# Solving a simple inhomogenous differential equation

solve the following starting value problem for $0\leq \delta < 1$

$\ddot{x}(t)+2\delta \dot{x}(t) + x(t)=\sin(t)$ with $x(0)=0, \dot{x}(t)=1$

and call the solution $x^\delta$ Show that $\lim_{\delta \to 0} x^\delta (t)=x^0(t)$ is.

Solution: The idea to solve that is, splitting it up into a homogenous problem and in a particular one. If think I'm fine if we just discuss solving the homogenous problem.

Let $0<\delta < 1$:

For the homogenous problem, we make the ansatz: $x(t)=e^{\lambda t}$ getting us the equation $\lambda^2+2\delta\lambda+1=0$

Now, I get the solution: $\lambda_{1,2} = -\delta \pm \sqrt{\delta ^2-1}$

Define $\omega :=\sqrt{\delta ^2-1}$

So we get:

$x(t)=Ae^{t(-\delta + i\omega)}+Be^{t(-\delta-i\omega)}$

$=Ae^{-t\delta}e^{i\omega}+Be^{-t\delta}e^{-i\omega}$

$=Ae^{-t\delta}[\cos(\omega)+i\sin(\omega)]+Be^{-t\delta}[\cos(-\omega)+i\sin(-\omega)]$

$=Ae^{-t\delta}[\cos(\omega)+i\sin(\omega)]+Be^{-t\delta}[\cos(\omega)-i\sin(\omega)]$

$=e^{t\delta}[\cos(\omega)(\underbrace{A+B}_{c_1}+sin(\omega)(\underbrace{iA-iB}_{c_2})]$

$=c_1e^{-t\delta}\cos(\omega)+c_2e^{-t\delta}\sin(\omega)$

Question: I think, if I have such a situation, where I have two solutions in the form of $a\pm ib$ I can always write $c_1e^{at}\cos(b) + c_2e^{at}\sin(b)$ right? I didn't really have had complex calculus. I just now barely about it and I know that I used the eulers form here.

Does this have some special name? This kind of differential equation? Can I somehoe kind of generalize this solution? Like what if this diff. eq. had an additional term $\dddot{x}$? So we would get $\lambda^3+\lambda^2+2\delta\lambda + 1 = 0$ How would I solve that?

Can I somehow easier see the resulted format of my solution or is this teh way to go?

• It should be $c_1e^{-δt}\cos(ωt)+c_2e^{-δt}\sin(ωt)$ with $ω=\sqrt{1-δ^2}$. – Lutz Lehmann Jan 16 '17 at 18:41
• ah, yeha sure. Sorry about that. Question remains :) – xotix Jan 16 '17 at 19:22
• I don't know if this is a better way, but my knee jerk reaction to seeing an $a\pm b i$ in an ODE has always been to write the general form as $y(x)=C_1e^{ax}cos(bx)+C_2e^{ax}sin(bx)$. Probably if you do ODEs everyday as a career this will be a bad strategy, but for the purposes of the physics classes I took, I would always default to that. As for your second questions, the $\lambda^3$ makes me worried. If you could factor by grouping you could still pull it off. In your case, I don't think factoring by grouping would lead to something nice. – emka Jan 16 '17 at 19:33
• with "factoring by grouping" you mean e.g. x^2 + 2x + 1 = (x+1)^2? – xotix Jan 16 '17 at 19:40
• I mean something like $x^3+x^2+2x+2=x^2(x+1)+2(x+1)=(x^2+2)(x+1)$ – emka Jan 16 '17 at 19:42

## 1 Answer

After finding $$y_h=c_1e^{-δt}\cos(ωt)+c_2e^{-δt}\sin(ωt)$$ as the solution of the homogeneous equation, try $$y_p=d_1\cos(t)+d_2\sin(t)$$ as particular solution for $δ\ne 0$. Then eliminate all constants by inserting the initial values. You should be able to group the terms that they look like difference quotients for the limit $δ\to0$ resp. $ω\to 1$.

If $p$ is a polynomial, then the linear differential operator $L=p(\frac{d}{dt})$ has the same factorization as $p$. Especially it is true that $$L\bigl(e^{λt}u(t)\bigr)=e^{λt}p\Bigl(λ+\frac{d}{dt}\Bigr)u(t)$$ so that for roots $p(λ)=0$ the order of the ODE can be reduced by simple integration.

• yeah, that's the appraoch I used for the particular problem. Thanks – xotix Jan 17 '17 at 13:15