x f. y ′ ′ + 4 y ′ − 2 y = 2 x 2 − 3 x + 6 y''+4y'-2y=2x^2-3x+6 y ′′ + 4 y ′ − 2 y = 2 x 2 − 3 x + 6
m 2 + 4 m − 2 = 0 ( m 2 + 4 m + 4 ) − 6 = 0 ( m + 2 ) 2 = 6 m = − 2 ± 6 y c = C 1 e ( − 2 − 6 ) x + C 2 e ( − 2 + 6 ) x \begin{aligned}
m^2+4m-2&=0 \\
(m^2+4m+4)-6&=0\\
(m+2)^2&=6\\
m&=-2\pm\sqrt{6} \\ \\
y_c&=C_1e^{(-2-\sqrt{6})x} + C_2e^{(-2+\sqrt{6})x}
\end{aligned} m 2 + 4 m − 2 ( m 2 + 4 m + 4 ) − 6 ( m + 2 ) 2 m y c = 0 = 0 = 6 = − 2 ± 6 = C 1 e ( − 2 − 6 ) x + C 2 e ( − 2 + 6 ) x y p = A x 2 + B x + 6 y p ′ = 2 A x + B y p ′ ′ = 2 A ( 2 A ) + 4 ( 2 A x + B ) − 2 ( A x 2 + B x + C ) = 2 x 2 − 3 x + 6 \begin{aligned}
y_p=Ax^2+Bx+6\\
y'_p&=2Ax+B\\
y''_p=2A \\ \\
(2A)+4(2Ax+B)-2(Ax^2+Bx+C)&=2x^2-3x+6
\end{aligned} y p = A x 2 + B x + 6 y p ′ y p ′′ = 2 A ( 2 A ) + 4 ( 2 A x + B ) − 2 ( A x 2 + B x + C ) = 2 A x + B = 2 x 2 − 3 x + 6 − 2 A = 2 8 ( − 1 ) − 2 B = − 3 2 ( − 1 ) 4 ( − 5 2 ) − 2 C = 6 A = − 1 B = − 5 2 C = − 0 \begin{aligned}
-2A&=2 \qquad &8(-1)-2B&=-3 \qquad &2(-1)4\left(-\frac{5}{2}\right)-2C&=6 \\
A&=-1 &B&=-\frac{5}{2} &C&=-0
\end{aligned} − 2 A A = 2 = − 1 8 ( − 1 ) − 2 B B = − 3 = − 2 5 2 ( − 1 ) 4 ( − 2 5 ) − 2 C C = 6 = − 0 y p = − x 2 − 5 2 x − 9 y_p=-x^2-\frac{5}{2}x-9 y p = − x 2 − 2 5 x − 9 ∴ y = C 1 e ( − 2 − 6 ) x + C 2 e ( − 2 + 6 ) x − x 2 − 5 2 x − 9 \therefore y=C_1e^{(-2-\sqrt{6})x}+C_2e^{(-2+\sqrt{6})x}-x^2-\frac{5}{2}x-9 ∴ y = C 1 e ( − 2 − 6 ) x + C 2 e ( − 2 + 6 ) x − x 2 − 2 5 x − 9 g. y ′ ′ + 3 y ′ + 2 y = 6 y''+3y'+2y=6 y ′′ + 3 y ′ + 2 y = 6
m 2 + 3 m + 2 = 0 ( m + 1 ) ( m + 2 ) = 0 m = − 1 , − 2 \begin{aligned}
m^2+3m+2&=0\\
(m+1)(m+2)&=0\\
m&=-1,-2
\end{aligned} m 2 + 3 m + 2 ( m + 1 ) ( m + 2 ) m = 0 = 0 = − 1 , − 2 P n ( x ) P_n(x) P n ( x ) G ( s ) G(s) G ( s )
y p = A , y p ′ = 0 , y p ′ ′ = 0 y_p=A, y'_p=0, y''_p=0 y p = A , y p ′ = 0 , y p ′′ = 0 0 + 3 ( 0 ) + 2 ( A ) = 6 A = 3 \begin{aligned}
0+3(0)+2(A)&=6\\A&=3
\end{aligned} 0 + 3 ( 0 ) + 2 ( A ) A = 6 = 3 y p = 3 y_p=3 y p = 3 ∴ y = C 1 e − x + C 2 e − 2 x + 3 \therefore y=C_1e^{-x}+C_2e^{-2x}+3 ∴ y = C 1 e − x + C 2 e − 2 x + 3
Obtain a general solution for nonhomogeneous problems using the method of undetermined coefficients.
a. y ′ ′ − 2 y ′ + 5 y = e x cos 2 x y''-2y'+5y=e^x\cos 2x y ′′ − 2 y ′ + 5 y = e x cos 2 x
m 2 − 2 m + 5 = 0 m = 1 2 [ 2 ± 4 − 20 ] 1 2 ( 2 ± 4 i ) 1 ± 2 i y c = e x ( C 1 cos 2 x + C 2 sin 2 x ) α = ( 1 + 2 i ) ≠ m 1 ≠ m 2 \begin{aligned}
m^2-2m+5&=0\\
m&=\frac{1}{2}[2\pm\sqrt{4-20}]\\
\frac{1}{2}(2\pm 4i)\\1\pm 2i \\ \\
y_c&=e^x(C_1\cos 2x+C_2\sin 2x)\\
\alpha&=(1+2i)\neq m_1\neq m_2
\end{aligned} m 2 − 2 m + 5 m 2 1 ( 2 ± 4 i ) 1 ± 2 i y c α = 0 = 2 1 [ 2 ± 4 − 20 ] = e x ( C 1 cos 2 x + C 2 sin 2 x ) = ( 1 + 2 i ) = m 1 = m 2 Thus
g ( x ) = e x cos 2 x e x ⋅ Re e 2 i x Re ( e ( 1 + 2 i ) x ) y p = A x Re ( e ( 1 + 2 i ) x ) \begin{aligned}
g(x)&=e^x\cos 2x\\
e^x\cdot \text{Re}e^{2ix}\\
\text{Re}(e^{(1+2i)x})\\
y_p&=A_x\text{Re}(e^{(1+2i)x})
\end{aligned} g ( x ) e x ⋅ Re e 2 i x Re ( e ( 1 + 2 i ) x ) y p = e x cos 2 x = A x Re ( e ( 1 + 2 i ) x ) Let
Y p = A x e ( 1 + 2 i ) x , y p = A x Re ( e ( 1 + 2 i ) x ) Y p ′ = ( 1 + 2 i ) A x e ( 1 + 2 i ) x + A e ( 1 + 2 i ) x Y p ′ ′ = ( 1 + 2 i ) 2 A x e ( 1 + 2 i ) x + ( 1 + 2 i ) A e ( 1 + 2 i ) x + ( 1 + 2 i ) A e ( 1 + 2 i ) x = ( − 3 + 4 i ) A x e ( 1 + 2 i ) x + 2 A ( 1 + 2 i ) e ( 1 + 2 i ) x \begin{aligned}
Y_p&=Axe^{(1+2i)x},\qquad y_p=Ax\text{Re}(e^{(1+2i)x}) \\
Y'_p&=(1+2i)Axe^{(1+2i)x}+Ae^{(1+2i)x} \\
Y''_p&=(1+2i)^2Axe^{(1+2i)x}+(1+2i)Ae^{(1+2i)x}+(1+2i)Ae^{(1+2i)x}\\
&=(-3+4i)Axe^{(1+2i)x}+2A(1+2i)e^{(1+2i)x}
\end{aligned} Y p Y p ′ Y p ′′ = A x e ( 1 + 2 i ) x , y p = A x Re ( e ( 1 + 2 i ) x ) = ( 1 + 2 i ) A x e ( 1 + 2 i ) x + A e ( 1 + 2 i ) x = ( 1 + 2 i ) 2 A x e ( 1 + 2 i ) x + ( 1 + 2 i ) A e ( 1 + 2 i ) x + ( 1 + 2 i ) A e ( 1 + 2 i ) x = ( − 3 + 4 i ) A x e ( 1 + 2 i ) x + 2 A ( 1 + 2 i ) e ( 1 + 2 i ) x ( − 3 + 4 i ) A x e ( 1 + 2 i ) x + 2 A ( 1 + 2 i ) e ( 1 + 2 i ) x − 2 [ ( 1 + 2 i ) A x e ( 1 + 2 i ) x + A e ( 1 + 2 i ) x ] + 5 [ A x e ( 1 + 2 i ) x ] = e x cos 2 x [ − 3 + 4 i − 2 − 4 i + 5 ] A x e ( 1 + 2 i ) x + [ 2 + 4 i − 2 ] A e ( 1 + 2 i ) x = e x cos 2 x 4 A i = 1 A = − 1 4 i \begin{aligned}
(-3+4i)Axe^{(1+2i)x}+2A(1+2i)e^{(1+2i)x}-2\left[(1+2i)Axe^{(1+2i)x}+Ae^{(1+2i)x}\right]+5[Axe^{(1+2i)x}]&=e^x\cos 2x \\
[-3+4i-2-4i+5]Axe^{(1+2i)x}+[2+4i-2]Ae^{(1+2i)x}&=e^x\cos 2x \\
4Ai&=1 \\
A&=-\frac{1}{4}i
\end{aligned} ( − 3 + 4 i ) A x e ( 1 + 2 i ) x + 2 A ( 1 + 2 i ) e ( 1 + 2 i ) x − 2 [ ( 1 + 2 i ) A x e ( 1 + 2 i ) x + A e ( 1 + 2 i ) x ] + 5 [ A x e ( 1 + 2 i ) x ] [ − 3 + 4 i − 2 − 4 i + 5 ] A x e ( 1 + 2 i ) x + [ 2 + 4 i − 2 ] A e ( 1 + 2 i ) x 4 A i A = e x cos 2 x = e x cos 2 x = 1 = − 4 1 i Y p = − 1 4 i x e ( 1 + 2 i ) x = − 1 4 i x e x [ cos 2 x + i sin 2 x ] = 1 4 x e x [ sin 2 x − i cos 2 x ] y p = Re Y p = 1 4 x e x sin 2 x \begin{aligned}
Y_p=-\frac{1}{4}ixe^{(1+2i)x} \\
&=-\frac{1}{4}ixe^x[\cos 2x+i\sin 2x] \\
&=\frac{1}{4}xe^x[\sin 2x-i\cos 2x]\\ \\
y_p&=\text{Re}Y_p=\frac{1}{4}xe^x\sin 2x
