What is the solution for x^3dx+(y+1)dy=0 ?

The solution for x^3dx+(y+1)dy=0 is y=(-4+sqrt(16-8(-c_{1)+x^4)})/4 ,y=(-4-sqrt(16-8(-c_{1)+x^4)})/4

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x^3dx+(y+1)dy=0

x^{3} d x + (y + 1) d y = 0

y = \frac{- 4 + 16 - 8 ( - c _{1} + x ^{4} )}{4}, y = \frac{- 4 - 16 - 8 ( - c _{1} + x ^{4} )}{4}

Solution steps

x^{3} d x + (y + 1) d y = 0

Solve separable ODE:

y = \frac{- 4 + 16 - 8 ( - c _{1} + x ^{4} )}{4}, y = \frac{- 4 - 16 - 8 ( - c _{1} + x ^{4} )}{4}

y = \frac{- 4 + 16 - 8 ( - c _{1} + x ^{4} )}{4}, y = \frac{- 4 - 16 - 8 ( - c _{1} + x ^{4} )}{4}

2 x y y^{^{'}} = x + 3

\frac{y y ^{^{'}}}{x} + e^{y} = 0

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

x^{2} y d y - e^{y} d x = 0

y^{^{'}} = \frac{y + 3}{7 t + 50}, y (2) = 0

What is the solution for x^3dx+(y+1)dy=0 ?
The solution for x^3dx+(y+1)dy=0 is y=(-4+sqrt(16-8(-c_{1)+x^4)})/4 ,y=(-4-sqrt(16-8(-c_{1)+x^4)})/4