Let y=y(x) solve (x^2+1)^2dydx+2x(x^2+1)y=1, y(0)=0.

Question

Let $y=y(x)$ solve $(x^2+1)^2\dfrac{dy}{dx}+2x(x^2+1)y=1$, $y(0)=0$. If $\sqrt a\,y(1)=\dfrac{\pi}{32}$, then $a$ is:

Accepted Answer

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

I.F.$=1+x^2$. $y(1+x^2)=\tan^{-1}x+C$; using $y(0)=0$: $C=0$.

At $x=1$: $2y(1)=\dfrac{\pi}{4} \implies y(1)=\dfrac{\pi}{8}$.

$\sqrt a\cdot\dfrac{\pi}{8}=\dfrac{\pi}{32} \implies \sqrt a=\dfrac14 \implies a=\dfrac{1}{16}$.

Solutions · 1