What is the solution for y^'= y/(y-x) ?

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

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y^'= y/(y-x)

y^{^{'}} = \frac{y}{y - x}

y = x + x^{2} + c_{1}, y = x - x^{2} + c_{1}

Solution steps

y^{'} (x) = \frac{y}{y - x}

Solve exact ODE:

y = x + x^{2} + c_{1}, y = x - x^{2} + c_{1}

y = x + x^{2} + c_{1}, y = x - x^{2} + c_{1}

n = 1 \sum \infty n^{- 2.4}

\int \frac{1}{1 + y ^{2}} d y

\int_{1}^{4} 8 x - 5 d x

\frac{\partial}{\partial x} (e^{x})

X^{2} + X - 12, X = 0

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