What is the solution for 1/(x^2+y^2)(xy^'-y)=0 ?

The solution for 1/(x^2+y^2)(xy^'-y)=0 is y=c_{1}x

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1/(x^2+y^2)(xy^'-y)=0

\frac{1}{x ^{2} + y ^{2}} (x y^{^{'}} - y) = 0

y = c_{1} x

Solution steps

\frac{1}{x ^{2} + y ^{2}} (x y^{'} (x) - y) = 0

Solve separable ODE:

y = c_{1} x

y = c_{1} x

\frac{d x}{d t} = \frac{sin ( t )}{sin ( x )}

\frac{d y}{d x} = 0.1 y (1 - 0.5 y), y (0) = 5000

\frac{d y}{d t} + 0.4 t y = 4 t

y^{^{'}} = (1 + 4 x) y

y^{^{'}} + 18 x y = 0

What is the solution for 1/(x^2+y^2)(xy^'-y)=0 ?
The solution for 1/(x^2+y^2)(xy^'-y)=0 is y=c_{1}x