Solution of initial value problem $u(t)=\frac{1}{v}\int^t_0 \sin(v(t−s))f(s) \, ds$ part 2 In my previous question (see here) a user answered me by using the leibniz rule.
I'll post my initial value problem here again so you can see it better...
Solution:
$$u(t)=\frac{1}{v} \int_0^t \sin(v(t-s))f(s)~ds.$$
Initial value problem:
$$u''(t) +v^2u(t)=f(t)\text{ and }u(0)=u'(0)=0.$$
I just wanted to know if I could solve this question by using just the Fundamental theorem of Calculus.
Because...
$$ f(x) = \int^x_0 f'(x) \, dt$$
So that means...
$$u(t)=\frac{1}{v} \int_0^t \sin(v(t-s))f(s)~ds \Rightarrow u'(t) =\frac{1}{v} \sin(v(t-s))f(s)$$
$$u'(t) =\int_0^t \frac{1}{v} \sin(v(t-s))f(s) \frac{d}{dt} = \int_0^t \cos(v(t-s))f(s) $$
So far so good?
Then I've tried the same for $u''(t)$
$$ u''(t) = \cos(v(t-s))f(s) = \int_0^t \cos(v(t-s))f(s) \frac{d}{dt} = -v \int_0^t \sin(v(t-s))f(s)$$
But $u''(t)$ should be $= f(t) -v \int_0^t \sin(v(t-s))f(s)$ otherwise the formula isn't true.
Where's my mistake?
 A: Your issue is $u''(t)$. Since $t$ is contained inside of the integral, you can't do that. In fact
$$ u'(t) = \frac{1}{v} \bigg[\int_0^t\sin(v(t-s))f(s)ds\bigg]' = \int_0^t \cos(v(t-s))f(s) ds$$
and so
$$ u''(t) = \bigg[\int_0^t \cos(v(t-s))f(s) ds\bigg]'=\cos(v(t-s))f(s)\bigg|_{s=t}-v\int_0^t \sin(v(t-s))f(s) ds=f(t)-v^2u(t).$$
A: Your mistake was in writing $u'(t)=\frac{1}{v}\sin(v(t-s))f(s)$. For starters, the integration variable $s$ should've been replaced with $t$ when differentiating the integral.
Another problem I noticed was that you differentiated $\int_{0}^{t}\sin(v(t-s))f(s)ds$ without taking into account that the integrand $\sin(v(t-s))f(s)$ depends on $t$. In general, the Fundamental Theorem of Calculus is only valid when the integrand does not depend on the differentiation variable ($t$ in this case). To find the correct derivative, you first need to rewrite $\sin(v(t-s))=\sin(vt-vs)$ using the $\sin$ angle difference formula.
\begin{align*}
u(t) &= \frac{1}{v}\int_{0}^{t}\sin(vt-vs)f(s)\text{ }ds\\
&= \frac{1}{v}\int_{0}^{t}\left[\sin(vt)\cos(vs)-\cos(vt)\sin(vs)\right]f(s)\text{ }ds\\
&= \frac{1}{v}\int_{0}^{t}\left[\sin(vt)\cos(vs)f(s)-\cos(vt)\sin(vs)f(s)\right]ds\\
&= \frac{1}{v}\int_{0}^{t}\sin(vt)\cos(vs)f(s)\text{ }ds-\frac{1}{v}\int_{0}^{t}\cos(vt)\sin(vs)f(s)\text{ }ds
\end{align*}
At this point, you may notice that $\sin(vt)$ and $\cos(vt)$ are constants in the eyes of $s$, so they can be factored out of their respective integrals.
\begin{align*}
u(t) &= \frac{1}{v}\int_{0}^{t}\sin(vt)\cos(vs)f(s)\text{ }ds-\frac{1}{v}\int_{0}^{t}\cos(vt)\sin(vs)f(s)\text{ }ds\\
&= \frac{\sin(vt)}{v}\int_{0}^{t}\cos(vs)f(s)\text{ }ds-\frac{\cos(vt)}{v}\int_{0}^{t}\sin(vs)f(s)\text{ }ds
\end{align*}
Computing $\frac{d}{dt}\left(\frac{\sin(vt)}{v}\int_{0}^{t}\cos(vs)f(s)ds\right)$ and $\frac{d}{dt}\left(\frac{\cos(vt)}{v}\int_{0}^{t}\sin(vs)f(s)ds\right)$ using the product rule gives $\cos(vt)\int_{0}^{t}\cos(vs)f(s)+\frac{\sin(vt)\cos(vt)f(t)}{v}$ and $-\sin(vt)\int_{0}^{t}\sin(vs)f(s)+\frac{\sin(vt)\cos(vt)f(t)}{v}$, respectively, so
\begin{align*}
u'(t) &= \cos(vt)\int_{0}^{t}\cos(vs)f(s)ds+\frac{\sin(vt)\cos(vt)f(t)}{v}-\left(-\sin(vt)\int_{0}^{t}\sin(vs)f(s)ds+\frac{\sin(vt)\cos(vt)f(t)}{v}\right)\\
&= \int_{0}^{t}\cos(vt)\cos(vs)f(s)\text{ }ds+\int_{0}^{t}\sin(vt)\sin(vs)f(s)\text{ }ds+\frac{\sin(vt)\cos(vt)f(t)}{v}-\frac{\sin(vt)\cos(vt)f(t)}{v}\\
&= \int_{0}^{t}\left[\cos(vt)\cos(vs)+\sin(vt)\sin(vs)\right]f(s)\text{ }ds\\
&= \int_{0}^{t}\cos(vt-vs)f(s)\text{ }ds
\end{align*}
The correct derivative $u'(t)=\int_{0}^{t}\cos(v(t-s))f(s)ds$ immediately follows. Similar reasoning also allows you to find $u''(t)$ with the Fundamental Theorem.
