What is the solution for x(xdy+ydx)+(2x^2+4)dy=0 ?

The solution for x(xdy+ydx)+(2x^2+4)dy=0 is y=(c_{1})/(\sqrt[6]{3x^2+4)}

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x(xdy+ydx)+(2x^2+4)dy=0

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

y = \frac{c _{1}}{6 3 x ^{2} + 4}

Solution steps

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

Solve separable ODE:

y = \frac{c _{1}}{6 3 x ^{2} + 4}

y = \frac{c _{1}}{6 3 x ^{2} + 4}

4 - x^{2} (\frac{d y}{d x}) = y^{2}, y = 5, x = 1

x y^{^{'}} + 8 y + 7 = 7 x^{2} y^{^{'}} + 7 y

y^{^{'}} = sin (y) \cdot tan (y)

\frac{d y}{d x} = (\frac{x + 1}{y})

y^{^{'}} = 4 + 14 x + 12 y + 42 x y

What is the solution for x(xdy+ydx)+(2x^2+4)dy=0 ?
The solution for x(xdy+ydx)+(2x^2+4)dy=0 is y=(c_{1})/(\sqrt[6]{3x^2+4)}