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

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

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