# Can I multiple a double-root solution to a 2nd order DE by a nonconstant?

I'm following online notes to (re)learn the standard knowledge for second order differential equations (DEs), with a focus for the moment on homogeneous DEs. My aim is to understanding from an engineering/applied perspective rather than hard core math perspective. I'm getting stuck on the case when the characteristic equation has double roots. "To find a second solution we will use the fact that a constant times a solution to a linear homogeneous differential equation is also a solution." The explanation then proceeds to multiply the first solution with a non-constant v(t) (likely the greek lower case "nu" rather than "v" per se).

How can I reconcile this with the stated multiplication by a constant? I intuitively get that multiplying by a constant still yields a solution to the original DE. For me to get why it is OK to multiply by a non-constant, however would involve a search for the theoretical justification, which is a detour that I'm hoping to avoid.

• The fact that a constant times a solution to a linear homogeneous differential equation is also a solution can be verified easily: just show that for any constant $c\in\mathbb{R}$ and solution $y(t)$ say, $cy(x)$ is also a solution. The point is that, this fact doesn't guarantee that you will get another solution if you multiply the solution with a non-constant function $v(t)$. It merely suggests to us the possibility of obtaining another solution by multiplying suitable function to the solution. Apr 2, 2017 at 0:15
• Hmm. OK, I missed that nuance in the narrative. I will trudge on and see if all of what follows makes sense. Thanks. Apr 2, 2017 at 0:30
• It all makes sense....would you be able to post your response as the answer? Apr 2, 2017 at 0:43

Let me offer a "perturbation" argument making this plausible. Consider the equation

$$y''(x) = 0.$$

The associated characteristic equation is $x^2 = 0$ and so it has a double root $x = 0$. Instead of solving the original equation, let us modify the equation slightly so that the problem doesn't occur and then see what happens. So instead of the original equation, consider the equation

$$y''(x) - \varepsilon^2 y(x) = 0$$

where we think of $\varepsilon$ as a small parameter. The associated characteristic equation is $x^2 - \varepsilon^2 = (x - \varepsilon)(x + \varepsilon)$ whose roots are $x_{1,2}(\varepsilon) = \pm \varepsilon$.

Hence, the modified equation has two linearly independent solutions

$$y_{1,\varepsilon}(x) = e^{x_1(\varepsilon) x} = e^{\varepsilon x}, y_{2,\varepsilon}(x) = e^{x_2(\varepsilon) x} = e^{-\varepsilon x}$$

and the general solution has the form

$$y_{\varepsilon}(x) = c_1(\varepsilon) y_{1,\varepsilon}(x) + c_2(\varepsilon)y_{2,\varepsilon}(x) = c_1(\varepsilon) e^{\varepsilon x}(x) + c_2(\varepsilon)e^{-\varepsilon x}$$

where $c_1(\varepsilon),c_2(\varepsilon)$ are arbitrary. As we take $\varepsilon \to 0$, the two different roots $x_{1,2}(\varepsilon)$ tend to the same root $x = 0$ and the two linearly independent solutions $y_{1,\varepsilon},y_{2,\varepsilon}$ tend (whatever that means) unfortunately to the same solution $y \equiv 1$ of the original equation.

How can we generate another solution? Instead of looking at the limit of $y_{1,\varepsilon},y_{2,\varepsilon}$ and get the same solution, we can look at the limit of $y_{\varepsilon}$ and try and choose the constants $c_1(\varepsilon),c_2(\varepsilon)$ so that we will get a solution other than $y \equiv 1$. An indeed, consider

$$y_{\varepsilon}(x) = \frac{1}{2\varepsilon} e^{\varepsilon x} - \frac{1}{2\varepsilon} e^{-\varepsilon x} = \frac{1}{2\varepsilon}(1 + \varepsilon x + \frac{\varepsilon^2 x^2}{2!} + \dots) - \frac{1}{2\varepsilon}(1 - \varepsilon x + \frac{\varepsilon^2 x^2}{2!} + \dots) \\ = x + \frac{1}{2\varepsilon}(e^{\varepsilon x} - e^{-\varepsilon x}).$$

Here, as $\varepsilon \to 0$ we have $y_{\varepsilon}(x) \to x$ and this gives us another solution $y(x) = x$ for the original equation.

Warning: What I described above is not a rigorous argument and there are quite a few problems in making it rigorous in full generality but this is an example of a perturbative argument which is quite common in physics. We use the extra freedom in the choice of constants and try and set them up so that we other solutions. This explains how the freedom of multiplying by constant can generate extra solutions.

• Thanks, levap. I'm going to have to chew on that explanation at a later date. I was really hoping to understand it at the level of undergraduate engineering, as the romp in to second order systems was itself a detour from another self-educational romp (into the subject area of coupled systems). However, I found that if I simply followed through the explanation that I linked to in my original post, the second solution from multiplying the first solution by a nonconstant actually makes sense. Apr 2, 2017 at 1:08