Question

solve the following initial value problem y" + y' -2y = -6sin2x-18cos2x y(0)=2,y'(0)=2

Answer 1

$\begin{array}{l} y^{\prime \prime}+y^{\prime}-2 y=-6 \sin 2 x-18 \cos 2 x \\[1 em] \begin{array}{c} \text { The general solution is } \\[1em] y(x)=y_{c}+y_{p}\\[1em] \end{array} \\[1em] \begin{array}{c} y_{c}=m^{2}+m-2=0\\[1em] (m+2)(m-1)=0 \\[1em] m_{1}=-2, m_{2}=1 \\[1em] y_{c}=c_{1} e^{m_{1} x}+c_{2} e^{m_{2} x}=c_{1} e^{-2 x}+c_{2} e^{x} \\[1em] y_{p}=A \cos 2 x+B \sin 2 x\\[1em] y_{1}^{\prime } p=-2 A \sin 2 x+2 B \cos 2 x \\[1em] y_{1}^{\prime\prime } p=-4 A \cos 2 x-4B \sin 2 x \\[1em] \text { substiute in the original equation } \end{array} \\[1em] \end{array} \\[1em] -4 A \cos 2 x-4 B \sin 2 x-2 A \sin 2 x+2B \cos 2 x-2 A \cos 2 x-2 B \sin 2x \\[1em] =-6 \sin 2 x-18 \cos 2 x \\[1em] -4 A+2 B-2 A=-18 \rightarrow 3 A-B=9 \\[1em] -4 B-2 A-2 B=-6 \quad \rightarrow A+3B=3 \\[1em] \text{Solve the two equations}\\[1em] \therefore A=3, B=0\\[1em] \therefore \quad y_{p}=3 \cos 2 x\\[1em] y(x)=c_{1} e^{-2x}+c_{2} e^{x}+3 \cos 2 x\\[1em] \because y(0)=2 \quad \rightarrow 2=c_{1}+c_{2}+3 \rightarrow c_{1}+c_{2}=-1\\[1em] y^{\prime}(x)=-2c_{1} e^{-2x}+c_{2} e^{x}-6 \sin 2 x\\[1em] \because y^{\prime}(0)=2 \rightarrow 2= -2 c_{1}+c_{2} \rightarrow -2 c_{1}+c_{2}=2 \\[1em] \text{Solve the two equations}\\[1em] \therefore c_{1}=-1 , \quad c_{2}=0\\[1em] y(x)=- e^{-2 x}+6 e^{x}+3 \cos 2 x= 3 \cos 2 x - e^{-2 x}$