What is the solution for (dx)/(dt)=(x^3-xt^2)/(t^3),x(1)=2 ?

The solution for (dx)/(dt)=(x^3-xt^2)/(t^3),x(1)=2 is x=(2t)/(sqrt(2-t^4))

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(dx)/(dt)=(x^3-xt^2)/(t^3),x(1)=2

\frac{d x}{d t} = \frac{x ^{3} - x t ^{2}}{t ^{3}}, x (1) = 2

x = \frac{2 t}{2 - t ^{4}}

Solution steps

\frac{d x}{d t} = \frac{x ^{3} - x t ^{2}}{t ^{3}}

Solve Bernoulli ODE:

x = \frac{2 t}{2 - t ^{4}}

x = \frac{2 t}{2 - t ^{4}}

\int 3 x^{2} + 2 x + 4 d x

\frac{d}{d x} (2.1 + e^{2} - \frac{π ^{2}}{20})

\frac{d}{d x} (x^{3} + 4 x^{2})

y^{^{'}} = \frac{x ^{4} y ^{4} + x ^{4}}{y ^{3}}

\frac{d}{d x} (\frac{3}{x + 3})

What is the solution for (dx)/(dt)=(x^3-xt^2)/(t^3),x(1)=2 ?
The solution for (dx)/(dt)=(x^3-xt^2)/(t^3),x(1)=2 is x=(2t)/(sqrt(2-t^4))