Answer to Question #237505 in Differential Equations for James

Question

Solve the differential equation of the following homogeneous equation.

xydx + (x^2+ y^2)dy = 0

Answer 1

Let us solve the differential equation $xydx + (x^2+ y^2)dy = 0.$ For this let us use the transformation $y=ux.$ Then $dy=udx+xdu,$ and hence we get the differential equation $ux^2dx + (x^2+ u^2x^2)(udx+xdu) = 0.$ It follows that $ux^2dx + x^2(1+u^2)udx+ (1+ u^2)x^3du= 0,$ and hence $x^2(2u+u^3)dx+ (1+ u^2)x^3du= 0.$

Consequently, we have the equation $\frac{dx}{x}+ \frac{(1+ u^2)du}{u(2+u^2)}= 0.$ It follows that $\int\frac{dx}{x}+ \int\frac{(1+ u^2)du}{u(2+u^2)}= \ln|C|.$

We conclude that $\ln|x|+ \frac{1}{2}\int(\frac{1}{u}+\frac{u}{2+u^2})du= \ln|C|,$ and hence $\ln|x|+ \frac{1}{2}\ln|u|+\frac{1}{4}\ln(2+u^2)= \ln|C|.$ Therefore, $4\ln|x|+ 2\ln|u|+\ln(2+u^2)= \ln|C_1|.$

We conclude $\ln(x^4u^2(2+u^2))= \ln|C_1|,$ and thus $x^4u^2(2+u^2)=C_1.$ Then $x^4(\frac{y}x)^2(2+(\frac{y}x)^2)=C_1.$ We conclude that the general solution of the equation is $y^2(2x^2+y^2)=C.$