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

Question

Let $y=y(x)$ be the solution curve of the differential equation $(1+{x}^{2})dy+(y-{	an }^{-1}x)dx=0,y(0)=1$. Then the value of $y(1)$ is :

Accepted Answer

Given

$(1+{x}^{2})\, dy+(y-{	an }^{-1}x)\, dx=0,\, y(0)=1,$

divide by $dx$:

$(1+{x}^{2})\frac{dy}{dx}+y-{	an }^{-1}x=0$

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

This is linear. Integrating factor:

$\mu (x)=\exp \, (\int \frac{1}{1+{x}^{2}}\, dx)={e}^{{	an }^{-1}x}.$

So

$\frac{d}{dx}(y{e}^{{	an }^{-1}x})={e}^{{	an }^{-1}x}\cdot \frac{{	an }^{-1}x}{1+{x}^{2}}.$

Let $t={	an }^{-1}x$, then $dt=\frac{dx}{1+{x}^{2}}$. Hence

$y{e}^{t}=\int t{e}^{t}\, dt+C={e}^{t}(t-1)+C.$

Therefore

$y=t-1+C{e}^{-t}={	an }^{-1}x-1+C{e}^{-{	an }^{-1}x}.$

Use $y(0)=1$ and ${	an }^{-1}0=0$:

$1=0-1+C\, \implies \, C=2.$

So

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

Correct option: B.

Solutions · 1