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

The solution for (x^2)dx+(2y)dy=0 is y=sqrt(-(x^3)/3+c_{1)},y=-sqrt(-(x^3)/3+c_{1)}

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(x^2)dx+(2y)dy=0

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

y = - \frac{x ^{3}}{3} + c_{1}, y = - - \frac{x ^{3}}{3} + c_{1}

Solution steps

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

Solve separable ODE:

y = - \frac{x ^{3}}{3} + c_{1}, y = - - \frac{x ^{3}}{3} + c_{1}

y = - \frac{x ^{3}}{3} + c_{1}, y = - - \frac{x ^{3}}{3} + c_{1}

\frac{d y}{d x} = \frac{x - 8}{y ^{3}}

y^{^{'}} = y^{2} (9 x + 5)

y^{^{'}} sec (x) = 5 y^{2}, y (\frac{π}{2}) = \frac{1}{2}

\frac{d Q}{d t} + \frac{Q}{20} = 0

\frac{d y}{d x} x y + 2 x + 3 x^{2} = 4

What is the solution for (x^2)dx+(2y)dy=0 ?
The solution for (x^2)dx+(2y)dy=0 is y=sqrt(-(x^3)/3+c_{1)},y=-sqrt(-(x^3)/3+c_{1)}