Answer to Question #224871 in Chemical Engineering for Lokika

Question

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

Answer 1

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