How to solve ${(1 + x)^2}y'' + (1 + x)y' + y = 4\sin \ln (1{\text{ + }}x)$? I don't know how to solve this equation:
${(1 + x)^2}y'' + (1 + x)y' + y = 4\sin \ln (1{\text{ + }}x)$
When it's homogeneous, I try to solve by power series tediously.
Is there any good way to consider this? Thanks.
 A: Let $u=\ln(1+x)$. Then 
$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}=\frac{1}{1+x}\frac{dy}{du}$$
$$\frac{d^2y}{dx^2}=-\frac{1}{(1+x)^2}\frac{dy}{du}+\frac{1}{(1+x)^2}\frac{d^2y}{du^2}$$
So
$$(1+x)^2y''+(1+x)y'+y=-\frac{dy}{du}+\frac{d^2y}{du^2}+\frac{dy}{du}+y=4\sin u$$
$$\frac{d^2y}{du^2}+y=4\sin u$$
A: HINT:
$$y''(x)(1+x)^2+y'(x)(1+x)+y(x)=4\sin(\ln(1+x))\Longleftrightarrow$$

Let $t=\ln(1+x)$, this gives $x=e^{t}-1$:

$$y''(x)e^{2t}+y'(x)e^t+y(x)=4\sin(\ln(e^t))\Longleftrightarrow$$

Using the chain rule, we can find:

$$y'(t)+e^{2t}\left[e^{-2t}y''(t)-e^{-2t}y'(t)\right]+y(t)=4\sin(\ln(e^t))\Longleftrightarrow$$
$$y''(t)+y(t)=4\sin(t)\Longleftrightarrow$$

The general solution will be the sum of the complementary solution and particular solution:

$$y''(t)+y(t)=0\Longleftrightarrow$$

Assume a solution will be proportional to $e^{\lambda t}$ for some constant $\lambda$.
Substitute $y(t)=e^{\lambda t}$:

$$\frac{\text{d}^2}{\text{d}t^2}\left(e^{\lambda t}\right)+e^{\lambda t}=0\Longleftrightarrow$$

Substitute $\frac{\text{d}^2}{\text{d}t^2}\left(e^{\lambda t}\right)=\lambda^2e^{\lambda t}$:

$$\lambda^2e^{\lambda t}+e^{\lambda t}=0\Longleftrightarrow$$
$$e^{\lambda t}\left[\lambda^2+1\right]=0\Longleftrightarrow$$

Since $e^{\lambda t}\ne0$ for any finite $\lambda$, the zeros must come from the polynomial:

$$\lambda^2+1=0\Longleftrightarrow$$
$$\lambda=\pm i$$
