What is the solution for (x^2+2xy)dx+xdy=0 ?

The solution for (x^2+2xy)dx+xdy=0 is y=-(2x-1)/4+(c_{1})/(e^{2x)}

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(x^2+2xy)dx+xdy=0

(x^{2} + 2 x y) d x + x d y = 0

y = - \frac{2 x - 1}{4} + \frac{c _{1}}{e ^{2 x}}

Solution steps

(x^{2} + 2 x y) d x + x d y = 0

Solve linear ODE:

y = - \frac{2 x - 1}{4} + \frac{c _{1}}{e ^{2 x}}

y = - \frac{2 x - 1}{4} + \frac{c _{1}}{e ^{2 x}}

\frac{\partial}{\partial x} (x^{4} + 2 x y^{9})

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

\frac{- x sin ( x ) - cos ( x )}{4 x ^{2}}

\int x^{2} - 6 x + 5 d x

\frac{\partial}{\partial x} (2 x^{2} + 5 y^{2})

What is the solution for (x^2+2xy)dx+xdy=0 ?
The solution for (x^2+2xy)dx+xdy=0 is y=-(2x-1)/4+(c_{1})/(e^{2x)}