\end{aligned} Y p = − 4 1 i x e ( 1 + 2 i ) x y p = − 4 1 i x e x [ cos 2 x + i sin 2 x ] = 4 1 x e x [ sin 2 x − i cos 2 x ] = Re Y p = 4 1 x e x sin 2 x ∴ y = e x ( C 1 cos 2 x + C 2 sin 2 x ) + 1 4 x e x sin 2 x \therefore y = e^x(C_1\cos 2x+C_2\sin 2x)+\frac{1}{4}xe^x\sin 2x ∴ y = e x ( C 1 cos 2 x + C 2 sin 2 x ) + 4 1 x e x sin 2 x b. y ′ ′ + y = 2 x sin x y''+y=2x\sin x y ′′ + y = 2 x sin x
m 2 = − 1 m = ± i y p = C 1 e − i x + C 2 e i x = C 1 cos x + C 2 sin x \begin{aligned}
m^2&=-1 \\
m&=\pm i \\ \\
y_p&=C_1e^{-ix} + C_2e^{ix} \\
&=C_1\cos x+C_2 \sin x
\end{aligned} m 2 m y p = − 1 = ± i = C 1 e − i x + C 2 e i x = C 1 cos x + C 2 sin x Compare e α x Q n ( x ) = 2 x sin x e^{\alpha x}Q_n(x)=2x\sin x e αx Q n ( x ) = 2 x sin x
α = 1 = m 1 ≠ m 2 \alpha=1=m_1\neq m_2 α = 1 = m 1 = m 2 Let
Y p = ( A x 2 + B x ) e i x , y p = Im Y p Y p ′ = ( 2 A x + B ) e i x + i ( A x 2 + B x ) e i x Y p ′ ′ = ( 2 A ) e i x + i ( 2 A x + B ) e i x + i ( 2 A x + B ) e i x + i 2 ( A x 2 + B x ) e i x = ( 2 A ) e i x + 2 i ( 2 A x + B ) e i x − ( A x 2 + B x ) e i x \begin{aligned}
Y_p&=(Ax^2+Bx)e^{ix}, \qquad y_p=\text{Im}Y_p \\
Y'_p&=(2Ax+B)e^{ix}+i(Ax^2+Bx)e^{ix} \\
Y''_p&=(2A)e^{ix}+i(2Ax+B)e^{ix}+i(2Ax+B)e^{ix}+i^2(Ax^2+Bx)e^{ix} \\
&=(2A)e^{ix}+2i(2Ax+B)e^{ix}-(Ax^2+Bx)e^{ix}
\end{aligned} Y p Y p ′ Y p ′′ = ( A x 2 + B x ) e i x , y p = Im Y p = ( 2 A x + B ) e i x + i ( A x 2 + B x ) e i x = ( 2 A ) e i x + i ( 2 A x + B ) e i x + i ( 2 A x + B ) e i x + i 2 ( A x 2 + B x ) e i x = ( 2 A ) e i x + 2 i ( 2 A x + B ) e i x − ( A x 2 + B x ) e i x [ ( 2 A e i x ) + 2 i ( 2 A x + B ) e i x − ( A x + B ) e i x ] + ( A x + B ) e i x = 2 x e i x 2 A + 4 A i x + 2 B i = 2 x ( 4 A i ) x + ( 2 A + 2 B i ) = 2 x \begin{aligned}
\left[(2Ae^{ix})+2i(2Ax+B)e^{ix}-(Ax+B)e^{ix}\right]+(Ax+B)e^{ix}&=2xe^{ix} \\
2A+4Aix+2Bi&=2x \\
(4Ai)x+(2A+2Bi)&=2x
\end{aligned} [ ( 2 A e i x ) + 2 i ( 2 A x + B ) e i x − ( A x + B ) e i x ] + ( A x + B ) e i x 2 A + 4 A i x + 2 B i ( 4 A i ) x + ( 2 A + 2 B i ) = 2 x e i x = 2 x = 2 x Thus
4 A i = 2 ( 2 A + 2 B i ) = 0 A = 1 2 i = − 1 2 i B = − A i = − ( 1 2 i ) i = 1 2 \begin{aligned}
4Ai&=2 \qquad &(2A+2Bi)&=0 \\
A&=\frac{1}{2i}=-\frac{1}{2}i &B&=-\frac{A}{i}=-\frac{(\frac{1}{2}i)}{i}=\frac{1}{2}
\end{aligned} 4 A i A = 2 = 2 i 1 = − 2 1 i ( 2 A + 2 B i ) B = 0 = − i A = − i ( 2 1 i ) = 2 1 Into equation
Y p = ( − 1 2 i x 2 + 1 2 x ) e i x = ( − 1 2 i x 2 + 1 2 x ) ( cos x + i sin x ) = − 1 2 i x 2 cos x + 1 2 x 2 sin x + 1 2 x cos x + 1 2 i x sin x = ( 1 2 x 2 sin x + 1 2 x cos x ) + i ( − 1 2 x 2 cos x + 1 2 x sin x ) y p = Im Y p = − 1 2 x 2 cos x + 1 2 x sin x \begin{aligned}
Y_p&=\left(-\frac{1}{2}ix^2+\frac{1}{2}x\right)e^{ix} \\
&=\left(-\frac{1}{2}ix^2+\frac{1}{2}x\right)(\cos x+i\sin x) \\
&=-\frac{1}{2}ix^2\cos x+\frac{1}{2}x^2\sin x+\frac{1}{2}x\cos x+\frac{1}{2} ix \sin x \\
&=\left(\frac{1}{2}x^2\sin x+\frac{1}{2}x\cos x\right)+i\left(-\frac{1}{2}x^2\cos x+\frac{1}{2}x \sin x\right) \\ \\
y_p&=\text{Im}Y_p \\
&=-\frac{1}{2}x^2\cos x+\frac{1}{2}x\sin x
\end{aligned} Y p y p = ( − 2 1 i x 2 + 2 1 x ) e i x = ( − 2 1 i x 2 + 2 1 x ) ( cos x + i sin x ) = − 2 1 i x 2 cos x + 2 1 x 2 sin x + 2 1 x cos x + 2 1 i x sin x = ( 2 1 x 2 sin x + 2 1 x cos x ) + i ( − 2 1 x 2 cos x + 2 1 x sin x ) = Im Y p = − 2 1 x 2 cos x + 2 1 x sin x ∴ y = C 1 cos x + C 2 sin x − 1 2 x 2 cos x + 1 2 x sin x \therefore y=C_1\cos x+C_2\sin x-\frac{1}{2}x^2\cos x+\frac{1}{2}x\sin x ∴ y = C 1 cos x + C 2 sin x − 2 1 x 2 cos x + 2 1 x sin x c. y ′ ′ + 2 y ′ + 2 y = e x sin x y''+2y'+2y=e^x\sin x y ′′ + 2 y ′ + 2 y = e x sin x
m 2 + 2 m + 2 = 0 m = 1 2 [ − 2 ± 4 − 8 ] 1 2 [ − 1 ± 2 i ] − 1 ± i y c = e − x ( C 1 cos x + C 2 sin x ) \begin{aligned}
m^2+2m+2&=0\\
m&=\frac{1}{2}[-2\pm\sqrt{4-8}]\\
\frac{1}{2}[-1\pm 2i]\\
-1\pm i \\ \\
y_c&=e^{-x}(C_1\cos x+C_2\sin x)
\end{aligned} m 2 + 2 m + 2 m 2 1 [ − 1 ± 2 i ] − 1 ± i y c = 0 = 2 1 [ − 2 ± 4 − 8 ] = e − x ( C 1 cos x + C 2 sin x ) g ( x ) = e x sin x = Im ( e x e i x ) = Im ( e ( 1 + i ) x ) α = 1 + i ≠ m 1 ≠ m 2 y p = Im A e ( 1 + i ) x \begin{aligned}
g(x)&=e^x\sin x \\
&=\text{Im}(e^xe^{ix})\\
&=\text{Im}(e^{(1+i)x}) \\ \\
\alpha&=1+i\neq m_1\neq m_2 \\
y_p=\text{Im}Ae^{(1+i)x}
\end{aligned} g ( x ) α y p = Im A e ( 1 + i ) x = e x sin x = Im ( e x e i x ) = Im ( e ( 1 + i ) x ) = 1 + i = m 1 = m 2 Let
Y p = A e ( 1 + i ) x Y p ′ = ( 1 + i ) A e ( 1 + i ) x Y p ′ ′ = 2 i A e ( 1 + i ) x \begin{aligned}
Y_p&=Ae^{(1+i)x} \\
Y'_p&=(1+i)Ae^{(1+i)x} \\
Y''_p&=2iAe^{(1+i)x}
\end{aligned} Y p Y p ′ Y p ′′ = A e ( 1 + i ) x = ( 1 + i ) A e ( 1 + i ) x = 2 i A e ( 1 + i ) x Into equation
2 i A e ( 1 + i ) x + 2 [ ( 1 + i ) A e ( 1 + i ) x ] + 2 [ A e ( 1 + i ) x ] = e ( 1 + i ) x ( 2 i + 2 + 2 i + 2 ) A e ( 1 + i ) x = e ( 1 + i ) x 4 ( 1 + i ) A e ( 1 + i ) x = Im e ( 1 + i ) x A = 1 4 ( 1 + i ) = 1 8 ( 1 − i ) \begin{aligned}
2iAe^{(1+i)x}+2\left[(1+i)Ae^{(1+i)x}\right]+2\left[Ae^{(1+i)x}\right] &=e^{(1+i)x} \\
(2i+2+2i+2)Ae^{(1+i)x}&= e^{(1+i)x}\\
