obtaining easy differential equation solution

Question

I passed by a question of a differential equation and i need help solving it thought its easy but am new with differential equations.

Let S be the solution of the differential equation :

$xy' -y= \frac{-x^2}{x^2+1}$. Deduce S.

i guess the first thing to do is to work out on $xy' -y= 0$, is that true?

Answer 1

the general form of first order linear differential equation is $$y'+P(x)y=Q(x)$$ so, your DE becomes $$y'-\frac{1}{x}y=-\frac{x}{x^2+1}$$ here $$P(x)=-\frac{1}{x}$$ and $$Q(x)=-\frac{x}{x^2+1}$$ $$\rho=e^{\int P(x)dx}=e^{-\frac{1}{x}dx}=\frac{1}{x}$$ and $$y.\rho=\int \rho.Q(x)dx$$ $$y.\frac{1}{x}=\int \frac{1}{x}(-\frac{x}{x^2+1})dx$$ $$y.\frac{1}{x}=-\tan^{-1}x+C_1$$ and then $$y=-x\tan^{-1}x+C_1x$$

Answer 2

Beside the method of integrating factors, you also have the method of variation of parameters.

As you say, the first thing is to solve the homegeneous equation. In you case, you then start with $x y'-y=0$ which leads to $y=C x$. But, for the complete equation, you consider that $C$ is a function of $x$.

Then $$x(C + C'x)-Cx=C'x^2=-\frac{x^2}{x^2+1}$$Which makes $$C'=-\frac{1}{x^2+1}$$ Integrating gives $$C=-\tan^{-1}(x)+K$$ and finally $$y=Cx=-x \tan^{-1}(x)+K x$$