Question

dy/dx + 4xy/x²+1 = 1/(x²+1)³ ,Solve it.

Answer 1

Answer:

Heya mate... May the above calculation help uh'''!!!! ^_^

Answer 2

Answer:

y = $\frac{tan^{-1}x}{(x^{2}+1)^{2}}\ +\ \frac{C}{(x^{2}+1)^{2}}$

Step-by-step explanation:

In this question,

$\frac{dy}{dx}\ +\ \frac{4xy}{x^{2}+1}\ =\ \frac{1}{(x^{2}+1)^{2}}$

It is a linear differential equation

Here P = $\frac{4x}{x^{2}+1}$

        Q = $\frac{1}{(x^{2}+1)^{3}}$

Integrating factor = $e^{\int\ P\ dx}$

Integrating factor = $e^{\int\ \frac{4x}{x^{2}+1}dx}$

On solving Integrating factor = $e^{2(logx^{2}+1)}$

Integrating factor = $e^{(x^{2}+1)^{2}}$

Therefore Integrating factor = $x^{2}+1$

Now,

According to the question

y × Integrating factor = $\int Q\times integrating factor\ dx$

$y(x^{2}+1)^{2}\ =\ \int\frac{(x^{2}+1)^{2}dx}{(x^{2}+1)^{3}}$

$y(x^{2}+1)^{2}\ =\ \int\frac{dx}{(x^{2}+1)}$

$y(x^{2}+1)^{2}\ =\ tan^{-1}x\ +\ C$

Therefore,y = $\frac{tan^{-1}x}{(x^{2}+1)^{2}}\ +\ \frac{C}{(x^{2}+1)^{2}}$

Where C is constant.