Q:

Find a particular solution to y" - y + y = 2 sin(3x)

Accepted Solution

A:
Answer with explanation:The given differential equation isy" -y'+y=2 sin 3x------(1)Let, y'=zy"=z'[tex]\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x[/tex]Substituting the value of , y, y' and y" in equation (1)z'-z+zx=2 sin 3 xz'+z(x-1)=2 sin 3 x-----------(1)This is a type of linear differential equation.Integrating factor      [tex]=e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}[/tex]Multiplying both sides of equation (1) by integrating factor and integrating we get[tex]\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I[/tex][tex]I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx[/tex]