Solve the equation:$y'+\frac{2y}{(x+1)^2}=\frac{1}{(x^2+1)^2}$

Question

Solve the equation:$$y'+\frac{2y}{(x+1)^2}=\frac{1}{(x^2+1)^2}$$ I don't know how to solve this equation,help me,thanks.

Answer 1

Let's proceed by finding an integrating factor. Let :

$$μ(x) = e^{\int \frac{2}{(x+1)^2}} = e^{-\frac{2}{x+1}}$$

and multiply both sides by it :

$$e^{-\frac{2}{x+1}}\bigg(y' + \frac{2y}{(x+1)^2}\bigg) = e^{-\frac{2}{x+1}}\frac{1}{(x^2+1)^2}$$

Substitute :

$$\frac{2e^{-\frac{2}{x+1}}}{(x+1)^2}= \frac{d}{dx}\bigg(e^{-\frac{2}{x+1}}\bigg)$$

so the initial differential equation becomes :

$$\Rightarrow e^{-\frac{2}{x+1}}\frac{dy(x)}{dx} + \frac{d}{dx}\bigg(e^{-\frac{2}{x+1}}\bigg)y(x) = \frac{e^{-\frac{2}{x+1}}}{(x^2+1)^2}$$

$$\Rightarrow \frac{d}{dx}\bigg(e^{-\frac{2}{x+1}}y(x)\bigg) = \frac{e^{-\frac{2}{x+1}}}{(x^2+1)^2}$$

$$\Rightarrow \int\frac{d}{dx}\bigg(e^{-\frac{2}{x+1}}y(x)\bigg)dx = \int\frac{e^{-\frac{2}{x+1}}}{(x^2+1)^2}dx$$

$$\Rightarrow e^{-\frac{2}{x+1}} y(x) = \frac{e^{-\frac{2}{x+1}}(x+1)^2}{4(x^2+1)} + c_1 \Leftrightarrow y(x) = \frac{(x+1)^2}{4(x^2+1)} + c_1e^{\frac{2}{x+1}}$$

As for the trajectories of all the solutions, here is a solution trajectory derived by sampling for different initial values of $y(0)$ :

$\qquad$ $\qquad$ $\qquad$ $\qquad$

Answer 2

an integrating factor is given by $$e^{-2/(x+1)}$$ and you can write $$\int \frac{d}{dx}\left(e^{-2/(x+1)}y(x)\right)dx=\int\frac{e^{-2/(x+1)}}{(x+1)^2}dx$$ the result is given by $$y \left( x \right) =1/4\,{\frac { \left( x+1 \right) ^{2}}{{x}^{2}+1}} +{{\rm e}^{2\, \left( x+1 \right) ^{-1}}}{\it \_C1} $$