Question

$\frac{dy}{dx}+y\cot x=y^2{\sin }^{2}x{\cos }^{2}x.$
Its solution is $\frac{1}{y}co\sec x=\frac{1}{3}{\cos }^{3}x+c$
If true enter 1 else enter 0.

Answer 1

$\frac{dy}{dx}+ycotx=y^{2}si{n}^{2}xco{s}^{2}x$
$y^{-2}\frac{dy}{dx}+\frac{cotx}{y}=si{n}^{2}xco{s}^{2}x$
Let
$y^{-1}=t$
Then
$-y^{-2}dy=dt$
Or
$-y^{-2}\frac{dy}{dx}=\frac{dt}{dx}$
Hence the differential equation transforms to
$\frac{dt}{dx}+tcotx=si{n}^{2}co{s}^{2}x$
$IF=e^{\int cotx.dx}$
$=e^{lnsinx}$
$=sinx$
Hence
$sinx\frac{dt}{dx}+tcosx=si{n}^{3}xco{s}^{2}x$
$sinxdt+tcosx.dx=si{n}^{3}xco{s}^{2}dx$
$d\left(\right(sinx\left)t\right)=si{n}^{3}xco{s}^{2}dx$
Let
$I=\int si{n}^{3}xco{s}^{2}dx$
Let
$co{s}^{2}x=t$
$-2cosx.sinxdx=dt$
$I=\int si{n}^{2}x.cosx(cosx.sinx)dx$
$=\int si{n}^{2}xcosx\left(\frac{-dt}{2}\right)$
$=\frac{-1}{2}[\int (1-t\left)\right(\sqrt{t}\left)dt\right]$
$=\frac{-1}{2}[\int \sqrt{t}-t^{3/2}dt]$
$=-\frac{1}{2}[\frac{2t\sqrt{t}}{3}-\frac{2t^2\sqrt{t}}{5}]$
$=\frac{t^2\sqrt{t}}{5}-\frac{t\sqrt{t}}{3}$
$=\frac{co{s}^{5}x}{5}-\frac{co{s}^{3}x}{3}$
Hence
$d\left(\right(sinx\left)t\right)=si{n}^{3}xco{s}^{2}dx$
$sinx.t=\frac{co{s}^{5}x}{5}-\frac{co{s}^{3}x}{3}+C$
$\frac{sinx}{y}=\frac{co{s}^{5}x}{5}-\frac{co{s}^{3}x}{3}+C$