Question #222288

Solve the equation dy/dx= -(x2+y2)/xy


1
Expert's answer
2021-08-05T16:28:17-0400

Let us solve the equation dydx=x2+y2xy\frac{dy}{dx}= -\frac{x^2+y^2}{xy} using the transformation y=ux.y=ux. Then dydx=dudxx+u,\frac{dy}{dx}=\frac{du}{dx}x+u, and hence we get the equation dudxx+u=1+u2u,\frac{du}{dx}x+u= -\frac{1+u^2}{u}, which is equivalent to dudxx=1+u2uu,\frac{du}{dx}x= -\frac{1+u^2}{u}-u, and hence to dudxx=1+2u2u.\frac{du}{dx}x= -\frac{1+2u^2}{u}.


It follows that


udu1+2u2=dxx\frac{udu}{1+2u^2}= -\frac{dx}{x}


udu1+2u2=dxx\int\frac{udu}{1+2u^2}= -\int\frac{dx}{x}


144udu1+2u2=dxx\frac{1}{4}\int\frac{4udu}{1+2u^2}= -\int\frac{dx}{x}


d(1+2u2)1+2u2=4dxx\int\frac{d(1+2u^2)}{1+2u^2}= -4\int\frac{dx}{x}


ln(1+2u2)=4lnx+lnC\ln(1+2u^2)= -4\ln|x|+\ln|C|


ln(1+2u2)+4lnx=lnC\ln(1+2u^2)+4\ln|x|=\ln|C|


ln(x4(1+2u2))=lnC\ln(x^4(1+2u^2))=\ln|C|


x4(1+2u2)=Cx^4(1+2u^2)=C


x4(1+2(yx)2)=Cx^4(1+2(\frac{y}{x})^2)=C


We conclude that the general solution of the equation dydx=x2+y2xy\frac{dy}{dx}= -\frac{x^2+y^2}{xy} is


x4+2x2y2=Cx^4+2x^2y^2=C



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