Answer to Question #229216 in Chemical Engineering for Lokika

Question #229216

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


1
Expert's answer
2021-08-26T01:05:35-0400

Given IVP:

y2yy3tanx=sinxcos2x,y(0)=1y^2y'-y^3\tan x=\sin x\cos^2x,\quad y(0)=1

Let us denote

z(x)=y3(x)z(x)=y^3(x)

We get new IVP

13zztanx=sinxcos2x,z(0)=1\frac{1}{3}z'-z\tan x=\sin x\cos^2x,\quad z(0)=1

Let

z=uvz=u\cdot v

Then

13(uv+uv)uvtanx=sinxcos2x\frac{1}{3}(u'v+uv')-uv\tan x=\sin x\cos^2x

Let

13vvtanx=0\frac{1}{3}v'-v\tan x=0

We obtain

dvv=3tanxdx\frac{dv}{v}=3\tan x dx

lnv=3lncosx=lncos3x\ln v=-3\ln \cos x=\ln \cos^{-3}x

v=cos3xv=\cos^{-3}x13uv=sinxcos2x\frac{1}{3}u'v=\sin x\cos^2x

13ucos3x=sinxcos2x\frac{1}{3}u'\cos^{-3}x=\sin x\cos^2x

u=3sinxcos5xu'=3\sin x\cos^5x

u=cos6x2+Cu=-\frac{\cos^6x}{2}+C

z(x)=uv=(cos6x2+C)cos3xz(x)=u\cdot v=(-\frac{\cos^6x}{2}+C)\cos^{-3}x

The initial condition gives

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

So

z(x)=(cos6x2+32)cos3xz(x)=\left(-\frac{\cos^6x}{2}+\frac{3}{2}\right)\cos^{-3}x

Finally

y(x)=(cos6x2+32)cos3x3y(x)=\sqrt[3]{\left(-\frac{\cos^6x}{2}+\frac{3}{2}\right)\cos^{-3}x}

y(x)=(cos3x2+32cos3x)3y(x)=\sqrt[3]{\left(-\frac{\cos^3x}{2}+\frac{3}{2\cos^3x}\right)}


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