107 775
Assignments Done
100%
Successfully Done
In July 2024

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
2021-08-26T01:05:35-0400

Given IVP:

Let us denote

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

We get new IVP

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

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)}"

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!