Answer to Question #163914 in Differential Equations for Jay

Question

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

Answer 1

Given: $(x^2+y^2)y'=y^2$

$\Rightarrow (x^2+y^2)\frac{dy}{dx}=y^2$

$\Rightarrow \frac{dx}{dy}=\frac{x^2+y^2}{y^2}=\frac{x^2}{y^2}+1$

Take the substitution $x=vy\Rightarrow \frac{dx}{dy}=v+y\frac{dv}{dy}$

Then the given differential equation becomes

$v+y\frac{dv}{dy}=\frac{v^2y^2}{y^2}+1=v^2+1$

$\Rightarrow y\frac{dv}{dy}=\frac{v^2y^2}{y^2}+1=v^2-v+1$

Using separation of variables, we get

$\frac{dv}{v^2-v+1}=\frac{dy}{y}$

Integrating on both sides, we get

$\int \frac{dv}{v^2-v+1}=\int \frac{dy}{y}$

$\int \frac{dv}{v^2-v+\frac{1}{4}-\frac{1}{4}+1}=\int \frac{dy}{y}$

$\int \frac{dv}{(v-\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}=\int \frac{dy}{y}$

$\frac{1}{\sqrt{3}/2}arctan(\frac{v-\frac{1}{2}}{\sqrt{3}/2})=ln(y)+c$

$\frac{2}{\sqrt{3}}arctan(\frac{2v-1}{\sqrt{3}})=ln(y)+c$

Substituting $v=\frac{x}{y}$ , we get

$\frac{2}{\sqrt{3}}arctan(\frac{2(x/y)-1}{\sqrt{3}})=ln(y)+c$