What is the solution for y^'=20t^4-20t^3+3-4t^{-2} ?

The solution for y^'=20t^4-20t^3+3-4t^{-2} is y=4t^5-5t^4+3t+4/t+c_{1}

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y^'=20t^4-20t^3+3-4t^{-2}

y^{^{'}} = 20 t^{4} - 20 t^{3} + 3 - 4 t^{- 2}

y = 4 t^{5} - 5 t^{4} + 3 t + \frac{4}{t} + c_{1}

Solution steps

y^{'} (t) = 20 t^{4} - 20 t^{3} + 3 - 4 t^{- 2}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int 20 t^{4} - 20 t^{3} + 3 - 4 t^{- 2} d t

\int 20 t^{4} - 20 t^{3} + 3 - 4 t^{- 2} d t = 4 t^{5} - 5 t^{4} + 3 t + \frac{4}{t} + c_{1}

y = 4 t^{5} - 5 t^{4} + 3 t + \frac{4}{t} + c_{1}

What is the solution for y^'=20t^4-20t^3+3-4t^{-2} ?
The solution for y^'=20t^4-20t^3+3-4t^{-2} is y=4t^5-5t^4+3t+4/t+c_{1}