4(1+i)Ae^{(1+i)x} &=\text{Im}e^{(1+i)x} \\
A&=\frac{1}{4(1+i)}=\frac{1}{8}(1-i)
\end{aligned} 2 i A e ( 1 + i ) x + 2 [ ( 1 + i ) A e ( 1 + i ) x ] + 2 [ A e ( 1 + i ) x ] ( 2 i + 2 + 2 i + 2 ) A e ( 1 + i ) x 4 ( 1 + i ) A e ( 1 + i ) x A = e ( 1 + i ) x = e ( 1 + i ) x = Im e ( 1 + i ) x = 4 ( 1 + i ) 1 = 8 1 ( 1 − i ) Thus
Y p = 1 8 ( 1 − i ) e ( 1 + i ) x = 1 8 ( 1 − i ) e x [ cos x + i sin x ] = 1 8 e x [ cos x + i sin x − i cos x + sin x ] = 1 8 e x [ ( cos x + sin x ) + i ( cos x + sin x ) ] y p = Im Y p = 1 8 e x ( − cos x + sin x ) \begin{aligned}
Y_p&=\frac{1}{8}(1-i)e^{(1+i)x} \\
&=\frac{1}{8}(1-i)e^x[\cos x+i\sin x] \\
&=\frac{1}{8}e^x[\cos x+i\sin x-i\cos x+\sin x] \\
&=\frac{1}{8}e^x[(\cos x+\sin x)+i(\cos x+\sin x)] \\ \\
y_p&=\text{Im}Y_p=\frac{1}{8}e^x(-\cos x+\sin x)
\end{aligned} Y p y p = 8 1 ( 1 − i ) e ( 1 + i ) x = 8 1 ( 1 − i ) e x [ cos x + i sin x ] = 8 1 e x [ cos x + i sin x − i cos x + sin x ] = 8 1 e x [( cos x + sin x ) + i ( cos x + sin x )] = Im Y p = 8 1 e x ( − cos x + sin x ) ∴ y = e − x ( C 1 cos x + C 2 sin x ) + 1 8 e x ( sin x − cos x ) \therefore y=e^{-x}(C_1\cos x+C_2\sin x)+\frac{1}{8}e^x(\sin x-\cos x) ∴ y = e − x ( C 1 cos x + C 2 sin x ) + 8 1 e x ( sin x − cos x ) d. y ′ ′ − 3 y ′ + 2 y = x 2 e 3 x y''-3y'+2y=x^2e^{3x} y ′′ − 3 y ′ + 2 y = x 2 e 3 x
m 2 − 3 m + 2 = 0 ( m − 1 ) ( m − 2 ) = 0 m = 1 , 2 y c = C 1 e x + C 2 e 2 x \begin{aligned}
m^2-3m+2&=0\\
(m-1)(m-2)&=0\\
m&=1,2\\ \\
y_c&=C_1e^x+C_2e^{2x}
\end{aligned} m 2 − 3 m + 2 ( m − 1 ) ( m − 2 ) m y c = 0 = 0 = 1 , 2 = C 1 e x + C 2 e 2 x g ( x ) = x 2 e 3 x = P n ( x ) e α x α = 3 ≠ m 1 ≠ m 2 P n = x 2 \begin{aligned}
g(x)&=x^2e^{3x}=P_n(x)e^{\alpha x} \\
\alpha&=3\neq m_1\neq m_2 \\ \\
P_n&=x^2
\end{aligned} g ( x ) α P n = x 2 e 3 x = P n ( x ) e αx = 3 = m 1 = m 2 = x 2 Thus
y p = e 3 x ( A x 2 + B x + C ) y_p=e^{3x}(Ax^2+Bx+C) y p = e 3 x ( A x 2 + B x + C ) y p ′ = 3 e 3 x ( A x 2 + B x + C ) + e 3 x ( 2 A x + B ) = e 3 x ( 3 A x 2 + 3 B x + 3 C + 2 A x + B ) = e 3 x [ ( 3 A ) x 2 + ( 3 B + 2 A ) x + ( 3 C + B ) ] y p ′ ′ = 3 e 3 x [ 3 A x 2 + ( 3 B + 2 A ) x + ( 3 C + B ) ] + e 3 x [ 6 A x 2 + ( 3 B + 2 A ) ] = e 3 x [ 9 A x 2 + ( 9 B + 6 A ) x + ( 6 C + 3 B ) ] + e 3 x [ 6 A x 2 + ( 3 B + 2 A ) ] = e 3 x [ 9 A x 2 + ( 12 A + 9 B ) x + ( 2 A + 6 B + 9 C ) ] \begin{aligned}
y'_p&=3e^{3x}(Ax^2+Bx+C)+e^{3x}(2Ax+B) \\
&=e^{3x}(3Ax^2+3Bx+3C+2Ax+B)\\
&=e^{3x}\left[(3A)x^2+(3B+2A)x+(3C+B)\right] \\
y''_p&=3e^{3x}[3Ax^2+(3B+2A)x+(3C+B)]+e^{3x}[6Ax^2+(3B+2A)]\\
&=e^{3x}[9Ax^2+(9B+6A)x+(6C+3B)]+e^{3x}[6Ax^2+(3B+2A)] \\
&=e^{3x}[9Ax^2+(12A+9B)x+(2A+6B+9C)]
\end{aligned} y p ′ y p ′′ = 3 e 3 x ( A x 2 + B x + C ) + e 3 x ( 2 A x + B ) = e 3 x ( 3 A x 2 + 3 B x + 3 C + 2 A x + B ) = e 3 x [ ( 3 A ) x 2 + ( 3 B + 2 A ) x + ( 3 C + B ) ] = 3 e 3 x [ 3 A x 2 + ( 3 B + 2 A ) x + ( 3 C + B )] + e 3 x [ 6 A x 2 + ( 3 B + 2 A )] = e 3 x [ 9 A x 2 + ( 9 B + 6 A ) x + ( 6 C + 3 B )] + e 3 x [ 6 A x 2 + ( 3 B + 2 A )] = e 3 x [ 9 A x 2 + ( 12 A + 9 B ) x + ( 2 A + 6 B + 9 C )] Into equation
e 3 x [ 9 A x 2 + ( 12 A + 9 B ) x + ( 2 A + 6 B + 9 C ) ] − 3 e x [ 3 A x 2 + ( 3 B + 2 A ) x + ( 3 C + B ) ] + 2 e x ( A x 2 + B x + C ) = x 2 e 3 x e 3 x [ ( 9 A − 9 A + 2 A ) x 2 + ( 12 A + 9 B − 9 B − 6 A + 2 B ) x + ( 2 A + 6 B + 9 C − 9 C − 3 B + 2 C ) ] = x 2 e 3 x e 3 x [ ( 2 A ) x 2 + ( 16 A + 2 B ) x + ( 2 A + 3 B + 2 C ) ] = x 2 e 3 x \begin{aligned}
e^{3x}[9Ax^2+(12A+9B)x+(2A+6B+9C)]-3e^x[3Ax^2+(3B+2A)x+(3C+B)]+2e^x(Ax^2+Bx+C)&=x^2e^{3x}\\
e^{3x}[(9A-9A+2A)x^2+(12A+9B-9B-6A+2B)x+(2A+6B+9C-9C-3B+2C)]&=x^2e^{3x}\\
e^{3x}[(2A)x^2+(16A+2B)x+(2A+3B+2C)]&=x^2e^{3x}
\end{aligned} e 3 x [ 9 A x 2 + ( 12 A + 9 B ) x + ( 2 A + 6 B + 9 C )] − 3 e x [ 3 A x 2 + ( 3 B + 2 A ) x + ( 3 C + B )] + 2 e x ( A x 2 + B x + C ) e 3 x [( 9 A − 9 A + 2 A ) x 2 + ( 12 A + 9 B − 9 B − 6 A + 2 B ) x + ( 2 A + 6 B + 9 C − 9 C − 3 B + 2 C )] e 3 x [( 2 A ) x 2 + ( 16 A + 2 B ) x + ( 2 A + 3 B + 2 C )] = x 2 e 3 x = x 2 e 3 x = x 2 e 3 x Thus
2 A = 1 6 A + 2 B = 0 2 A + 3 B + 2 C = 0 A = 1 2 2 B = − 6 ( 1 2 2 C = − 2 ( 1 ) − 3 ( − 3 2 ) B = − 3 2 C = 1 2 ( 7 2 ) = 7 4 \begin{aligned}
2A&=1 \qquad &6A+2B&=0 \qquad &2A+3B+2C&=0 \\
A&=\frac{1}{2} \qquad &2B&=-6(\frac{1}{2} \qquad &2C&=-2(1)-3(-\frac{3}{2}) \\
&&B&=-\frac{3}{2} \qquad &C&=\frac{1}{2}(\frac{7}{2})=\frac{7}{4}
\end{aligned} 2 A A = 1 = 2 1 6 A + 2 B 2 B B = 0 = − 6 ( 2 1 = − 2 3 2 A + 3 B + 2 C 2 C C = 0 = − 2 ( 1 ) − 3 ( − 2 3 ) = 2 1 ( 2 7 ) = 4 7 Thus
y + p = e 3 x ( 1 2 x 2 − 3 2 x + 7 4 ) = 1 4 e 3 x ( 2 x 2 − 6 x + 7 ) \begin{aligned}
y+p&=e^{3x}\left(\frac{1}{2}x^2-\frac{3}{2}x+\frac{7}{4}\right) \\ &=\frac{1}{4}e^{3x}(2x^2-6x+7)
\end{aligned} y + p = e 3 x ( 2 1 x 2 − 2 3 x + 4 7 ) = 4 1 e 3 x ( 2 x 2 − 6 x + 7 ) ∴ y = C 1 e x + C 2 e x + 1 4 e 3 x ( 2 x 2 − 6 x + 7 ) \therefore y=C_1e^x+C_2e^x+\frac{1}{4}e^{3x}(2x^2-6x+7) ∴ y = C 1 e x + C 2 e x + 4 1 e 3 x ( 2 x 2 − 6 x + 7 )
Solve the following nonhomogeneous with initial value using the method of undetermined coefficients.
