What is the solution for ty^'+2y= 1/(y^2) ?

The solution for ty^'+2y= 1/(y^2) is y=(\sqrt[3]{-c_{1}+t^6})/(\sqrt[3]{2)t^2}

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ty^'+2y= 1/(y^2)

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

y = \frac{3 - c _{1} + t ^{6}}{3 2 t ^{2}}

Solution steps

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

Solve separable ODE:

y = \frac{3 - c _{1} + t ^{6}}{3 2 t ^{2}}

y = \frac{3 - c _{1} + t ^{6}}{3 2 t ^{2}}

\frac{d v}{d t} = 9 v

\frac{d h}{d t} = - \frac{3}{100} h - 17

x y^{2} d x - 4 x^{2} y d y = 0

y^{^{'}} - x y = 0, y (0) = 1

\frac{d y}{d x} - \frac{2 x}{x ^{2} + 2} y = 0

What is the solution for ty^'+2y= 1/(y^2) ?
The solution for ty^'+2y= 1/(y^2) is y=(\sqrt[3]{-c_{1}+t^6})/(\sqrt[3]{2)t^2}