Answer to Question #225919 in Chemical Engineering for Lokika

Question

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

Answer 1

$ye^xdx+\left(2y+e^x\right)dy=0\\ \mathrm{An\:ODE\:}M\left(x,\:y\right)+N\left(x,\:y\right)y'=0\mathrm{\:is\:in\:exact\:form\:if\:the\:following\:holds:}\\ 1.\:\:\mathrm{There\:exists\:a\:function\:}\Psi \left(x,\:y\right)\mathrm{\:such\:that\:}\Psi _x\left(x,\:y\right)=M\left(x,\:y\right),\:\quad \Psi _y\left(x,\:y\right)=N\left(x,\:y\right)\\ 2.\:\:\Psi \left(x,\:y\right)\mathrm{\:has\:continuous\:partial\:derivatives:\quad }\frac{\partial M\left(x,\:y\right)}{\partial y}=\frac{\partial ^2\Psi \left(x,\:y\right)}{\partial y\partial x}=\frac{\partial ^2\Psi \left(x,\:y\right)}{\partial x\partial y}=\frac{\partial N\left(x,\:y\right)}{\partial x}\\ \mathrm{Let\:}y\mathrm{\:be\:the\:dependent\:variable.\:Divide\:by\:}dx\mathrm{:}\\ ye^x+\left(2y+e^x\right)\frac{dy}{dx}=0\\ \mathrm{Substitute\quad }\frac{dy}{dx}\mathrm{\:with\:}y'\:\\ ye^x+\left(2y+e^x\right)y'\:=0\\ \mathrm{If\:the\:conditions\:are\:met,\:then\:}\Psi _x+\Psi _y\cdot \:y'=\frac{d\Psi \left(x,\:y\right)}{dx}=0\\ \mathrm{The\:general\:solution\:is\:}\Psi \left(x,\:y\right)=C\\ \mathrm{True}\\ Ψ\left(x,\:y\right)=c_2\\ e^xy+y^2+c_1=c_2\\ \mathrm{Combine\:the\:constants}\\ e^xy+y^2=c_1\\ y=\frac{-e^x+\sqrt{e^{2x}+4c_1}}{2},\:y=\frac{-e^x-\sqrt{e^{2x}+4c_1}}{2}\\ y=\frac{-e^x+\sqrt{e^{2x}+c_1}}{2},\:y=\frac{-e^x-\sqrt{e^{2x}+c_1}}{2}$