Answer to Question #126608 in Differential Equations for Hakam

Question

Solve the differential equation
y" + 4y' + 4y = 2cos2x + e^-x, y(0) = -1, y'(0) = 2

Answer 1

$~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~y^{\prime \prime}+4y^{\prime}+4 y= 2\cos 2x+ e^{-x}\\[1 em] \text{ Second order linear non-homogeneous differential equation . } \\[1 em] ~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~y=y_{h}+y_{p}\\[1 em] y_{h}~~~ \text{is the solution to the homogeneous .}\\[1 em] y^{\prime \prime}+4 y^{\prime}+4 y=0\\[1 em] r^{2}+4 r+4=0 \\[1 em] \begin{array}{l} (r+2)^2=0 \\[1 em] \therefore r_{1}= r_{2}=-2 \end{array}\\[1 em] ~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~\therefore y_{h}=c_{1} e^{- 2x}+c_{2}x e^{-2x}\\[1 em] y_{p} \text{ is the particular solution is any function} \\[1 em] \text{ That satisfies the non-homogeneous equation .} \\[1 em] \begin{array}{l} y_{p}=A e^{-x}+B \cos 2 x+C \sin 2 x \\[1 em] y_{p}^{\prime} =-A e^{-x} -2 B \sin 2 x+2 C \cos 2 x \\[1 em] y_{p}^{\prime\prime} =A e^{-x}-4 B \cos 2 x-4 C \sin 2 x \\[1 em] \text { substite in The original equation } \\[1 em] A e^{-x}-4B \cos 2 x-4 C \sin 2 x \\[1 em] -4 A e^{-x}-8 B \sin 2 x+8 C \cos 2 x \\[1 em] +4 A e^{-x}+4 B \cos 2 x+4 C \sin 2 x \\[1 em] =2 \cos 2 x+e^{-x} \\[1 em] \therefore A e^{-x}-8 B \sin 2 x+8 C \cos 2 x=2 \cos 2 x+e^{-x} \\[1 em] \therefore A=1,-8 B=0 \rightarrow B=0,8 C=2 \rightarrow C=\frac{1}{4} \\[1 em] \therefore y_{p}=e^{-x}+\frac{1}{4} \sin 2 x \\[1 em] \end{array} \\[1 em] \begin{array}{l} \therefore y=y_{h}+y_{p} \\[1 em] y=c_{1} e^{-2 x}+c_{2} x e^{-2 x}+\frac{1}{4} \sin 2 x+e^{-x} \\[1 em] y^{\prime}=-2 c_{1}e^{-2 x}+c_{2}\left(-2 x e^{-2 x}+e^{-2 x}\right)\\[1 em] ~~~~~~~~+\frac{1}{2} \cos 2 x-e^{-x} \\[1 em] y(0)=-1 \\[1 em] \therefore-1=c_{1}+1 \rightarrow c_{1}=-2 \\[1 em] y^{\prime}(0)=2 \\[1 em] \therefore \quad 2=-2 c_{1}+c_{2}+\frac{1}{2}-1 \\[1 em] \therefore \quad 2=4+c_{2}-\frac{1}{2} \Rightarrow c_{2}=-\frac{3}{2} \\[1 em] \therefore y=-2 e^{-2 x}-\frac{3}{2} x e^{-2 x}+\frac{1}{4} \sin 2 x+e^{-x} \end{array}$