Answer to Question #229216 in Chemical Engineering for Lokika

Question

Solve the initial value problem y^{2}y -y^{3}tanx=sinxcos^{2}x,y(0)=1

Answer 1

Given IVP:

$y^2y'-y^3\tan x=\sin x\cos^2x,\quad y(0)=1$

Let us denote

$z(x)=y^3(x)$

We get new IVP

$\frac{1}{3}z'-z\tan x=\sin x\cos^2x,\quad z(0)=1$

Let

$z=u\cdot v$

Then

$\frac{1}{3}(u'v+uv')-uv\tan x=\sin x\cos^2x$

Let

$\frac{1}{3}v'-v\tan x=0$

We obtain

$\frac{dv}{v}=3\tan x dx$

$\ln v=-3\ln \cos x=\ln \cos^{-3}x$

$v=\cos^{-3}x$ $\frac{1}{3}u'v=\sin x\cos^2x$

$\frac{1}{3}u'\cos^{-3}x=\sin x\cos^2x$

$u'=3\sin x\cos^5x$

$u=-\frac{\cos^6x}{2}+C$

$z(x)=u\cdot v=(-\frac{\cos^6x}{2}+C)\cos^{-3}x$

The initial condition gives

$z(0)=(-1/2+C)=1,\quad C=3/2$

So

$z(x)=\left(-\frac{\cos^6x}{2}+\frac{3}{2}\right)\cos^{-3}x$

Finally

$y(x)=\sqrt[3]{\left(-\frac{\cos^6x}{2}+\frac{3}{2}\right)\cos^{-3}x}$

$y(x)=\sqrt[3]{\left(-\frac{\cos^3x}{2}+\frac{3}{2\cos^3x}\right)}$