Green's Function for first order Linear ODE

Hi there,
I've been completing a problem of a question based on ODE and Green's Functions. The question asks me to come up with a general solution for the IVP ODE $L[y] = y\prime + y\cdot sin(t)=\delta(t - \tau), y(0) = 0$ by imposing a jump condition on $y(t)$ at $t = \tau$ and solving the differential equation for $t < \tau$ and $t > \tau$ . Then identify the Green's function $G(t,\tau)$ for the IVP ODE $L[y] = y\prime + y\cdot sin(t)=q(t), y(0) = 0$
I was wondering whether someone could read through my working and see whether I have the correct Green's function.

Lets intergrate $L[y] = y\prime + y\cdot sin(t)=\delta(t - \tau)$ from times $\tau + \varepsilon$ to $\tau - \varepsilon$
$\delta(t - \tau)$ will equal 1 over this interval, as it is the step function. The area under it's curve will always equal 1.
We take the limit as $\varepsilon$ approaches 0. This makes the intergal of $y\cdot sin(t)$ equal to zero. So we are left with $[y]$ evaluated from $\tau^+$ to $\tau^-$ . This equals 1.

To find the general solution, I need to solve the IVP for $t < \tau$ and $t > \tau$ . Start with $t < \tau$ .
The solution to the ODE is $y(t) = A\cdot \exp^{cos(t)}$ . Since the IVP states $y(0) = 0$ , $0 = A\cdot \exp^{cos(0)}$ This means that A=0. Therefore, $y(t) = 0$ for all $t < \tau$ .

For $t > \tau$ , we need a new IVP. It is found from [y] = 1 from evaluating [y] from $\tau^+$ to $\tau^-$ . Expanding this, $y(\tau^+) - y(\tau^-) = 1$ . We found that $y(\tau^-)$ = 0 from above, therefore $y(\tau^+) = 1$ . This is the new IVP for $t > \tau$ .
The ODE is $y(t) = B\cdot \exp^{cos(t)}$ . From the IVP, $1 = B\cdot \exp^{cos(\tau)}$ . This means $B = \exp^{-cos(\tau)}$ .
Therefore, $y(t) = \exp^{-cos(\tau)} \cdot \exp^{cos(t)}$ .

Finally, the solution to the ODE:
$y(t) = 0$ for all $t < \tau$ .
$y(t) = \exp^{cos(t) - cos(\tau)}$ for all $t > \tau$
This means my Green's function is $\exp^{cos(t) - cos(\tau)}$
Am I right?

Thank you for your time.