Answer to Question #223867 in Chemical Engineering for Lokika

Question

Solve d^{2}y/dx^{2}+2dy/dx+y=xsinx

Answer 1

$$y''(x)+2y'(x)+y(x)=x\sin x$$

The characteristic equation of the homogeneous DE

$$k^2+2k+1=0$$

Roots

$$k_1=k_2=-1$$

Hence, the general solution of homogeneous DE

$$y_h(x)=C_1e^{-x}+C_2xe^{-x}$$

The partial solution of inhomogeneous DE

$$y_i(x)=(A+Bx)\sin x+(C+Dx)\cos x$$

We get

$$y'_i(x)=(A+Bx)\cos x-(C+Dx)\sin x\\+B\sin x+D\cos x$$

$$y''_i(x)=-(A+Bx)\sin x-(C+Dx)\cos x\\+2B\cos x-2D\sin x$$

$$-(A+Bx)\sin x-(C+Dx)\cos x\\+2B\cos x-2D\sin x+2((A+Bx)\cos x\\-(C+Dx)\sin x+B\sin x+D\cos x)\\+(A+Bx)\sin x+(C+Dx)\cos x\\=x \sin x$$

So, the partial solution of inhomogeneous DE

$$y_i(x)=\frac{1}{2}(\cos x-x\cos x+\sin x)$$

Finally,

$$y(x)=C_1e^{-x}+C_2xe^{-x}\\+\frac{1}{2}(\cos x-x\cos x+\sin x)$$