Answer to Question #238615 in Differential Equations for James

Question

Solve the differential equation of the following homogeneous equation.

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

Answer 1

$$y'=-\dfrac{xy}{x^2+y^2}$$

Substitute $y=ux$

$y'=u+xu'$

$$u+xu'=-\dfrac{x^2u}{x^2+x^2u^2}$$

$$xu'=-\dfrac{2u+u^3}{1+u^2}$$

$$\dfrac{1+u^2}{2u+u^3}du=-\dfrac{dx}{x}$$

Integrate

$$\int \dfrac{1+u^2}{2u+u^3}du=-\int \dfrac{dx}{x}$$

$$\dfrac{1+u^2}{2u+u^3}=\dfrac{A}{u}+\dfrac{Bu+C}{2+u^2}$$

$$=\dfrac{2A+Au^2+Bu^2+Cu}{u(2+u^2)}$$

$u^2: A+B=1$

$u^1:C=0$

$u^0:2A=1$

$$A=\dfrac{1}{2}, B=-\dfrac{1}{2}$$

$$\int \dfrac{1+u^2}{2u+u^3}du=\dfrac{1}{2}\int \dfrac{du}{u}-\dfrac{1}{2}\int\dfrac{u}{2+u^2}du$$

$$=\dfrac{1}{2}\ln(|u|)-\dfrac{1}{4}\ln(2+u^2)+C_1$$

$$\dfrac{1}{2}\ln(|u|)-\dfrac{1}{4}\ln(2+u^2)=-\ln|x|+\ln C$$

$$\dfrac{x\sqrt{u}}{\sqrt[4]{2+u^2}}=C$$

$$\dfrac{x\sqrt{y}}{\sqrt[4]{2x^2+y^2}}=C$$