What is the solution for (dy)/(dx)=(40)/(x^{-5)} ?

The solution for (dy)/(dx)=(40)/(x^{-5)} is y=(20x^6)/3+c_{1}

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(dy)/(dx)=(40)/(x^{-5)}

\frac{d y}{d x} = \frac{40}{x ^{- 5}}

y = \frac{20 x ^{6}}{3} + c_{1}

Solution steps

\frac{d y}{d x} = \frac{40}{x ^{- 5}}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = \frac{40}{x ^{- 5}}

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

then

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

y = \int \frac{40}{x ^{- 5}} d x

\int \frac{40}{x ^{- 5}} d x = \frac{20 x ^{6}}{3} + c_{1}

y = \frac{20 x ^{6}}{3} + c_{1}

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What is the solution for (dy)/(dx)=(40)/(x^{-5)} ?
The solution for (dy)/(dx)=(40)/(x^{-5)} is y=(20x^6)/3+c_{1}