What is the solution for (dy}{dx}=\frac{sqrt(x^2-y^2)+y)/x ?

The solution for (dy}{dx}=\frac{sqrt(x^2-y^2)+y)/x is y=sin(ln(x)+c_{1})x

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(dy}{dx}=\frac{sqrt(x^2-y^2)+y)/x

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

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

Solution steps

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

Solve using substitution:

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

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

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + 3 y})

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

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

f (x) = 2 x + \frac{1936}{x}

f (x) = 110 x - 16 x^{2}, at x = 4

What is the solution for (dy}{dx}=\frac{sqrt(x^2-y^2)+y)/x ?
The solution for (dy}{dx}=\frac{sqrt(x^2-y^2)+y)/x is y=sin(ln(x)+c_{1})x