Solving $y'' + 2y' + 2y = 2\delta' + 2\delta$ without Laplace transform

Question

Im trying to solve the following differential equation:

$$y'' + 2y' + 2y = 2\delta' + 2\delta$$

I did this by first setting $ y(t) = z(t)\theta(t)$ and finding the causal solution to the problem. From this i got the following solution:

$$y(t) = 2e^{-t}\cos(t)\theta(t) $$

However this is only the solution to the homogenous equation and when it comes to finding a particular solution im stuck. I'm somewhat aware that it can be solved with Laplace transform but since i haven't studied it yet i can't use it.

Answer 1

Your solution is $y=2u+2u'$ where $u$ is the solution to $$ u''+2u'+2u=δ, $$ presumably with $u(t)=0$ for $t<0$. This in turn can be obtained by the factorization $u(t)=θ(t)v(t)$ where $v$ solves $$ v''+2v'+2v=0, ~~~v(0)=0,~~~v'(0)=1. $$ This now is a standard task.

Answer 2

The solution to $y'' + 2y' + 2y = 0$ is $y(t) = e^{-t}(A \cos(t) + B \sin(t))$. Let's add an extra term for the particular solution.

$$y(t) = e^{-t}(A \cos(t) + B \sin(t)) + f(t)$$ $$y'(t) = e^{-t}((-A+B)\cos(t) + (-A-B)\sin(t)) + f'(t)$$ $$y''(t) = e^{-t}(-2B \cos(t) + 2A\sin(t)) + f''(t)$$

Plugging this into the original equation, the exponential/sinusoidal terms all cancel out, and we're left with:

$$f'' + 2f' + 2f = 2\delta' + 2\delta$$

But $\delta$ is a weird function to work with, so let's let $g$ be a second antiderivative of $f$ (i.e., $f = g''$), and integrate.

$$g'''' + 2g''' + 2g'' = 2\delta' + 2\delta$$ $$g''' + 2g'' + 2g' = 2\delta + 2H + C$$ $$g'' + 2g' + 2g = 2H + 2tH + Ct + D$$

If $t > 0$, then $H = 1$, and $g'' + 2g' + 2g = (C + 2) + D + 2$, which has the particular solution $g(t) = (\frac{C}{2}+1)t + \frac{D-C}{2}$.

If $t < 0$, then $H = 0$, and $g'' + 2g' + 2g = Ct + D$, which has the particular solution $g(t) = \frac{C}{2}t + \frac{D-C}{2}$.

These can be combined into a single function by using the step function.

$$g(t) = \frac{C}{2}t + \frac{D-C}{2} + tH(t)$$

Now differentiate twice, applying the Product Rule when needed.

$$g'(t) = \frac{C}{2} + H(t) + t\delta(t)$$ $$g''(t) = f(t) = \delta(t) + \delta(t) + t\delta'(t) = 2\delta(t) + t\delta'(t)$$

We now have the solution:

$$y(t) = e^{-t}(A \cos(t) + B \sin(t)) + 2\delta(t) + t\delta'(t)$$

Answer 3

As mentioned by Dan, the homogeneous equation has solution $y(t) = e^{-t}(A\cos(t) + B\sin(t))$ for arbitrary constants $A$ and $B$. We shall denote the two basis functions as $y_1(t) = e^{-t}\cos(t)$ and $y_2(t) = e^{-t}\sin(t)$. Using a standard method to solve second-order linear ODE (see Particular solution of second order differential equation), I shall show how one can use the "Variation of Parameters" here.

As a recap, the general solution using Variation of Parameters (see https://tutorial.math.lamar.edu/classes/de/VariationofParameters.aspx) is given by $$y_p(t) = -y_1(t)\int^t \frac{y_2(s)f(s)}{W(y_1,y_2)(s)}\mathrm{d}s + y_2(t)\int^t \frac{y_1(s)f(s)}{W(y_1,y_2)(s)}\mathrm{d}s$$ where $f(t)$ is the forcing function of the second-order linear ODE $\mathcal{L}y(t) = f(t)$, and $W(y_1,y_2)(t)$ is the Wronskian, defined by $$W(y_1,y_2)(t) = y_1(t)y_2'(t) - y_2(t)y_1'(t). $$

All that's left is to compute the relevant terms. Substituting $y_1(t)$ and $y_2(t)$ as deduced above, we have $$\begin{aligned}W(y_1,y_2)(t) &= y_1(t)y_2'(t) - y_2(t)y_1'(t) \\ &= e^{-t}\cos(t)(e^{-t}\sin(t))' - e^{-t}\sin(t)(e^{-t}\cos(t))' \\ &= e^{-2t}.\end{aligned} $$

Let's now compute the two integrals in the formula given above. Before we begin, we note that for any smooth function $g(t)$, we have $$\int^t g(s)\delta(s) \mathrm{d}s = \int^t g(0)\delta(s)\mathrm{d}s = g(0) \int^t \delta(s)\mathrm{d}s = g(0)\theta(t)$$ and using integration by parts, we have $$\int^t g(s)\delta'(s) \mathrm{d}s = g(t)\delta(t) - \int^t g'(s)\delta(s) \mathrm{d}s = g(0)\delta(t) - g'(0)\theta(t).$$ Here, we have used the fact that $\delta(t)g(t) = \delta(t)g(0)$ since the values of $g$ outside of $t = 0$ does not matter as $\delta = 0$ there. (These are the physicists type of argument, in which from the way the question is phrased, would be more suitable as a solution to this question!)

The first is given by $$\begin{aligned}\int^t \frac{y_2(s)f(s)}{W(y_1,y_2)(s)}\mathrm{d}s &= \int^t e^{2s}e^{-s}\sin(s)(2\delta'(s) + 2\delta(s)) \mathrm{d}s \\ &= 2 \int^t e^{s}\sin(s)\delta'(s)\mathrm{d}s + 2\int^t e^{s}\sin(s)\delta(s) \mathrm{d}s \\ &= -2(e^t \sin(t))'|_{t=0}\theta(t) = -2 \theta(t)\end{aligned}$$ where I have ignored the other terms that vanishes at $t = 0$ in the last line. Similarly, the second is given by $$\begin{aligned}\int^t \frac{y_1(s)f(s)}{W(y_1,y_2)(s)}\mathrm{d}s &= 2 \int^t e^{s}\cos(s)\delta'(s)\mathrm{d}s + 2\int^t e^{s}\cos(s)\delta(s) \mathrm{d}s \\ &= 2(e^t \cos{t}|_{t=0} - (e^t \cos{t})'|_{t=0} ) \theta(t) + 2(e^t \cos{t}|_{t=0})\delta(t) \\ &= 2 \delta(t).\end{aligned}$$

Combining all the above computations, we get $y_p(t) = 2\theta(t)y_1(t) + 2 \delta(t)y_2(t)$. Thus, $$\boxed{y_p(t) = 2e^{-t}(\cos(t)\theta(t) + \delta(t)\sin(t)) = 2e^{-t}(\sin(t)\theta(t))'}. $$

In retrospect, $2\theta(t)y_1(t)$ looks like what you have written in your solution so I'm guessing there might be a shorter/slicker solution to this. Nonetheless, this illustrates that standard calculus techniques still work for this (though from a rigorous mathematical point of view, one has to rely on distribution theory).