Convolution and Total Response Differential Equations

Question

Convolution with differential equations is extremely confusing to me.

The two following questions were asked in class and we were asked to think about them. I want to work them out but I don't know how.

For both problems, express the total response of the given initial value problem using a convolution integral to represent the forced response.

1) $y'' - 3y' + 2y = cos(\alpha t)$ $y(0)=1, y'(0) = 0$

and

2) $y^{(4)} - y = g(t); y(0) = 0, y'(0) = 0, y''(0) =0; y'''(0)=0$

I've tried doing the laplace transform for both problems and I just can't get the answer to look like the solutions given.

Answer 1

For the case $1)$ you can use the following trick:

• Define $z(t) = y-y_0 $ such that $z(0) = 0$ and $z'(0) = y'_0 = 0$.

• Rewrite the problem for $z$:

  $$ z'' - 3z' + 2 z = \cos{\alpha t} - 2 \equiv f(t), \quad z(0)=z'(0) = 0, \quad t > 0. \tag{1}$$

• Laplace-transform both sides to have:

  $$ Z(s) = \frac{1}{s^2-3 s+2} \, \mathcal{L}_s [f[t]] = \mathcal{L}_s[q(t)]\mathcal{L}_s[f(t)]. \tag{2} $$

• Compute $q(t)$ as the inverse Laplace transform of $1/(s^2 -3s+2)$. It turns out that $q(t) = e^t(e^t-1)$. Then, we can use the convolution theorem to conclude that:

  $$\color{blue}{ z(t) = q(t)*f(t) = \int^t_0 q(\tau)f(t-\tau) \, \mathrm{d}\tau} \tag{3}$$

• Finally, $y(t)$ is given by $y(t) = z(t)+y_0$.

The process for the second problem is identical.

Hope this helps!

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Note that under homogenous initial conditions, $G(s) = \frac{1}{s^2-3s+2}$ is defined as the transfer function of the dynamical system, given that: $\frac{Z(s)}{U(s)} = G(s)$.