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

The solution for y^'=(y+x)/(y-x) is y=(2x+sqrt(8x^2+c_{1)})/2 ,y=(2x-sqrt(8x^2+c_{1)})/2

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y^'=(y+x)/(y-x)

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

y = \frac{2 x + 8 x ^{2} + c _{1}}{2}, y = \frac{2 x - 8 x ^{2} + c _{1}}{2}

Solution steps

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

Solve exact ODE:

y = \frac{2 x + 8 x ^{2} + c _{1}}{2}, y = \frac{2 x - 8 x ^{2} + c _{1}}{2}

y = \frac{2 x + 8 x ^{2} + c _{1}}{2}, y = \frac{2 x - 8 x ^{2} + c _{1}}{2}

y = \frac{3}{x}, y = \frac{3}{x ^{2}}, x = 7

\int \frac{1}{X} d X

f (x) = x + x + 1

\frac{d}{d x} (- λ x)

x \to 0 lim (\frac{x + sin ( x )}{x})

What is the solution for y^'=(y+x)/(y-x) ?
The solution for y^'=(y+x)/(y-x) is y=(2x+sqrt(8x^2+c_{1)})/2 ,y=(2x-sqrt(8x^2+c_{1)})/2