What is the solution for (x+2y)y^'=5x-y ?

The solution for (x+2y)y^'=5x-y is y=(-x+sqrt(11x^2+c_{1)})/2 ,y=-(x+sqrt(11x^2+c_{1)})/2

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(x+2y)y^'=5x-y

(x + 2 y) y^{^{'}} = 5 x - y

y = \frac{- x + 11 x ^{2} + c _{1}}{2}, y = - \frac{x + 11 x ^{2} + c _{1}}{2}

Solution steps

(x + 2 y) y^{'} (x) = 5 x - y

Solve exact ODE:

y = \frac{- x + 11 x ^{2} + c _{1}}{2}, y = - \frac{x + 11 x ^{2} + c _{1}}{2}

y = \frac{- x + 11 x ^{2} + c _{1}}{2}, y = - \frac{x + 11 x ^{2} + c _{1}}{2}

\int \frac{sin ( t )}{t} d t

\frac{d}{d x} (\frac{4}{1 + 4 x})

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\frac{\partial}{\partial x} (\frac{x + 4}{y + 1})

\frac{d}{d x} (\frac{2 x}{3 y ^{2}})

What is the solution for (x+2y)y^'=5x-y ?
The solution for (x+2y)y^'=5x-y is y=(-x+sqrt(11x^2+c_{1)})/2 ,y=-(x+sqrt(11x^2+c_{1)})/2