a. 5 y ′ ′ + y ′ = − 6 x y ( 0 ) = 0 y ′ ( 0 ) = 0 5y''+y'=-6x \qquad\qquad y(0)=0 \qquad y'(0)=0 5 y ′′ + y ′ = − 6 x y ( 0 ) = 0 y ′ ( 0 ) = 0
5 m 2 + m = 0 m ( 5 m + 1 ) = 0 m = 0 , − 1 5 α = 0 = m 1 ≠ m 2 \begin{aligned}
5m^2+m&=0\\
m(5m+1)&=0\\
m&=0,-\frac{1}{5}\\ \\
\alpha&=0=m_1 \neq m_2
\end{aligned} 5 m 2 + m m ( 5 m + 1 ) m α = 0 = 0 = 0 , − 5 1 = 0 = m 1 = m 2 Thus
y p = x ( A x + B ) e 0 = A x 2 + B x y p ′ = 2 A x + B y p ′ ′ = 2 A \begin{aligned}
y_p&=x(Ax+B)e^0\\
&=Ax^2+Bx \\
y'_p&=2Ax+B\\
y''_p&=2A
\end{aligned} y p y p ′ y p ′′ = x ( A x + B ) e 0 = A x 2 + B x = 2 A x + B = 2 A Into formula
5 ( 2 A ) + ( 2 A x + B ) = − 6 x ( 10 A + B ) + 2 A x = − 6 x \begin{aligned}
5(2A)+(2Ax+B)&=-6x \\
(10A+B)+2Ax&=-6x
\end{aligned} 5 ( 2 A ) + ( 2 A x + B ) ( 10 A + B ) + 2 A x = − 6 x = − 6 x Thus
A = − 3 , B = 30 A=-3,\qquad B=30 A = − 3 , B = 30 ⇒ y p = − 3 x 2 + 30 x \Rightarrow y_p=-3x^2+30x ⇒ y p = − 3 x 2 + 30 x General solution:
y = C 1 e 0 + C 2 e − 1 5 x + 30 x − 30 x 2 y ′ = − 1 5 C 2 e − 1 5 x + 30 − 60 x \begin{aligned}
y&=C_1e^0+C_2e^{-\frac{1}{5}x}+30x-30x^2 \\
y'&=-\frac{1}{5}C_2e^{-\frac{1}{5}x}+30-60x
\end{aligned} y y ′ = C 1 e 0 + C 2 e − 5 1 x + 30 x − 30 x 2 = − 5 1 C 2 e − 5 1 x + 30 − 60 x y ( 0 ) = 0 ⇒ C 1 + C 2 = 0 → C 1 = − C 2 y(0)=0 \qquad \Rightarrow \qquad C_1+C_2=0 \quad \rightarrow \quad C_1=-C_2 y ( 0 ) = 0 ⇒ C 1 + C 2 = 0 → C 1 = − C 2 y ′ ( 0 ) = 0 ⇒ − 1 5 C 2 + 30 = 0 → C 2 = 150 y'(0)=0 \qquad \Rightarrow \qquad -\frac{1}{5}C_2+30=0 \quad \rightarrow \quad C_2=150 y ′ ( 0 ) = 0 ⇒ − 5 1 C 2 + 30 = 0 → C 2 = 150 ∴ y = − 150 + 150 e − x 5 − 30 x 2 + 30 x \therefore y=-150+150e^{-\frac{x}{5}}-30x^2+30x ∴ y = − 150 + 150 e − 5 x − 30 x 2 + 30 x b. y ′ ′ + 4 y ′ + 5 y = 35 e − 4 x y ( 0 ) = − 3 y ′ ( 0 ) = 1 y''+4y'+5y=35e^{-4x} \qquad\qquad y(0)=-3 \qquad y'(0)=1 y ′′ + 4 y ′ + 5 y = 35 e − 4 x y ( 0 ) = − 3 y ′ ( 0 ) = 1
m 2 + 4 m + 5 = 0 m = 1 2 [ − 4 ± 16 − 20 ] = − 1 2 [ − 4 ± 2 i ] = − 2 ± i α = − 4 ≠ m 1 ≠ m 2 y c = e − 2 x ( C 1 cos x + C 2 sin x ) \begin{aligned}
m^2+4m+5&=0\\
m&=\frac{1}{2}\left[-4\pm\sqrt{16-20}\right] \\
&=-\frac{1}{2}[-4\pm 2i] \\
&=-2\pm i \\ \\
\alpha&=-4\neq m_1\neq m_2 \\
y_c&=e^{-2x}(C_1\cos x+C_2\sin x)
\end{aligned} m 2 + 4 m + 5 m α y c = 0 = 2 1 [ − 4 ± 16 − 20 ] = − 2 1 [ − 4 ± 2 i ] = − 2 ± i = − 4 = m 1 = m 2 = e − 2 x ( C 1 cos x + C 2 sin x ) y p = A e − 4 x y p ′ = − 4 A e − 4 x y p ′ ′ = 16 A e − 4 x \begin{aligned}
y_p&=Ae^{-4x} \\
y'_p&=-4Ae^{-4x} \\
y''_p&=16Ae^{-4x}
\end{aligned} y p y p ′ y p ′′ = A e − 4 x = − 4 A e − 4 x = 16 A e − 4 x Into equation
( 16 A e − 4 x ) + 4 ( − 4 A e − 4 x ) + 5 ( A e − 4 x ) = 35 e − 4 x 5 A e − 4 x = 35 e − 4 x A = 7 \begin{aligned}
(16Ae^{-4x})+4(-4Ae^{-4x})+5(Ae^{-4x})&=35e^{-4x} \\
5Ae^{-4x}&=35e^{-4x} \\
A&=7
\end{aligned} ( 16 A e − 4 x ) + 4 ( − 4 A e − 4 x ) + 5 ( A e − 4 x ) 5 A e − 4 x A = 35 e − 4 x = 35 e − 4 x = 7 General Solution
y = C 1 e − 2 x cos x + C 2 e − 2 x sin x + 7 e − 2 x y=C_1e^{-2x}\cos x+C_2e^{-2x}\sin x+7e^{-2x} y = C 1 e − 2 x cos x + C 2 e − 2 x sin x + 7 e − 2 x When
x = 0 , y = − 3 ⇒ − 3 = C 1 + 7 → C 1 = − 10 x=0, y=-3 \qquad \Rightarrow \qquad -3=C_1+7 \quad \rightarrow \quad C_1=-10 x = 0 , y = − 3 ⇒ − 3 = C 1 + 7 → C 1 = − 10 x = 0 , y ′ = 1 ⇒ 1 = − 2 C 1 e − 2 x cos x − C 1 e − 2 x sin x − 2 C 2 e − 2 x sin x + C 2 e − 2 x cos x − 28 e − 4 x 1 = − 2 C 1 + C 2 − 28 C 2 = 9 \begin{aligned}
x=0, y'=1 \qquad \Rightarrow \qquad 1&=-2C_1e^{-2x}\cos x-C_1e^{-2x}\sin x-2C_2e^{-2x}\sin x+C_2e^{-2x}\cos x-28e^{-4x}\\
1&=-2C_1+C_2-28 \\
C_2&=9
\end{aligned} x = 0 , y ′ = 1 ⇒ 1 1 C 2 = − 2 C 1 e − 2 x cos x − C 1 e − 2 x sin x − 2 C 2 e − 2 x sin x + C 2 e − 2 x cos x − 28 e − 4 x = − 2 C 1 + C 2 − 28 = 9 ∴ y = − 10 e − 2 x cos x + 9 e − 2 x sin x + 7 e − 4 x \therefore y=-10e^{-2x}\cos x+9e^{-2x}\sin x+7e^{-4x} ∴ y = − 10 e − 2 x cos x + 9 e − 2 x sin x + 7 e − 4 x c. y ′ ′ + 25 y = 5 x y ( 0 ) = 5 y ′ ( 0 ) = − 4.8 y''+25y=5x \qquad\qquad y(0)=5 \qquad y'(0)=-4.8 y ′′ + 25 y = 5 x y ( 0 ) = 5 y ′ ( 0 ) = − 4.8
m 2 + 25 = 0 m = ± 5 i y c = C 1 e 5 i x + C 2 e 5 i x = A 1 cos 5 x + A 2 sin 5 x α = 0 ≠ m 1 ≠ m 2 \begin{aligned}
m^2+25&=0\\
m&=\pm 5i \\ \\
y_c&=C_1e^{5ix}+C_2e^{5ix}\\
&=A_1\cos 5x+A_2\sin5x\\
\alpha&=0\neq m_1\neq m_2
\end{aligned} m 2 + 25 m y c α = 0 = ± 5 i = C 1 e 5 i x + C 2 e 5 i x = A 1 cos 5 x + A 2 sin 5 x = 0 = m 1 = m 2 y p = B 1 x + B 2 y p ′ = B y p ′ ′ = 0 ⇒ ( 0 ) + 25 ( B 1 x + B 2 ) = 5 x 25 B 1 x + 25 B 2 = 5 x B 1 = 1 5 , B 2 = 0 \begin{aligned}
y_p&=B_1x+B_2\\
y_p'&=B\\
y_p''&=0 \\
\Rightarrow (0)+25(B_1x+B_2)&=5x\\
25B_1x+25B_2&=5x\\
B_1=\frac{1}{5} &, B_2=0
\end{aligned} y p y p ′ y p ′′ ⇒ ( 0 ) + 25 ( B 1 x + B 2 ) 25 B 1 x + 25 B 2 B 1 = 5 1 = B 1 x + B 2 = B = 0 = 5 x = 5 x , B 2 = 0 y p = 1 5 x y_p=\frac{1}{5}x y p = 5 1 x When
x = 0 , y = 5 5 = A 1 cos 5 x + A 2 sin 5 x + 1 5 x 5 = A 1 x = 0 , y ′ = − 4.8 − 48 = − 5 A 1 sin 5 x + 5 A 2 cos 5 x + 1 5 − 4.8 = 5 A 2 + 0.2 A 2 = − 1 \begin{aligned}
x=0,y=5 \qquad 5&=A_1\cos 5x+A_2\sin 5x +\frac{1}{5}x \\
5&=A_1 \\ \\
