What is the solution for 2xdy-2ydx=sqrt(x^2+4y^2)dx ?

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

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2xdy-2ydx=sqrt(x^2+4y^2)dx

2 x d y - 2 y d x = x^{2} + 4 y^{2} d x

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

Solution steps

2 x d y - 2 y d x = x^{2} + 4 y^{2} d x

Solve using substitution:

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

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

\int \frac{x ^{2}}{25 + x ^{6}} d x

\frac{\partial}{\partial x} (e^{- 4 t} cos (π x))

y = \frac{x}{x + 4}, (1, 0.2)

\frac{d}{d y} (- 2 y)

(1 - 3 sin (4 t))^{2}

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