I have the following math problem from my intro to dif. eq. class: (so don't just give an answer)
If the convolution $$ (y*f)(t) = \int_0^t y(t-v)f(v)\,dv$$ then show that $$ (y*f)'(t) = (y'*f)(t)+y(0)f(t) $$
So here's what I did: \begin{align*} (y*f)'(t) &= \frac{d}{dt} \def\i#1{\int_0^t #1 \, dv} \i{y(t-v)f(v)}\\ &= \i{\frac{d}{dt}\bigl(y(t-v)f(v)\bigr)} \tag 1\\ &= \i{y'(t-v)f(v) + y(t-v)f'(v)} \tag 2\\ &= \i{y'(t-v)f(v)} + \i{y(t-v)f'(v)}\\ &= y(t-t)f(t)-y(t-0)f(0) - \i{y(t-v)f'(v)} + \i{t(t-v)f'(v)} \tag 3\\ &= y(0)f(t)-y(t)f(0) \tag 4 \end{align*} where in
- Leibniz's Rule
- product rule
- integration by parts
- integrals cancel.
So this is clearly wrong according to the question so where do I go wrong?
EDIT:
Also how can I better understand the meaning of $$ (y'*f)(t) $$ in comparison to $$ (y*f)(t) $$
Note that, according to the chain rule, we have $$ \frac{d}{dt}\int_0^t y(t-v)f(v)\,dv = y(t-t)f(t) + \int_0^t y'(t-v)f(v) \, dv = y(0)f(t) + (y' * f)(t) $$
Addendum: You did two mistakes: (1) You used Leibniz's rule wrong. You did not take into account that one integration limit depends on $t$, note that Leibniz's rule states that $$ \frac{d}{dt} \int_{a(t)}^{b(t)} F(t,v)\, dv = \int_{a(t)}^{b(t)} \frac{\partial}{\partial t} F(t,v)\, dt + f(t,b(t))b'(t) - f(t,a(t))a'(t) $$ which gives the above (what I wrote above).
(2) You did apply the product rule wrong, note that you are taking derivatives with respect to $t$, not $v$ and $$ \frac{\partial}{\partial t}\bigl(f(t-v)g(v)\bigr) = f'(t-v)g(v) $$ as $g(v)$ does not depend on $t$.