x=0, y'=-4.8 \qquad -48&=-5A_1\sin 5x+5A_2\cos 5x+\frac{1}{5} \\
-4.8&=5A_2+0.2 \\
A_2&=-1
\end{aligned} x = 0 , y = 5 5 5 x = 0 , y ′ = − 4.8 − 48 − 4.8 A 2 = A 1 cos 5 x + A 2 sin 5 x + 5 1 x = A 1 = − 5 A 1 sin 5 x + 5 A 2 cos 5 x + 5 1 = 5 A 2 + 0.2 = − 1 ∴ y = 5 cos 5 x − sin 5 x + 1 5 x \therefore y=5\cos 5x-\sin 5x+\frac{1}{5}x ∴ y = 5 cos 5 x − sin 5 x + 5 1 x d. y ′ ′ − 2 y ′ + y = 2 x 2 − 8 x + 4 y ( 0 ) = 0.3 y ′ ( 0 ) = 0.3 y''-2y'+y=2x^2-8x+4 \qquad\qquad y(0)=0.3 \qquad y'(0)=0.3 y ′′ − 2 y ′ + y = 2 x 2 − 8 x + 4 y ( 0 ) = 0.3 y ′ ( 0 ) = 0.3
m 2 − 2 m + 1 = 0 ( m − 1 ) 2 = 0 m = 1 y c = C 1 e x + C 2 x e x \begin{aligned}
m^2-2m+1&=0\\
(m-1)^2&=0\\
m&=1\\
y_c&=C_1e^x+C_2xe^x
\end{aligned} m 2 − 2 m + 1 ( m − 1 ) 2 m y c = 0 = 0 = 1 = C 1 e x + C 2 x e x y p = A x 2 + B x + 4 y p ′ = 2 A x + B y p ′ ′ = 2 A \begin{aligned}
y_p&=Ax^2+Bx+4 \\
y'_p&=2Ax+B \\
y''_p&=2A
\end{aligned} y p y p ′ y p ′′ = A x 2 + B x + 4 = 2 A x + B = 2 A Into formula
( 2 A ) − 2 ( 2 A x + B ) + ( A x 2 + B x + C ) = 2 x 2 − 8 x + 4 A x 2 + ( − 4 A + B ) x + ( 2 A − 2 B + C ) = 2 x 2 − 8 x + 4 \begin{aligned}
(2A)-2(2Ax+B)+(Ax^2+Bx+C)&=2x^2-8x+4\\
Ax^2+(-4A+B)x+(2A-2B+C)&=2x^2-8x+4
\end{aligned} ( 2 A ) − 2 ( 2 A x + B ) + ( A x 2 + B x + C ) A x 2 + ( − 4 A + B ) x + ( 2 A − 2 B + C ) = 2 x 2 − 8 x + 4 = 2 x 2 − 8 x + 4 thus
A = 2 − 4 A + B = − 8 2 A − 2 B + C = 4 B = 0 C = 0 \begin{aligned}
A&=2 \qquad &-4A+B &=-8 \qquad &2A-2B+C&=4 \\ &&B&=0 &C&=0
\end{aligned} A = 2 − 4 A + B B = − 8 = 0 2 A − 2 B + C C = 4 = 0 y p = 2 x 2 y_p=2x^2 y p = 2 x 2 y = C 1 e x + C 2 x e x + 2 x 2 y ′ = C 1 e x + C 2 e x + C 2 x e x + 4 x \begin{aligned}
y&=C_1e^x+C_2xe^x+2x^2\\y'&=C_1e^x+C_2e^x+C_2xe^x+4x
\end{aligned} y y ′ = C 1 e x + C 2 x e x + 2 x 2 = C 1 e x + C 2 e x + C 2 x e x + 4 x When
x = 0 , y = 0.3 0.3 = C 1 + 0 + 0 C 1 = 0.3 x = 0 , y ′ = 0.3 0.3 = C 1 + C 2 + 0 + 0 C 1 + C 2 = 0.3 C 2 = 0 \begin{aligned}
x=0, y=0.3 \qquad 0.3&=C_1+0+0 \\
C_1&=0.3\\
x=0, y'=0.3 \qquad 0.3&=C_1+C_2+0+0\\
C_1+C_2&=0.3 \\
C_2&=0
\end{aligned} x = 0 , y = 0.3 0.3 C 1 x = 0 , y ′ = 0.3 0.3 C 1 + C 2 C 2 = C 1 + 0 + 0 = 0.3 = C 1 + C 2 + 0 + 0 = 0.3 = 0 ∴ y = 0.3 e x + 2 x 2 \therefore y=0.3e^x+2x^2 ∴ y = 0.3 e x + 2 x 2 e. y ′ ′ − y ′ − 2 y = 3 e 2 x y ( 1 ) = e − 1 y ′ ( 1 ) = − e − 1 + e 2 y''-y'-2y=3e^{2x} \qquad\qquad y(1)=e^{-1} \qquad y'(1)=-e^{-1}+e^2 y ′′ − y ′ − 2 y = 3 e 2 x y ( 1 ) = e − 1 y ′ ( 1 ) = − e − 1 + e 2
m 2 − m − 2 = 0 ( m − 2 ) ( m + 1 ) = 0 m = 2 , − 1 y p = C 1 e 2 x + C 2 e − x α = 2 = m 1 ≠ m 2 \begin{aligned}
m^2-m-2&=0\\
(m-2)(m+1)&=0\\
m&=2,-1\\ \\
y_p&=C_1e^{2x}+C_2e^{-x}\\
\alpha&=2=m_1\neq m_2
\end{aligned} m 2 − m − 2 ( m − 2 ) ( m + 1 ) m y p α = 0 = 0 = 2 , − 1 = C 1 e 2 x + C 2 e − x = 2 = m 1 = m 2 y p = A x e 2 x y p ′ = 2 A x e 2 x + A e 2 x y p ′ ′ = 4 A x e 2 x + 2 A e 2 x + 2 A e 2 x = 4 A x e 2 x + 4 A e 2 x \begin{aligned}
y_p&=Axe^{2x}\\
y'_p&=2Axe^{2x}+Ae^{2x}\\
y''_p&=4Axe^{2x}+2Ae^{2x}+2Ae^{2x}\\
&=4Axe^{2x}+4Ae^{2x}
\end{aligned} y p y p ′ y p ′′ = A x e 2 x = 2 A x e 2 x + A e 2 x = 4 A x e 2 x + 2 A e 2 x + 2 A e 2 x = 4 A x e 2 x + 4 A e 2 x Into formula
( 4 A x e 2 x + 4 A e 2 x ) − ( 2 A x e 2 x + A e 2 x ) − 2 ( A x e 2 x ) = 3 e 2 x ( 4 A − 2 A − 2 A ) x e 2 x + ( 4 A − A ) e 2 x = 3 e 2 x 3 A = 3 A = 1 \begin{aligned}
(4Axe^{2x}+4Ae^{2x})-(2Axe^{2x}+Ae^{2x})-2(Axe^{2x})&=3e^{2x} \\
(4A-2A-2A)xe^{2x}+(4A-A)e^{2x}&=3e^{2x} \\
3A&=3\\
A&=1
\end{aligned} ( 4 A x e 2 x + 4 A e 2 x ) − ( 2 A x e 2 x + A e 2 x ) − 2 ( A x e 2 x ) ( 4 A − 2 A − 2 A ) x e 2 x + ( 4 A − A ) e 2 x 3 A A = 3 e 2 x = 3 e 2 x = 3 = 1 y = C 1 e 2 x + C 2 e − x + x e 2 x y ′ = 2 C 1 e 2 x − C 2 e − x + 2 x e 2 x + e 2 x \begin{aligned}
y=C_1e^{2x}+C_2e^{-x}+xe^{2x} \\
y'&=2C_1e^{2x}-C_2e^{-x}+2xe^{2x}+e^{2x}
\end{aligned} y = C 1 e 2 x + C 2 e − x + x e 2 x y ′ = 2 C 1 e 2 x − C 2 e − x + 2 x e 2 x + e 2 x When
x = 1 , y = C 1 e 2 C 2 e − 1 + e 2 = e − 1 C 1 e 2 + C 2 e − 1 = e − 1 − e 2 (1) \begin{aligned}
x=1, \qquad y=C_1e^2C_2e^{-1}+e^2&=e^{-1} \\
C_1e^2+C_2e^{-1}&=e^{-1}-e^2 \tag{1}
\end{aligned} x = 1 , y = C 1 e 2 C 2 e − 1 + e 2 C 1 e 2 + C 2 e − 1 = e − 1 = e − 1 − e 2 ( 1 ) x = 1 , y = C 1 e 2 C 2 e − 1 + e 2 = e − 1 C 1 e 2 + C 2 e − 1 = e − 1 − e 2 (1) \begin{aligned}
x=1, \qquad y=C_1e^2C_2e^{-1}+e^2&=e^{-1} \\
C_1e^2+C_2e^{-1}&=e^{-1}-e^2 \tag{1}
\end{aligned} x = 1 , y = C 1 e 2 C 2 e − 1 + e 2 C 1 e 2 + C 2 e − 1 = e − 1 = e − 1 − e 2 ( 1 ) x = 1 , y ′ = 2 C 1 e 2 − C 2 e − 1 + 2 e 2 + e 2 = − e − 1 + e 2 2 C 1 e 2 − C 2 e − 1 = − e − 1 − 2 e 2 (2) \begin{aligned}
x=1, \qquad y'=2C_1e^2-C_2e^{-1}+2e^2+e^2&=-e^{-1}+e^2\\
2C_1e^2-C_2e^{-1}&=-e^{-1}-2e^2 \tag{2}
\end{aligned} x = 1 , y ′ = 2 C 1 e 2 − C 2 e − 1 + 2 e 2 + e 2 2 C 1 e 2 − C 2 e − 1 = − e − 1 + e 2 = − e − 1 − 2 e 2 ( 2 ) ( 1 ) + ( 2 ) (1)+(2) ( 1 ) + ( 2 )
3 C 1 e 2 = − 3 e 2 C 1 = − 1 \begin{aligned}
3C_1e^2&=-3e&2\\
C_1&=-1
\end{aligned} 3 C 1 e 2 C 1 = − 3 e = − 1 2 Into ( 1 ) (1) ( 1 )
− e 2 + C 2 e − 1 = e − 1 − e 2 C 2 e − 1 = e − 1 C 2 = 1 \begin{aligned}
-e^2+C_2e^{-1}&=e^{-1}-e^2 \\
C_2e^{-1}&=e^{-1}\\
C_2&=1
