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Answer to Question #224869 in Chemical Engineering for Lokika

Question #224869

Solve the initial value problem y²y'-y^3 tanx = sin x cos² x, y(0)=1

1
2021-08-17T12:14:01-0400

"y^2y'\\:-y^3\\tan \\left(x\\right)=\\sin \\left(x\\right)\\cos ^2\\left(x\\right)\\\\\n\\mathrm{A\\:first\\:order\\:Bernoulli\\:ODE\\:has\\:the\\:form\\:of\\:}y'+p\\left(x\\right)y=q\\left(x\\right)y^n\\\\\ny'\\:-\\tan \\left(x\\right)y=\\sin \\left(x\\right)\\cos ^2\\left(x\\right)y^{-2}\\\\\n\\mathrm{The\\:general\\:solution\\:is\\:obtained\\:by\\:substituting\\:}v=y^{1-n}\\mathrm{\\:and\\:solving\\:}\\frac{1}{1-n}v'+p\\left(x\\right)v=q\\left(x\\right)\\\\\n\u03bd=-\\frac{\\cos ^3\\left(x\\right)}{2}+\\frac{c_1}{\\cos ^3\\left(x\\right)}\\\\\ny^3=-\\frac{\\cos ^3\\left(x\\right)}{2}+\\frac{c_1}{\\cos ^3\\left(x\\right)}\\\\\ny=\\frac{2^{\\frac{2}{3}}\\sqrt[3]{-\\cos ^6\\left(x\\right)+3}}{2\\cos \\left(x\\right)}\\\\\ny=\\frac{2^{\\frac{2}{3}}\\sqrt[3]{-\\cos ^6\\left(x\\right)+3}}{2\\cos \\left(x\\right)}"

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