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

Solution steps

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

Solve separable ODE:

(x^{2} + 4) y^{^{'}} = \frac{x}{2 y}

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

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

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

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