Integral equation with convolution I am tasked with finding the convolution
$$g(t)=2t+5\int_{0}^{t}\sin(\tau)g(t-\tau)\,\mathrm{d}\tau$$
I can see that this is the convolution of $\sin(t)$ and $g(t)$ but I don't know how to solve it because I'm solving for $g$ on both sides. What do I do here? 
Thanks for any help. 
 A: Taking the Laplace transform and recalling the convolution property, rearranging yields
$$\widetilde{g} (s^2 - 4) = 2 + \frac{2}{s^2}$$
Employing partial fractions, this becomes
$$\widetilde{g} = \frac{2}{s^2-4} - \frac{1}{22^2} + \frac{1}{2(s^2-4)}$$
Whence taking the inverse transform we get
$$g (t) = \frac{5}{4} \sinh (2t) - \frac{1}{2} t^2$$
A: By the properties of Laplace transform you have
$$\mathcal{L}g=2\mathcal{L}(t)+5\mathcal{L} \big(\sin(t) \big) \mathcal{L}g.$$
If you calculate the transforms of $t$ and $\sin⁡(t)$ then you can solve for $\mathcal{L}(g)$ and get
$$\mathcal{L}g=\frac2{s^2}\left(1-\frac5{1+s^2}\right)^{-1}=\frac{5}{8(s-2)}-\frac{5}{8(s+2)}-\frac{1}{2s^2}$$
which implies
$$g(t)=\frac58e^{2t}-\frac58e^{-2t}-\frac12t.$$
A: Changing the integration variable gives the equivalent integral equation
$$
g(t)=2t+5\int_0^t\sin(t-τ)g(τ)dτ
$$
You read of $g(0)=0$ and
$$
g'(t)=2+5\int_0^t\cos(t-τ)g(τ)dτ
$$
and after that $g'(0)=2$ and
$$
g''(t)=5g(t)-5\int_0^t\sin(t-τ)g(τ)dτ=4g(t)+2t
$$
This last differential equation has the solution
$$
g(t)=\frac54\sinh(2t)-\frac12t.
$$
