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

The solution for xy^'=sqrt(x^2-y^2)+y is y=sin(ln(x)+c_{1})x

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xy^'=sqrt(x^2-y^2)+y

x y^{^{'}} = x^{2} - y^{2} + y

y = sin (ln (x) + c_{1}) x

Solution steps

x y^{'} (x) = x^{2} - y^{2} + y

Solve using substitution:

y = sin (ln (x) + c_{1}) x

y = sin (ln (x) + c_{1}) x

\frac{d y}{d x} + y^{2} = y

\int \frac{1}{x ^{2} - 3 x - 10} d x

f (x) = 7

f (x) = - 9 x

\int \frac{ln ^{8} ( x )}{x} d x

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