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Question Number 67939 by ramirez105 last updated on 02/Sep/19

Commented by mr W last updated on 02/Sep/19

$$sir,yougot{y}^2(y+2x)=C,butthis$$$${doesn}^{\prime }tsatisfytheoriginalequ.$$$$is$$$$\frac{dy}{y}=-\frac{dx}{2x+3y}\Rightarrow \int \frac{dy}{y}=-\int \frac{dx}{2x+3y}+c$$$$allowed?sinceyisafunctionfrom$$$$x,notaconstantw.r.t.x.$$

Commented by mathmax by abdo last updated on 02/Sep/19

$$ydx+(2x+3y)dy=0\Rightarrow (2x+3y)dy=-ydx\Rightarrow$$$$\frac{dy}{y}=-\frac{dx}{2x+3y}\Rightarrow \int \frac{dy}{y}=-\int \frac{dx}{2x+3y}+c\Rightarrow$$$$ln\mid y\mid =-\frac{1}{2}ln\mid 2x+3y\mid +c\Rightarrow \mid y\mid =\frac{K}{\sqrt{\mid 2x+y\mid }}\Rightarrow$$$$y^2=\frac{k^2}{\mid 2x+y\mid }but2x+ymustbe>0\Rightarrow y^2=\frac{k^2}{2x+y}\Rightarrow$$$$2{xy}^2+y^3-k^2=0\Rightarrow y^3+2{xy}^2-k^2=0resttoslveequation$$$$atform{t}^3+{at}^2+k=0....becontinued....$$

Answered by mr W last updated on 02/Sep/19

$$ydx+(2x+3y)dy=0$$$$y+(2x+3y)\frac{dy}{dx}=0$$$$\frac{y}{x}+(2+3\frac{y}{x})\frac{dy}{dx}=0$$$$lety=tx$$$$\frac{dy}{dx}=t+x\frac{dt}{dx}$$$$t+(2+3t)(t+x\frac{dt}{dx})=0$$$$t+2t+3t^2+x(2+3t)\frac{dt}{dx}=0$$$$x(2+3t)\frac{dt}{dx}=-3t(t+1)$$$$\frac{2+3t}{t(t+1)}dt=-\frac{3dx}{x}$$$$\int \frac{2+3t}{t(t+1)}dt=-3\int \frac{dx}{x}$$$$\int [\frac{2}{t}+\frac{1}{t+1}]dt=-3\int \frac{dx}{x}$$$$2\ln t+\ln (t+1)=-3\ln x+c$$$$\ln t^2(t+1)x^3=c$$$$\Rightarrow t^2(t+1)x^3=C$$$$\Rightarrow y^2(y+x)=C$$$$check:$$$$2{yy}^{\prime }(y+x)+y^2(y^{\prime }+1)=0$$$$(3y^2+2xy)y^{\prime }+y^2=0$$$$(3y+2x)y^{\prime }+y=0$$$$\Rightarrow (3y+2x)dy+ydx=0$$

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