1
$\begingroup$

$mg'' + kg = \delta(t)$, where $f(t) = \delta(t)$ is an impulse.

I am supposed to show that $y(t) = \int_{0}^{t} g(t-s)f(s) ds$ is the particular solution. In other words, $y(t)$ satisfies $my'' + ky = f(t)$. I also know that initial conditions are $g(0) = 0$ and $g'(0) = 1/m$.

What I have so far:

By the Leibniz Rule, I know that

$y'(t) = \int_{0}^{t}g'(t-s)f(s)ds + g(0)f(t)$.

since $g(0) = 0$,

$y'(t) = \int_{0}^{t}g'(t-s)f(s)ds$

By another application of the Leibniz Rule I get

$y''(t) = \int_{0}^{t}g''(t-s)f(s)ds + g'(0)f(t)$

Now $my'' + ky = m\int_{0}^{t}g''(t-s)f(s)ds + mg'(0)f(t) + k\int_{0}^{t}g(t-s)f(s)ds$

I know that $mg'(0)f(t)$ simplifies to $f(t)$ because of the initial conditions, but I don't see how the integrals cancel out. Any help or hints would be appreciated. Just for context, I'm trying to teach myself differential equations. The solutions manual agrees with my integrals, but states they cancel and I don't understand why!

  • 0
    $g'$ should have a jump of size $1/m$ at $t=0$, not the initial value. Or should one read that $g'(0^-)=0$ and $g'(0^+)=\frac1m$? – 2017-02-10

2 Answers 2

0

By the initial equation, combining the integrals and extracting $f(s)$ leaves the factor $$mg''(t−s)+kg(t−s)$$ which is zero for $t-s\ge 0$, that is inside all of the integration interval.

0

Sorry all! For some reason I got so focused on problem solving right away that I kept skipping over an important additional piece of information in the problem. It is also given that $mg'' + kg = 0$ for $g(t)$. That is the reason that the integrals end up adding up to $0$. I knew, when I initially worked on this, that it held for sinusoidal functions, but didn't think I could generalize (which you can't). That added conditions is what allows this to work out. Please let me know if I'm wrong on this, but I think I understand now.