What is the solution for ((t^2))/x dt=dx ?

The solution for ((t^2))/x dt=dx is t(x)=\sqrt[3]{(3x^2)/2+c_{1}}

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((t^2))/x dt=dx

\frac{( t ^{2} )}{x} d t = d x

t (x) = 3 \frac{3 x ^{2}}{2} + c_{1}

Solution steps

\frac{( t ( x ) ^{2} )}{x} d t (x) = d x

Solve separable ODE:

t (x) = 3 \frac{3 x ^{2}}{2} + c_{1}

t (x) = 3 \frac{3 x ^{2}}{2} + c_{1}

\frac{d y}{d x} y = \frac{4 sin ( x )}{9 - 8 sin ( x )}

\frac{d y}{d t} = - k (y - 24)

2 x y^{^{'}} - y^{3} = 0

\frac{d y}{d x} y = - 3 x^{7} + \frac{3}{5} x^{4} + 568 x + 12

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

What is the solution for ((t^2))/x dt=dx ?
The solution for ((t^2))/x dt=dx is t(x)=\sqrt[3]{(3x^2)/2+c_{1}}