Question

Find a particular solution of the differential equation $(1+x^2)dy+2xydx=\cot xdx$, given that $y=0$ if $x=\frac{\pi }{2}$

Answer 1

$\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{cotx}{1+x^2}$
$I.F=e^{\int \frac{2x}{1+x^2}.dx}$
$=e^{ln(1+x^2)}$
$=1+x^2$
Hence, multiplying the entire equation by $1+x^2$, gives us
$(1+x^2)\frac{dy}{dx}+2xy=cotx$
$(1+x^2)dy+2xy.dx=cotxdx$
$d\left(\right(1+x^2\left)y\right)=cotx.dx$
$\int d(y+x^{2}y)=\int cotxdx$
$y(1+x^2)=ln|sinx|+C$
At $x=\frac{\pi }{2}$, $y=0$.
Hence
$ln\left(1\right)+C=0$
Or
$C=0$
Therefore the particular solution of the above differential equation is
$y=\frac{ln|sinx|}{1+x^2}$