What is the solution for ty^'-y=(-t^3)/(y^2),y(1)=2 ?

The solution for ty^'-y=(-t^3)/(y^2),y(1)=2 is y=t\sqrt[3]{-3ln(t)+8}

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ty^'-y=(-t^3)/(y^2),y(1)=2

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

y = t 3 - 3 ln (t) + 8

Solution steps

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

Solve Bernoulli ODE:

y = t 3 - 3 ln (t) + 8

y = t 3 - 3 ln (t) + 8

\int_{0}^{3} (3 t - 5) d t

n = 1 \sum \infty n^{2} e^{n^{3}}

(\frac{x + 2}{x - 2})^{2}

\int_{1}^{2} \frac{1}{x ^{4}} d x

x \to 7 lim (\frac{x}{( x - 7 ) ^{2}})

What is the solution for ty^'-y=(-t^3)/(y^2),y(1)=2 ?
The solution for ty^'-y=(-t^3)/(y^2),y(1)=2 is y=t\sqrt[3]{-3ln(t)+8}