What is the solution for (xy+4y^2+2x^2)dx-(x^2)dy=0 ?

The solution for (xy+4y^2+2x^2)dx-(x^2)dy=0 is y=(tan(sqrt(2)(2ln(x)+c_{1}))x)/(sqrt(2))

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(xy+4y^2+2x^2)dx-(x^2)dy=0

(x y + 4 y^{2} + 2 x^{2}) d x - (x^{2}) d y = 0

y = \frac{tan ( 2 ( 2 ln ( x ) + c _{1} ) ) x}{2}

Solution steps

(x y + 4 y^{2} + 2 x^{2}) d x - (x^{2}) d y = 0

Solve using substitution:

y = \frac{tan ( 2 ( 2 ln ( x ) + c _{1} ) ) x}{2}

y = \frac{tan ( 2 ( 2 ln ( x ) + c _{1} ) ) x}{2}

y = e^{9}

\int + e^{- 3 x} sin (b x) d x

\frac{d}{d x} (arcsin (\frac{x}{3}))

\int \frac{4 x + 1}{4 x} d x

\frac{d}{d x} (\frac{1 - 3 x}{1 + 3 x})

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