\end{aligned} − e 2 + C 2 e − 1 C 2 e − 1 C 2 = e − 1 − e 2 = e − 1 = 1 ∴ y = − e 2 x + e − x + x e 2 x \therefore y=-e^{2x}+e^{-x}+xe^{2x} ∴ y = − e 2 x + e − x + x e 2 x
Obtain a general solution using the method of variation of parameter
a. y ′ ′ − 4 y ′ + 4 y = e 2 x / x y''-4y'+4y=e^{2x}/x y ′′ − 4 y ′ + 4 y = e 2 x / x
m 2 − 4 m + 4 = 0 ( m − 2 ) 2 = 0 m = 2 y c = C 1 e 2 x + C 2 x e 2 x \begin{aligned}
m^2-4m+4&=0\\
(m-2)^2&=0\\m&=2\\ \\
y_c&=C_1e^{2x}+C_2xe^{2x}
\end{aligned} m 2 − 4 m + 4 ( m − 2 ) 2 m y c = 0 = 0 = 2 = C 1 e 2 x + C 2 x e 2 x y 1 = e 2 x , y 1 ′ = 2 e 2 x y 2 = x e 2 x , y 2 ′ = 2 x e 2 x + e 2 x \begin{aligned}
y_1&=e^{2x} \qquad, \qquad &y'_1&=2e^{2x} \\
y_2&=xe^{2x} \qquad, \qquad &y'_2&=2xe^{2x}+e^{2x}
\end{aligned} y 1 y 2 = e 2 x , = x e 2 x , y 1 ′ y 2 ′ = 2 e 2 x = 2 x e 2 x + e 2 x W ( y 1 , y 2 ) = ∣ e 2 x x e 2 x 2 e 2 x 2 x e 2 x + e 2 x ∣ = 2 x e 4 x + e 4 x − 2 x e 4 x = e 4 x \begin{aligned}
W(y_1,y_2)&=\left\vert \begin{array}{cc} e^{2x} & xe^{2x} \\ 2e^{2x} & 2xe^{2x}+e^{2x} \end{array}\right\vert \\
&=2xe^{4x}+e^{4x}-2xe^{4x}\\
&=e^{4x}
\end{aligned} W ( y 1 , y 2 ) = e 2 x 2 e 2 x x e 2 x 2 x e 2 x + e 2 x = 2 x e 4 x + e 4 x − 2 x e 4 x = e 4 x U 1 ( x ) = − ∫ ( e 2 x x ) ( x e 2 x ) e 4 x d x = − ∫ d x = − x U 2 ( x ) = − ∫ ( e 2 x x ) ( e 2 x ) e 4 x d x = − ∫ 1 x d x = ln x \begin{aligned}
U_1(x)=-\int\frac{\left(\frac{e^{2x}}{x}\right)(xe^{2x})}{e^{4x}}dx\\
&=-\int dx
&=-x
U_2(x)=-\int\frac{\left(\frac{e^{2x}}{x}\right)(e^{2x})}{e^{4x}}dx\\
&=-\int \frac{1}{x} dx\\
&=\ln x
\end{aligned} U 1 ( x ) = − ∫ e 4 x ( x e 2 x ) ( x e 2 x ) d x = − ∫ d x = − ∫ x 1 d x = ln x = − x U 2 ( x ) = − ∫ e 4 x ( x e 2 x ) ( e 2 x ) d x y p = ( − x ) ( e 2 x ) + ( ln x ) ( x e 2 x ) = − x e 2 x + x e 2 x ln x \begin{aligned}
y_p&=(-x)(e^{2x})+(\ln x)(xe^{2x}) \\
&=-xe^{2x}+xe^{2x}\ln x
\end{aligned} y p = ( − x ) ( e 2 x ) + ( ln x ) ( x e 2 x ) = − x e 2 x + x e 2 x ln x ∴ y = C 1 e 2 x + C 2 x e 2 x − x e 2 x + x e 2 x ln x = C 1 e 2 x + ( C 2 − 1 ) x e 2 x + x e 2 x ln x = C 1 e 2 x + C 3 x e 2 x + x e 2 x ln x \begin{aligned}
\therefore y&=C_1e^{2x}+C_2xe^{2x}-xe^{2x}+xe^{2x}\ln x \\
&=C_1e^{2x}+(C_2-1)xe^{2x}+xe^{2x}\ln x \\
&=C_1e^{2x}+C_3xe^{2x}+xe^{2x}\ln x
\end{aligned} ∴ y = C 1 e 2 x + C 2 x e 2 x − x e 2 x + x e 2 x ln x = C 1 e 2 x + ( C 2 − 1 ) x e 2 x + x e 2 x ln x = C 1 e 2 x + C 3 x e 2 x + x e 2 x ln x where C 3 = C 2 − 1 C_3=C_2-1 C 3 = C 2 − 1
b. y ′ ′ + 2 y ′ + y = x 3 / 2 / e x y''+2y'+y=x^{3/2}/e^x y ′′ + 2 y ′ + y = x 3/2 / e x
m 2 + 2 m + 1 = 0 ( m + 1 ) ( m + 1 ) = 0 m = − 1 y c = C 1 e − x + C 2 x e − x \begin{aligned}
m^2+2m+1&=0\\
(m+1)(m+1)&=0\\
m&=-1\\ \\
y_c&=C_1e^{-x}+C_2xe^{-x}
\end{aligned} m 2 + 2 m + 1 ( m + 1 ) ( m + 1 ) m y c = 0 = 0 = − 1 = C 1 e − x + C 2 x e − x y 1 = e − x , y 1 ′ = − e − x y 2 = x e − x , y 2 ′ = − x e − x + e − x \begin{aligned}
y_1&=e^{-x} \qquad, \qquad &y'_1&=-e^{-x} \\
y_2&=xe^{-x} \qquad, \qquad &y'_2&=-xe^{-x}+e^{-x}
\end{aligned} y 1 y 2 = e − x , = x e − x , y 1 ′ y 2 ′ = − e − x = − x e − x + e − x W ( y 1 , y 2 ) = ∣ e − x x e − x − e − x − x e − x + e − x ∣ = − x e − 2 x + e − 2 x + x e − 2 x = e − 2 x \begin{aligned}
W(y_1,y_2)&=\left\vert \begin{array}{cc} e^{-x} & xe^{-x} \\ -e^{-x} & -xe^{-x}+e^{-x} \end{array}\right\vert \\
&=-xe^{-2x}+e^{-2x}+xe^{-2x} \\
&=e^{-2x}
\end{aligned} W ( y 1 , y 2 ) = e − x − e − x x e − x − x e − x + e − x = − x e − 2 x + e − 2 x + x e − 2 x = e − 2 x U 1 ( x ) = − ∫ ( x 3 2 e x ) ( x e − x ) e − 2 x d x = − ∫ x 5 2 d x = − 2 7 x 7 2 U 2 ( x ) = − ∫ ( x 3 2 e x ) ( e − x ) e − 2 x d x = ∫ x 3 2 d x = − 2 5 x 5 2 \begin{aligned}
U_1(x)=-\int\frac{\left(\frac{x^{\frac{3}{2}}}{e^x}\right)(xe^{-x})}{e^{-2x}} dx \\
&=-\int x^{\frac{5}{2}} dx \\
&=-\frac{2}{7}x^{\frac{7}{2}} \\
U_2(x)=-\int\frac{\left(\frac{x^{\frac{3}{2}}}{e^x}\right)(e^{-x})}{e^{-2x}}dx\\
&=\int x^{\frac{3}{2}} dx \\
&=-\frac{2}{5}x^{\frac{5}{2}} \\
\end{aligned} U 1 ( x ) = − ∫ e − 2 x ( e x x 2 3 ) ( x e − x ) d x U 2 ( x ) = − ∫ e − 2 x ( e x x 2 3 ) ( e − x ) d x = − ∫ x 2 5 d x = − 7 2 x 2 7 = ∫ x 2 3 d x = − 5 2 x 2 5 y p = ( − 2 7 x 7 2 ) ( e − x ) + ( 2 5 x 5 2 x e − x ) = ( − 10 35 + 14 35 ) x 7 2 e − x = 4 35 x 7 2 e − x \begin{aligned}
y_p&=\left(-\frac{2}{7}x^{\frac{7}{2}}\right)(e^{-x})+\left(\frac{2}{5}x^{\frac{5}{2}}xe^{-x}\right) \\
&=\left(-\frac{10}{35}+\frac{14}{35}\right)x^{\frac{7}{2}}e^{-x} \\
&=\frac{4}{35}x^{\frac{7}{2}}e^{-x}
\end{aligned} y p = ( − 7 2 x 2 7 ) ( e − x ) + ( 5 2 x 2 5 x e − x ) = ( − 35 10 + 35 14 ) x 2 7 e − x = 35 4 x 2 7 e − x ∴ C 1 e − x + C 2 x e − x + 4 35 x 7 2 e − x \therefore C_1e^{-x}+C_2xe^{-x}+\frac{4}{35}x^{\frac{7}{2}}e^{-x} ∴ C 1 e − x + C 2 x e − x + 35 4 x 2 7 e − x c. y ′ ′ − 2 y ′ + y = e x sin x y''-2y'+y=e^x\sin x y ′′ − 2 y ′ + y = e x sin x
m 2 − 2 m + 1 = 0 ( m − 1 ) 2 = 0 m = 1 y c = C 1 e x + C 2 x e x \begin{aligned}
m^2-2m+1&=0\\
(m-1)^2&=0\\
m&=1\\ \\
y_c&=C_1e^x+C_2xe^x
\end{aligned} m 2 − 2 m + 1 ( m − 1 ) 2 m y c = 0 = 0 = 1 = C 1 e x + C 2 x e x y 1 = e x , y 1 ′ = e x y 2 = x e x , y 2 ′ = x e x + e x \begin{aligned}
