Answer to Question #163913 in Differential Equations for Jay

Question

(x^2+y^2)y'=y^2

Answer 1

Solution.

$(x^2+y^2)y'=y^2.$

Make a replacement

$u(x)=\frac{y(x)}{x},$ from here $y(x)=xu(x),$ and $y'=u(x)+xu'(x).$

Substitute the obtained expressions into the equation:

$(x^2+x^2u^2)(u+xu')=x^2u^2,$

$x^2u+x^2u^3+x^3u'+x^3u^2u'-x^2u^2=0,$

$x^3u'(1+u^2)+x^2(u^3-u^2+u)=0.$

Divide both parts of the equation by $u^3-u^2+u$ and write the equation in the form:

$\frac{(u^2+1)u'}{u^3-u^2+u}=-\frac{1}{x}.$

Thus we obtain a separable differential equation. Let's solve it:

$\frac{dx(u^2+1)\frac{du}{dx}}{u^3-u^2+u}=-\frac{dx}{x},$ $\frac{du(u^2+1)}{u^3-u^2+u}=-\frac{dx}{x},$ $\int\frac{u^2+1}{u^3-u^2+u}du=\int-\frac{1}{x}dx,$ $\int\frac{1}{u^2-u+1}du+\int\frac{1}{u}du=\int-\frac{1}{x}dx,$ $\frac{2\arctan{\frac{2u-1}{\sqrt{3}}}}{\sqrt{3}}+\ln{|u|}=-\ln{|x|}+C.$

where C is a constant.

Returning to the replacement we obtain the solution of the given equation:

$\frac{2\arctan{\frac{2\frac{y(x)}{x}-1}{\sqrt{3}}}}{\sqrt{3}}+\ln(|\frac{y(x)}{x}|)=-\ln{|x|}+C,$

or

$\frac{2\sqrt{3}\arctan{\frac{\sqrt{3}(2\frac{y(x)}{x}-1)}{3}}}{3}+\ln(|y(x)|)=C.$

Answer.

$\frac{2\sqrt{3}\arctan{\frac{\sqrt{3}(2\frac{y(x)}{x}-1)}{3}}}{3}+\ln(|y(x)|)=C.$