# Easy heaviside differential equation with sin(u(t))=y’

The problem is the following:

$$y”* \theta(t) = sin(\theta(t))$$

This is how I have tried to solve it:

$$y”* \theta(t) = y’ * \theta(t) ‘ = y’ * \delta(t) = y’$$ (since everything convoluted with the derivative of the heaviside unit step function is the other factor)

This means that:

$$y’ = sin(\theta(t))$$

and

$$y’ = \theta(t) sin(1)$$ (since theta(t) = 1 for t>0, but it is equivalent with the last step).

$$y=t \cdot sin(1)\cdot \theta(t) + C$$

However, the correct answer to the problem is:

$$y(t) = t \cdot sin (1) \cdot θ(t) + C_1t + C_2$$

Which is very similar to mine, but how did they get those extra constants?