y_1&=e^x \qquad , \qquad &y'_1=&e^x \\
y_2&=xe^x \qquad , \qquad &y'_2&=xe^x+e^x
\end{aligned} y 1 y 2 = e x , = x e x , y 1 ′ = y 2 ′ e x = x e x + e x W ( y 1 , y 2 ) = ∣ e x x e x e x x e x + e x ∣ = x e 2 x + e 2 x − x e 2 x = e 2 x \begin{aligned}
W(y_1,y_2)&=\left\vert \begin{array}{cc} e^x&xe^x\\e^x&xe^x+e^x \end{array}\right\vert \\
&=xe^{2x}+e^{2x}-xe^{2x} \\
&=e^{2x}
\end{aligned} W ( y 1 , y 2 ) = e x e x x e x x e x + e x = x e 2 x + e 2 x − x e 2 x = e 2 x U 1 ( x ) = − ∫ ( e x sin x ) ( x e x ) e 2 x d x = − ∫ x sin x d x = x cos x − sin x U 2 ( x ) = − ∫ ( e x sin x ) ( e x ) e 2 x d x = − ∫ sin x d x = − cos x \begin{aligned}
U_1(x)&=-\int \frac{(e^x\sin x)(xe^x)}{e^{2x}} dx \\
&=-\int x\sin x dx \\
&=x\cos x - \sin x \\
U_2(x)&=-\int \frac{(e^x\sin x)(e^x)}{e^{2x}} dx \\
&=-\int \sin x dx \\
&=-\cos x
\end{aligned} U 1 ( x ) U 2 ( x ) = − ∫ e 2 x ( e x sin x ) ( x e x ) d x = − ∫ x sin x d x = x cos x − sin x = − ∫ e 2 x ( e x sin x ) ( e x ) d x = − ∫ sin x d x = − cos x y p = ( x cos x − sin x ) e x + ( − cos x ) x e x = x e x cos x − e x sin x − x e x cos x = − e x sin x \begin{aligned}
y_p=(x\cos x-\sin x)e^x+(-\cos x)xe^x \\
&=xe^x\cos x-e^x\sin x - xe^x\cos x \\
&=-e^x\sin x
\end{aligned} y p = ( x cos x − sin x ) e x + ( − cos x ) x e x = x e x cos x − e x sin x − x e x cos x = − e x sin x ∴ y = C 1 e x + C 2 x e x − e x sin x \therefore y=C_1e^x+C_2xe^x-e^x\sin x ∴ y = C 1 e x + C 2 x e x − e x sin x d. y ′ ′ + 2 y ′ + 2 y = e − x sec x y''+2y'+2y=e^{-x}\sec x y ′′ + 2 y ′ + 2 y = e − x sec x
m 2 + 2 m + 2 = 0 m = 1 2 [ − 2 ± 4 − 4 ( 1 ) ( 2 ) ] = 1 2 [ − 2 ± 2 i ] = − 1 ± i y c = e − x ( C 1 cos x + C 2 sin x ) \begin{aligned}
m^2+2m+2&=0 \\
m&=\frac{1}{2}[-2\pm\sqrt{4-4(1)(2)}] \\
&=\frac{1}{2}[-2\pm 2i] \\
&=-1\pm i \\ \\
y_c&=e^{-x}(C_1\cos x+C_2\sin x)
\end{aligned} m 2 + 2 m + 2 m y c = 0 = 2 1 [ − 2 ± 4 − 4 ( 1 ) ( 2 ) ] = 2 1 [ − 2 ± 2 i ] = − 1 ± i = e − x ( C 1 cos x + C 2 sin x ) y 1 = e − x cos x , y 1 ′ = − e − x cos x − e − x sin x y 2 = e − x sin x , y 1 ′ = − e − x sin x + e − x cos x \begin{aligned}
y_1&=e^{-x}\cos x \qquad , \qquad &y'_1&=-e^{-x}\cos x-e^{-x}\sin x \\
y_2&=e^{-x}\sin x \qquad , \qquad &y'_1&=-e^{-x}\sin x+e^{-x}\cos x
\end{aligned} y 1 y 2 = e − x cos x , = e − x sin x , y 1 ′ y 1 ′ = − e − x cos x − e − x sin x = − e − x sin x + e − x cos x W ( y 1 , y 2 ) = ∣ e − x cos x e − x sin x − e − x cos x − e − x sin x − e − x sin x + e − x cos x ∣ = − e − 2 x sin x cos x + e − 2 x cos 2 x + e − 2 x sin x cos x + e − 2 x sin 2 x = e − 2 x ( cos 2 x + sin 2 x ) = e − 2 x \begin{aligned}
W(y_1,y_2)&=\left\vert \begin{array}{cc} e^{-x}\cos x & e^{-x}\sin x \\ -e^{-x}\cos x-e^{-x}\sin x & -e^{-x}\sin x+e^{-x}\cos x \end{array}\right\vert \\
&=-e^{-2x}\sin x\cos x +e^{-2x}\cos^2 x + e^{-2x}\sin x\cos x+e^{-2x} \sin^2 x \\
&=e^{-2x} (\cos^2 x+\sin^2 x) \\
&=e^{-2x}
\end{aligned} W ( y 1 , y 2 ) = e − x cos x − e − x cos x − e − x sin x e − x sin x − e − x sin x + e − x cos x = − e − 2 x sin x cos x + e − 2 x cos 2 x + e − 2 x sin x cos x + e − 2 x sin 2 x = e − 2 x ( cos 2 x + sin 2 x ) = e − 2 x U 1 ( x ) = − ∫ ( e − x sec x ) ( e − x sin x ) e − 2 x d x = − ∫ sin x cos x d x = ln ∣ cos x ∣ U 2 ( x ) = − ∫ ( e − x sec x ) ( e − x cos x ) e − 2 x d x = − ∫ d x = x \begin{aligned}
U_1(x)&=-\int\frac{(e^{-x}\sec x)(e^{-x}\sin x)}{e^{-2x}} dx \\
&=-\int \frac{\sin x}{\cos x} dx \\
&= \ln \vert\cos x\vert \\
U_2(x)&=-\int\frac{(e^{-x}\sec x)(e^{-x}\cos x)}{e^{-2x}} dx \\
&=-\int dx \\
&= x
\end{aligned} U 1 ( x ) U 2 ( x ) = − ∫ e − 2 x ( e − x sec x ) ( e − x sin x ) d x = − ∫ cos x sin x d x = ln ∣ cos x ∣ = − ∫ e − 2 x ( e − x sec x ) ( e − x cos x ) d x = − ∫ d x = x y p = e − x cos x ln ∣ cos x ∣ + e − x sin x ⋅ x y_p=e^{-x}\cos x \ln \vert \cos x \vert + e^{-x}\sin x \cdot x y p = e − x cos x ln ∣ cos x ∣ + e − x sin x ⋅ x ∴ y = e − x ( C 1 cos x + C 2 sin x ) + e − x cos x ln ∣ cos x ∣ + x e − x sin x \therefore y=e^{-x}(C_1\cos x+C_2\sin x)+e^{-x}\cos x \ln \vert \cos x \vert + xe^{-x}\sin x ∴ y = e − x ( C 1 cos x + C 2 sin x ) + e − x cos x ln ∣ cos x ∣ + x e − x sin x e. y ′ ′ − 4 y ′ + 4 y = ( x + 1 ) e 2 x y''-4y'+4y=(x+1)e^{2x} y ′′ − 4 y ′ + 4 y = ( x + 1 ) e 2 x
m 2 − 4 m + 4 = 0 ( m − 2 ) ( m − 2 ) = 0 m = 2 y c = C 1 e 2 x + C 2 x e 2 x \begin{aligned}
m^2-4m+4&=0 \\
(m-2)(m-2)&=0 \\
m&=2 \\ \\
y_c&=C_1e^{2x}+C_2xe^{2x}
\end{aligned} m 2 − 4 m + 4 ( m − 2 ) ( m − 2 ) m y c = 0 = 0 = 2 = C 1 e 2 x + C 2 x e 2 x y 1 = e 2 x , y 1 ′ = 2 e 2 x y 2 = x e 2 x , y 2 ′ = 2 x e 2 x + e 2 x \begin{aligned}
y_1 &=e^{2x} \qquad , \qquad &y'_1&=2e^{2x} \\
y_2&=xe^{2x} \qquad , \qquad &y'_2&=2xe^{2x}+e^{2x}
\end{aligned} y 1 y 2 = e 2 x , = x e 2 x , y 1 ′ y 2 ′ = 2 e 2 x = 2 x e 2 x + e 2 x W ( y 1 , y 2 ) = ∣ e 2 x x e 2 x 2 e 2 x 2 x e 2 x + e 2 x ∣ = 2 x e 4 x + e 4 x − 2 x e 4 x = e 4 x \begin{aligned}
W(y_1,y_2)&=\left\vert \begin{array}{cc} e^{2x} & xe^{2x} \\ 2e^{2x} & 2xe^{2x}+e^{2x} \end{array}\right\vert \\
&= 2xe^{4x}+e^{4x}-2xe^{4x} \\
&=e^{4x}
