What is the solution for (dy)/(dt)=sin(2pi(t+3)) ?

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

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(dy)/(dt)=sin(2pi(t+3))

\frac{d y}{d t} = sin (2 π (t + 3))

y = - \frac{1}{2 π} cos (2 π (t + 3)) + c_{1}

Solution steps

\frac{d y}{d t} = sin (2 π (t + 3))

Substitute

\frac{d y}{d t}

with

y^{'} (t)

y^{^{'}} (t) = sin (2 π (t + 3))

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

then

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

y = \int sin (2 π (t + 3)) d t

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

y = - \frac{1}{2 π} cos (2 π (t + 3)) + c_{1}

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