What is the solution for (dx)/(dt)=sin(2pi(t+3)),x(3)=1 ?

The solution for (dx)/(dt)=sin(2pi(t+3)),x(3)=1 is x=-1/(2pi)cos(2pi(t+3))+1+1/(2pi)

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(dx)/(dt)=sin(2pi(t+3)),x(3)=1

\frac{d x}{d t} = sin (2 π (t + 3)), x (3) = 1

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

Solution steps

\frac{d x}{d t} = sin (2 π (t + 3))

Substitute

\frac{d x}{d t}

with

x^{'} (t)

x^{^{'}} (t) = sin (2 π (t + 3))

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

then

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

x = \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}

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

Apply initial conditions:

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

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

What is the solution for (dx)/(dt)=sin(2pi(t+3)),x(3)=1 ?
The solution for (dx)/(dt)=sin(2pi(t+3)),x(3)=1 is x=-1/(2pi)cos(2pi(t+3))+1+1/(2pi)