Assume i have a function $y(t)$ that is differentiable twice such that there exists $y'(t)$ and $y''(t)$
Im having a hard time calculating the convolution of $2t+1$ and $y'(t)$
What i did so far was write $$ (2t+1)\star y'(t) = \intop_0^t[2(t-\tau)+1]y(\tau)d\tau = \intop_0^t2ty'(\tau)d\tau-2\intop_0^t\tau y'(\tau)d\tau + \intop_0^ty'(\tau)d\tau $$
and got this so far $$ \intop_0^t[2(t-\tau)+1]y(\tau)d\tau = 2ty(\tau)\bigg|_0^t+y(\tau)\bigg|_0^t-2\intop_0^t\tau y'(\tau)d\tau $$
and im stuck at the last integral $$ \intop_0^t\tau y'(\tau)d\tau $$
Is there another way to solve this kind of problem?
This convolution was part of a bigger question where i had an ODE with convolutions to solve that was of the kind $$ sint\star y'' + (2t+1)\star y' + sint y = 0 $$
The 2nd convolution is the one i got stuck at and how i eventually solved it was by leaving the integral and differentiating both sides of the equation so that i could use the fundumental theorem of calculus to get rid of the anoying integral like so:
$$ 1)\ \ sint\star y'' = \intop_0^tsin(t-\tau)y''(\tau)d\tau = y'(\tau)sin(t-\tau)\bigg|_0^t+\intop y'(\tau)cos(t-\tau)d\tau \\ 2) \ \ sint\star y = \intop_0^tsin(t-\tau)y(\tau)d\tau = y(\tau)cos(t-\tau)\bigg|_0^t - \intop_0^ty'(\tau)cos(t-\tau)d\tau $$
So we get $$ sint\star y'' + sint\star y = -sin(t)y'(0) + y(t)-y(0)cos(t) $$
Then we go and solve the middle part of the equation (that i was having a problem solving)
$$ (2t+1)\star y' = \intop_0^t[2(t-\tau)+1]y'(\tau)d\tau = \intop_0^t2ty'(\tau)d\tau + \intop_0^ty'(\tau)d\tau - 2\intop_0^t\tau y'(\tau)d\tau \\ =2ty(\tau)\big|_0^t + y(\tau)\big|_0^t - \intop_0^t\tau y'(\tau)d\tau $$
We can plug in and differentiate both side to get a first order linear ODE
$$ -sin(t)y'(0) + y(t)-y(0)cos(t) + 2ty(t)-2ty(0) + y(t) - y(0) = \intop_0^t\tau y'(\tau)d\tau $$
After we differentiate we get the problem was the Right had side so i will only show how i handled it:
By the fundumental theorem of calculus: $$ \frac{d}{dt}\intop_0^t\tau y'(\tau)d\tau = ty'(t) $$