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# Answer to Question #225919 in Chemical Engineering for Lokika

Question #225919

The solution of ye^{x}dx + (2y + e^{x})dy=0, y(0) = -1 is A) y = e^{x}. B) y=e^{-x} C) y = 1 D) y = -1

1
2021-08-20T01:37:06-0400

"ye^xdx+\\left(2y+e^x\\right)dy=0\\\\\n\\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{:}\\\\\nye^x+\\left(2y+e^x\\right)\\frac{dy}{dx}=0\\\\\n\\mathrm{Substitute\\quad }\\frac{dy}{dx}\\mathrm{\\:with\\:}y'\\:\\\\\nye^x+\\left(2y+e^x\\right)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\\mathrm{True}\\\\\n\u03a8\\left(x,\\:y\\right)=c_2\\\\\ne^xy+y^2+c_1=c_2\\\\\n\\mathrm{Combine\\:the\\:constants}\\\\\ne^xy+y^2=c_1\\\\\ny=\\frac{-e^x+\\sqrt{e^{2x}+4c_1}}{2},\\:y=\\frac{-e^x-\\sqrt{e^{2x}+4c_1}}{2}\\\\\ny=\\frac{-e^x+\\sqrt{e^{2x}+c_1}}{2},\\:y=\\frac{-e^x-\\sqrt{e^{2x}+c_1}}{2}"

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