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

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

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

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

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

Solution steps

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

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}

\int \frac{1}{x} + 1 d x

\int 3 r^{2} d r

f (x) = x^{2} + 5, at x = 2

y = x^{3} - 4 x, (2, 0)

y^{^{''}} + 25 y = - 30 sin (5 t)

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