Physical Models confusion diff eq.

Question

Just to give an example:

If given the equation $$y'' + 5y' + 6y = g(t)$$ where

$$ g(t) = \begin{cases} 5, && 0 < t < \pi/2 \\ 0, && t> \pi/2 \end{cases} $$ if this is the case and you’ve got to solve for a certain physical model, where the non homogenous term is not always continuous but has a jump discontinuity how do you solve for a reasonable solution?

I’m slightly confused on what’s being asked I’m assuming I need to use variation of parameters but in what way? Because I used a method that give $\lambda_{1,2} = 3, 2$ respectively but my actual problem has imaginary roots with an $\alpha$ term in it. So I’m not entirely sure how to approach it if somebody could elaborate that would be great! Please and thank you!

Answer 1

Solve the equation $y'' +5y'+6y = 5$ on $[0,\pi/2]$ and then use the final values $y(\pi/2)$ and $y'(\pi/2)$ as initial conditions for a solution of $y''+5y' + 6y=0$ on $[\pi/2,\infty).$

Answer 2

First of all, you seem to not know what to do with complex roots. [Here's a guide on how to deal with them]https://en.wikipedia.org/wiki/Characteristic_equation_(calculus)#Complex_roots). The short answer is the imaginary parts become trig functions due to Euler's formula $$ e^{i\alpha t} = \cos(\alpha t) + i\sin(\alpha t) $$

Since you didn't tell us what your actual equation is, I'm going to use a different one that's more relevant to the problem $$ y'' + 4y = 0 $$

The characteristic polynomial is $\lambda^2 + 4 = 0$, or $\lambda = \pm 2i$. The fundamental solution is then $$ y(t) = a\cos 2t + b\sin 2t $$

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Next step is the non-homogeneous equation. Whenever you have a piece-wise problem, an intuitive approach is to look for a piece-wise solution. This involves solving two equations

\begin{cases} y'' + 4y = 5, && 0 < t < \pi/2 \\ y'' + 4y = 0, && t > \pi/2 \end{cases}

The second piece is already homogeneous, so the solution is the same as before. I'm assuming you already know the method undetermined coefficients to solve the first piece. In this case the particular solution is just a constant $y_p(t) = 5$

Putting it all together we get

$$ y(t) = \begin{cases} 5 + a_1\cos 2t + b_1\sin 2t, && 0 < t < \pi/2 \\ a_2\cos 2t + b_2\sin 2t, && t > \pi/2 \end{cases} $$

The actual constants will depend on the initial conditions. Again, you didn't give any, so let's suppose they are $y(0) = 0, y'(0) = 1$. Applying this to the first piece gives $$ y(0) = 5 + a_1 = 0 \\ y'(0) = 5 + 2b_1 = 1 $$ or $a_1 = -5, b_1 = -2$

We want to have the function and first derivative be continuous at the "jump" point. Computing these two values on the first pieces gives $$ y(\pi/2) = 5 - a_1 = 10 \\ y'(\pi/2) = 5 - 2b_1 = 9 $$

You can use these conditions to solve for $a_2, b_2$, giving the final solution

$$ y(t) = \begin{cases} 5 - 5\cos 2t - 2\sin 2t, && 0 < t < \pi/2 \\ 10\cos 2t - \frac92 \sin 2t, && t > \pi/2 \end{cases} $$