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# Answer to Question #224871 in Chemical Engineering for Lokika

Question #224871

Solve (2y³xe^y+ y² + y)dx + (y³x²e^y-xy- 2x)dy = 0.

1
2021-08-18T01:53:02-0400

"\\mathrm{An\\:ODE\\:}M\\left(x,\\:y\\right)+N\\left(x,\\:y\\right)y'=0\\mathrm{\\:is\\:in\\:exact\\:form\\:if\\:the\\:following\\:holds:}\\\\\n1.\\:\\:\\mathrm{There\\:exists\\:a\\:function\\:}\\Psi \\left(x,\\:y\\right)\\mathrm{\\:such\\:that\\:}\\Psi _x\\left(x,\\:y\\right)=M\\left(x,\\:y\\right),\\:\\quad \\Psi _y\\left(x,\\:y\\right)=N\\left(x,\\:y\\right)\\\\\n2.\\:\\:\\Psi \\left(x,\\:y\\right)\\mathrm{\\:has\\:continuous\\:partial\\:derivatives:\\quad }\\frac{\\partial M\\left(x,\\:y\\right)}{\\partial y}=\\frac{\\partial ^2\\Psi \\left(x,\\:y\\right)}{\\partial y\\partial x}=\\frac{\\partial ^2\\Psi \\left(x,\\:y\\right)}{\\partial x\\partial y}=\\frac{\\partial N\\left(x,\\:y\\right)}{\\partial x}\\\\\n\\mathrm{Let\\:}y\\mathrm{\\:be\\:the\\:dependent\\:variable.\\:Divide\\:by\\:}dx\\mathrm{:}\\\\\n2y^3xe^y+y^2+y+\\left(y^3x^2e^y-xy-2x\\right)\\frac{dy}{dx}=0\\\\\n\\mathrm{Substitute\\quad }\\frac{dy}{dx}\\mathrm{\\:with\\:}y'\\:\\\\\n2y^3xe^y+y^2+y+\\left(y^3x^2e^y-xy-2x\\right)y'\\:=0\\\\\n\\frac{2e^yxy^2+y+1}{y^2}+\\frac{e^yx^2y^3-xy-2x}{y^3}y'\\:=0\\\\\n\\mathrm{If\\:the\\:conditions\\:are\\:met,\\:then\\:}\\Psi _x+\\Psi _y\\cdot \\:y'=\\frac{d\\Psi \\left(x,\\:y\\right)}{dx}=0\\\\\n\\mathrm{The\\:general\\:solution\\:is\\:}\\Psi \\left(x,\\:y\\right)=C\\\\\n\u03a8\\left(x,\\:y\\right)=c_2\\\\\n\\mathrm{Roots\\:of}\\:\\frac{x}{y^2}+\\frac{x}{y}+x^2e^y=c_1"

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