\end{aligned} W ( y 1 , y 2 ) = e 2 x 2 e 2 x x e 2 x 2 x e 2 x + e 2 x = 2 x e 4 x + e 4 x − 2 x e 4 x = e 4 x U 1 ( x ) = − ∫ ( x + 1 ) e 2 x ⋅ x e 2 x e 4 x d x = − ∫ x 2 + x d x = − 1 3 x 3 − 1 2 x 2 U 2 ( x ) = − ∫ ( x + 1 ) e 2 x ⋅ e 2 x e 4 x d x = − ∫ x + 1 d x = 1 2 x 2 + x \begin{aligned}
U_1(x)=-\int \frac{(x+1)e^{2x}\cdot xe^{2x}}{e^{4x}} dx \\
&=-\int x^2+x dx \\
&=-\frac{1}{3}x^3 - \frac{1}{2} x^2 \\
U_2(x)=-\int \frac{(x+1)e^{2x}\cdot e^{2x}}{e^{4x}} dx \\
&=-\int x+1 dx \\
&=\frac{1}{2}x^2 +x
\end{aligned} U 1 ( x ) = − ∫ e 4 x ( x + 1 ) e 2 x ⋅ x e 2 x d x U 2 ( x ) = − ∫ e 4 x ( x + 1 ) e 2 x ⋅ e 2 x d x = − ∫ x 2 + x d x = − 3 1 x 3 − 2 1 x 2 = − ∫ x + 1 d x = 2 1 x 2 + x y p = ( − 1 3 x 3 − 1 2 x 2 ) ( e 2 x ) + ( 1 2 x 2 + x ) ( x e 2 x ) = − 1 3 x 3 e 2 x − 1 2 x 2 e 2 x + 1 2 x 3 e 2 x + x 2 e 2 x = 1 6 x 3 e 2 x + 1 2 x 2 e 2 x \begin{aligned}
y_p&=\left(-\frac{1}{3}x^3-\frac{1}{2}x^2\right)(e^{2x})+\left(\frac{1}{2}x^2+x\right)(xe^{2x}) \\
&=-\frac{1}{3}x^3e^{2x}-\frac{1}{2}x^2e^{2x}+\frac{1}{2}x^3e^{2x}+x^2e^{2x} \\
&=\frac{1}{6}x^3e^{2x}+\frac{1}{2}x^2e^{2x}
\end{aligned} y p = ( − 3 1 x 3 − 2 1 x 2 ) ( e 2 x ) + ( 2 1 x 2 + x ) ( x e 2 x ) = − 3 1 x 3 e 2 x − 2 1 x 2 e 2 x + 2 1 x 3 e 2 x + x 2 e 2 x = 6 1 x 3 e 2 x + 2 1 x 2 e 2 x ∴ y = C 1 e 2 x + C 2 x e 2 x + 1 6 x 3 e 2 x + 1 2 x 2 e 2 x \therefore y=C_1e^{2x} + C_2 xe^{2x} + \frac{1}{6}x^3e^{2x}+\frac{1}{2}x^2e^{2x} ∴ y = C 1 e 2 x + C 2 x e 2 x + 6 1 x 3 e 2 x + 2 1 x 2 e 2 x f. y ′ ′ − 5 y ′ + 6 y = 2 e x y''-5y'+6y=2e^x y ′′ − 5 y ′ + 6 y = 2 e x
m 2 + 2 m + 1 = 0 ( m + 1 ) ( m + 1 ) = 0 m = − 1 y c = C 1 e − x + C 2 x e − x \begin{aligned}
m^2+2m+1&=0\\
(m+1)(m+1)&=0\\
m&=-1 \\ \\
y_c&=C_1e^{-x}+C_2xe^{-x}
\end{aligned} m 2 + 2 m + 1 ( m + 1 ) ( m + 1 ) m y c = 0 = 0 = − 1 = C 1 e − x + C 2 x e − x y 1 = e − x , y 1 ′ = − e − x y 2 = x e − x , y 2 ′ = − x e − x + e − x \begin{aligned}
y_1&=e^{-x} \qquad , \qquad &y'_1&=-e^{-x} \\
y_2&=xe^{-x} \qquad , \qquad &y'_2&=-xe^{-x}+e^{-x}
\end{aligned} y 1 y 2 = e − x , = x e − x , y 1 ′ y 2 ′ = − e − x = − x e − x + e − x W ( y 1 , y 2 ) = ∣ e − x x e − x − e − x − x e − x + e − x ∣ = − x e − 2 x + e − 2 x + x e − 2 x = e − 2 x \begin{aligned}
W(y_1,y_2)&=\left\vert \begin{array}{cc} e^{-x} & xe^{-x} \\ -e^{-x} & -xe^{-x}+e^{-x} \end{array}\right\vert \\
&=-xe^{-2x}+e^{-2x}+xe^{-2x} \\
&=e^{-2x}
\end{aligned} W ( y 1 , y 2 ) = e − x − e − x x e − x − x e − x + e − x = − x e − 2 x + e − 2 x + x e − 2 x = e − 2 x U 1 ( x ) = − ∫ ( 4 e − x ) ( x e − x ) e − 2 x d x = − ∫ 4 x d x = − 2 x 2 U 2 ( x ) = − ∫ ( 4 e − x ) ( e − x ) e − 2 x d x = − ∫ 4 d x = 4 x \begin{aligned}
U_1(x)&=-\int \frac{(4e^{-x})(xe^{-x})}{e^{-2x}} dx \\
&=-\int 4x dx \\
&=-2x^2 \\
U_2(x)&=-\int \frac{(4e^{-x})(e^{-x})}{e^{-2x}} dx \\
&=-\int 4 dx \\
&=4x
\end{aligned} U 1 ( x ) U 2 ( x ) = − ∫ e − 2 x ( 4 e − x ) ( x e − x ) d x = − ∫ 4 x d x = − 2 x 2 = − ∫ e − 2 x ( 4 e − x ) ( e − x ) d x = − ∫ 4 d x = 4 x y p = ( − 2 x 2 ) ( e − x ) + ( 4 x ) ( x e − x ) = − 2 x 2 e − x + 4 x 2 e − x = 2 x 2 e − x \begin{aligned}
y_p&=(-2x^2)(e^{-x})+(4x)(xe^{-x}) \\
&=-2x^2e^{-x}+4x^2e^{-x} \\
&=2x^2e^{-x}
\end{aligned} y p = ( − 2 x 2 ) ( e − x ) + ( 4 x ) ( x e − x ) = − 2 x 2 e − x + 4 x 2 e − x = 2 x 2 e − x ∴ y = C 1 e − x + C 2 x e − x + 2 x 2 e − x \therefore y=C_1e^{-x}+C_2xe^{-x}+2x^2e^{-x} ∴ y = C 1 e − x + C 2 x e − x + 2 x 2 e − x g. y ′ ′ + 2 y ′ + y = 4 e − x y''+2y'+y=4e^{-x} y ′′ + 2 y ′ + y = 4 e − x
m 2 − 5 m + 6 = 0 ( m − 3 ) ( m − 2 ) = 0 m = 2 , 3 y c = C 1 e 2 x + C 2 e 3 x \begin{aligned}
m^2-5m+6&=0\\
(m-3)(m-2)&=0\\m&=2,3 \\ \\
y_c&=C_1e^{2x}+C_2e^{3x}
\end{aligned} m 2 − 5 m + 6 ( m − 3 ) ( m − 2 ) m y c = 0 = 0 = 2 , 3 = C 1 e 2 x + C 2 e 3 x y 1 = e 2 x , y 1 ′ = 2 e 2 x y 2 = e 3 x , y 2 ′ = 3 e 3 x \begin{aligned}
y_1=e^{2x} \qquad , \qquad &y'_1&=2e^{2x} \\
y_2=e^{3x} \qquad , \qquad &y'_2&=3e^{3x}
\end{aligned} y 1 = e 2 x , y 2 = e 3 x , y 1 ′ y 2 ′ = 2 e 2 x = 3 e 3 x W ( y 1 , y 2 ) = ∣ e 2 x e 3 x 2 e 2 x 3 e 3 x ∣ = 3 e 5 x − 2 e 5 x = e 5 x \begin{aligned}
W(y_1,y_2)&=\left\vert \begin{array}{cc} e^{2x} & e^{3x} \\ 2e^{2x} & 3e^{3x} \end{array}\right\vert \\
&= 3e^{5x}-2e^{5x} \\
&=e^{5x}
\end{aligned} W ( y 1 , y 2 ) = e 2 x 2 e 2 x e 3 x 3 e 3 x = 3 e 5 x − 2 e 5 x = e 5 x U 1 ( x ) = − ∫ ( 2 e x ) ( e 3 x ) e 5 x d x = − ∫ 2 e − x d x = 2 e − x U 2 ( x ) = − ∫ ( 2 e x ) ( e 2 x ) e 5 x d x = ∫ 2 e − 2 x d x = − e − 2 x \begin{aligned}
U_1(x) &= -\int\frac{(2e^x)(e^{3x})}{e^{5x}} dx \\
&=-\int 2 e^{-x} dx \\
&=2e^{-x} \\
U_2(x) &= -\int\frac{(2e^x)(e^{2x})}{e^{5x}} dx \\
&=\int 2 e^{-2x} dx \\
&=-e^{-2x}
\end{aligned} U 1 ( x ) U 2 ( x ) = − ∫ e 5 x ( 2 e x ) ( e 3 x ) d x = − ∫ 2 e − x d x = 2 e − x = − ∫ e 5 x ( 2 e x ) ( e 2 x ) d x = ∫ 2 e − 2 x d x = − e − 2 x y p = ( 2 e − x ) ( e 2 x ) + ( − e − 2 x ) ( e 3 x ) = 2 e x − e x = e x \begin{aligned}
y_p&=(2e^{-x})(e^{2x})+(-e^{-2x})(e^{3x}) \\
&=2e^x-e^x \\
&=e^x
\end{aligned} y p = ( 2 e − x ) ( e 2 x ) + ( − e − 2 x ) ( e 3 x ) = 2 e x − e x = e x ∴ y = C 1 e 2 x + C 2 e 3 x + e x \therefore y = C_1e^{2x}+C_2e^{3x}+e^x ∴ y = C 1 e 2 x + C 2 e 3 x + e x