What is the solution for yy^'=(t^2)/((1+t^3)) ?

The solution for yy^'=(t^2)/((1+t^3)) is y=sqrt((2ln(1+t^3)+c_{1))/3},y=-sqrt((2ln(1+t^3)+c_{1))/3}

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yy^'=(t^2)/((1+t^3))

y y^{^{'}} = \frac{t ^{2}}{( 1 + t ^{3} )}

y = \frac{2 ln ( 1 + t ^{3} ) + c _{1}}{3}, y = - \frac{2 ln ( 1 + t ^{3} ) + c _{1}}{3}

Solution steps

y y^{'} (t) = \frac{t ^{2}}{( 1 + t ^{3} )}

Solve separable ODE:

y = \frac{2 ln ( 1 + t ^{3} ) + c _{1}}{3}, y = - \frac{2 ln ( 1 + t ^{3} ) + c _{1}}{3}

y = \frac{2 ln ( 1 + t ^{3} ) + c _{1}}{3}, y = - \frac{2 ln ( 1 + t ^{3} ) + c _{1}}{3}

\int x^{3} cosh (x) d x

\int \frac{3 x ^{3} - 8 x ^{2} + 10}{x ( x - 1 ) ^{3}} d x

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

\frac{d}{d x} (a t)

y^{^{'''}} + 14 y^{^{''}} + 49 y^{^{'}} = 0

What is the solution for yy^'=(t^2)/((1+t^3)) ?
The solution for yy^'=(t^2)/((1+t^3)) is y=sqrt((2ln(1+t^3)+c_{1))/3},y=-sqrt((2ln(1+t^3)+c_{1))/3}