Question

The solution of the differential equation
$\frac{dy}{dx}+\frac{2yx}{1+x^2}=\frac{1}{{(1+x^2)}^2}$

• A. $y(1+x^2)=c+{\tan }^{-1}x$
• B. $\frac{y}{1+x^2}=c+{\tan }^{-1}x$
• C. $y\log (1+x^2)=c+{\tan }^{-1}x$
• D. $y(1+x^2)=c+{\sin }^{-1}x$

Answer 1

The correct option is A $y(1+x^2)=c+{\tan }^{-1}x$

Given, $\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{1}{{(1+x^2)}^2}$
On comparing with $\frac{dy}{dx}+Py=Q$, we get
$P=\frac{2x}{1+x^2},Q=\frac{1}{{(1+x^2)}^2}$
Now, $IF=e^{\int Pdx}$
$=e^{\int \frac{2x}{1+x^2}dx}=e^{\log (1+x^2)}$
$=(1+x^2)$
$\therefore$ Solution of the given differential equation is
$y(1+x^2)=\int \frac{1}{{(1+x^2)}^2}\cdot (1+x^2)dx+c$
or $y(1+x^2)=\int \frac{1}{1+x^2}dx+c$
So, $y(1+x^2)={\tan }^{-1}x+c$