What is the solution for (dy)/(dt)=3t^2cos(t^3) ?

The solution for (dy)/(dt)=3t^2cos(t^3) is y=sin(t^3)+c_{1}

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(dy)/(dt)=3t^2cos(t^3)

\frac{d y}{d t} = 3 t^{2} cos (t^{3})

y = sin (t^{3}) + c_{1}

Solution steps

\frac{d y}{d t} = 3 t^{2} cos (t^{3})

Substitute

\frac{d y}{d t}

with

y^{'} (t)

y^{^{'}} (t) = 3 t^{2} cos (t^{3})

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

then

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

y = \int 3 t^{2} cos (t^{3}) d t

\int 3 t^{2} cos (t^{3}) d t = sin (t^{3}) + c_{1}

y = sin (t^{3}) + c_{1}

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What is the solution for (dy)/(dt)=3t^2cos(t^3) ?
The solution for (dy)/(dt)=3t^2cos(t^3) is y=sin(t^3)+c_{1}