What is the solution for x^2ydx+(x^3+3xy^2)dy=0 ?

The solution for x^2ydx+(x^3+3xy^2)dy=0 is

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x^2ydx+(x^3+3xy^2)dy=0

x^{2} y d x + (x^{3} + 3 x y^{2}) d y = 0

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

Solution steps

x^{2} y d x + (x^{3} + 3 x y^{2}) d y = 0

Solve using substitution:

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

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

t y^{^{'}} + (1 + t) y = t

y^{^{''}} + 7 y^{^{'}} + 12 y = 2 e^{t}

t^{2} y^{^{''}} - 2 t y^{^{'}} + 2 y = 4 t^{2}

\int (a + x)^{3} d x

8 e^{- 3 s}

What is the solution for x^2ydx+(x^3+3xy^2)dy=0 ?
The solution for x^2ydx+(x^3+3xy^2)dy=0 is