Question #229216

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

Expert's answer

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)}"

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