Let y=y(x) be the solution of the differential equation [(1+x^2),dy + (y-^-1x),dx=0, y(0)=…

Question

Let $y=y(x)$ be the solution of the differential equation
$$(1+x^2)\,dy + (y-\tan^{-1}x)\,dx=0,\qquad y(0)=1.$$
Then $y(1)$ is equal to:

Accepted Answer

$$\frac{dy}{dx} + \frac{1}{1+x^2}y = \frac{\tan^{-1}x}{1+x^2}.$$

Integrating factor $\mu(x) = e^{\tan^{-1}x}$.
$$\frac{d}{dx}\big(ye^{\tan^{-1}x}\big) = e^{\tan^{-1}x}\cdot\frac{\tan^{-1}x}{1+x^2}.$$

Let $t=\tan^{-1}x$, $dt=\dfrac{dx}{1+x^2}$:
$$ye^t = \int te^t\,dt+C = e^t(t-1)+C \implies y = t-1+Ce^{-t} = \tan^{-1}x -1+Ce^{-\tan^{-1}x}.$$

$y(0)=1$: $1=0-1+C \implies C=2$.

$$y(1) = \frac{\pi}{4}-1+2e^{-\pi/4} = 2e^{\pi/4}+\frac{\pi}{4}-1.$$

Solutions · 1