What is the solution for (x+8y)y^'=2x-y ?

The solution for (x+8y)y^'=2x-y is y=(-x+sqrt(17x^2+c_{1)})/8 ,y=(-x-sqrt(17x^2+c_{1)})/8

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(x+8y)y^'=2x-y

(x + 8 y) y^{^{'}} = 2 x - y

y = \frac{- x + 17 x ^{2} + c _{1}}{8}, y = \frac{- x - 17 x ^{2} + c _{1}}{8}

Solution steps

(x + 8 y) y^{'} (x) = 2 x - y

Solve exact ODE:

y = \frac{- x + 17 x ^{2} + c _{1}}{8}, y = \frac{- x - 17 x ^{2} + c _{1}}{8}

y = \frac{- x + 17 x ^{2} + c _{1}}{8}, y = \frac{- x - 17 x ^{2} + c _{1}}{8}

\frac{d}{d x} (x e^{x y} cos (2 x))

\frac{d}{d x} (1 - x^{2} - y^{2})

\int_{0}^{8} y + 1 d y

x \to 0 + lim (tan (\frac{4}{x}))

f (x) = x^{4} - x^{2}

What is the solution for (x+8y)y^'=2x-y ?
The solution for (x+8y)y^'=2x-y is y=(-x+sqrt(17x^2+c_{1)})/8 ,y=(-x-sqrt(17x^2+c_{1)})/8