How to solve ${(1 + x)^2}y'' + (1 + x)y' + y = 4\sin \ln (1{\text{ + }}x)$?

Question

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.

Answer 1

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

$$(1+x)^2y''+(1+x)y'+y=-\frac{dy}{du}+\frac{d^2y}{du^2}+\frac{dy}{du}+y=4\sin u$$

$$\frac{d^2y}{du^2}+y=4\sin u$$

Answer 2

HINT:

$$y''(x)(1+x)^2+y'(x)(1+x)+y(x)=4\sin(\ln(1+x))\Longleftrightarrow$$

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

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

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Using the chain rule, we can find:

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

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

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

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Assume a solution will be proportional to $e^{\lambda t}$ for some constant $\lambda$.

Substitute $y(t)=e^{\lambda t}$:

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

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

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

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

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