Intuition for multiplying by x for independent solutions

Question

I'm taking a differential equations course, and I was wondering what the intuition behind multiplying by x to get a linearly independent solution to a 2nd order homogeneous linear ODE is. Consider the DE: $$ y^{\prime\prime} + 4y^{\prime} + 4y = 0 $$ Evaluating the characteristic polynomial: $$ \lambda^2 + 4\lambda + 4 = 0\\ (\lambda+2)^2=0, \lambda=-2\ \text{(double root)}\\ y=c_1e^{-2x}+c_2xe^{-2x} $$ I'm trying to get a hold of why multiplying by $x$ makes sense. I understand the derivation via variation of coefficients, but I don't get why it makes sense. Why does multiplying a solution to a differential equation by $x$ to get another solution that is linearly independent make sense? Is it a guess, or is there some deeper intuition for why this makes sense in the case of a double root? Does multiplying a function of $x$ by $x$ always produce a new linearly independent function?

My teacher mentioned something about it being the first term of some taylor series expansion for $f(\epsilon)$ when nudging $x$ by $\epsilon$ in the first solution to the DE, but I didn't really understand that explanation. Is there more to that explanation?

Answer 1

Here's an explanation as to why all solutions of the differential equation $$\sum_{k=0}^n \binom{n}{k}(-\lambda)^{n-k} y^{(k)}=0$$ must be of the form $$c_1e^{\lambda x}+c_2xe^{\lambda x}+c_3x^2e^{\lambda x}+\cdots + c_n x^{n-1}e^{\lambda x}$$ for some constants $c_1,c_2,c_3\ldots, c_n$. Note that the characteristic polynomial of the above differential equation is $(x-\lambda)^n$, so hopefully this helps with your inquiry (in particular by setting $n=2$).

We first prove a short lemma: If $y$ is a $n$-times differentiable function, then $$\frac{d^n}{dx^n}\left(ye^{-\lambda x}\right)=e^{-\lambda x}\sum_{k=0}^n\binom{n}{k}(-\lambda)^{n-k}y^{(k)}.$$ We'll prove this by induction. The base case $n=0$ is easily verifiable. Now, if we assume the statement is true for some particular $n$, then \begin{align*} \frac{d^{n+1}}{dx^{n+1}}\left(ye^{-\lambda x}\right)&=\frac{d}{dx}\left[e^{-\lambda x}\sum_{k=0}^n\binom{n}{k}(-\lambda)^{n-k}y^{(k)}\right]\\ &=e^{-\lambda x}\left[\sum_{k=0}^n\binom{n}{k}(-\lambda)^{n-k}y^{(k+1)}\right]+e^{-\lambda x}\left[\sum_{k=0}^n\binom{n}{k}(-\lambda)^{n+1-k}y^{(k)}\right]\\ &=e^{-\lambda x}\sum_{k=0}^{n+1}\binom{n+1}{k}(-\lambda)^{n+1-k}y^{(k)}. \end{align*} Hence, the lemma is proven. If $y$ satisfies the differential equation above, then by the lemma we have $$\frac{d^n}{dx^n}\left(ye^{-\lambda x}\right)=0.$$ Since the only functions whose $n$th derivative is zero are polynomials of degree strictly less than $n$, it follows that $$ye^{-\lambda x}=c_1+c_2x+c_3x^2+\cdots + c_n x^{n-1}$$ for some constants $c_1,c_2,c_3\ldots, c_n$. This yields the desired result.

Answer 2

Suppose we have the ODE $$ y'' - (2r+\varepsilon)y' +r(r+\varepsilon) y = 0 , $$ of which the characteristic equation is obviously $(\lambda-r)(\lambda-r-\varepsilon)=0$, so it has general solution $Ae^{rx} + Be^{(r+\varepsilon )x}$. Suppose also that we have initial conditions $ y(0) = a $ and $ y'(0) = b $. Then, by our general solution we find $$ A+B = a , \qquad rA+(r+\varepsilon)B = b \implies \varepsilon B = b-ra , $$ so the specific solution is $$ y = \frac{a(r+\varepsilon)-b}{\varepsilon} e^{rx} + \frac{b-ar}{\varepsilon} e^{(r+\varepsilon)x} . $$ Written like this, it doesn't look very promising, but considering that we are interested in the limit as $\varepsilon \to 0$, let's instead collect the powers of $\varepsilon$: $$ y = \left( a + (b-ar)\frac{e^{\varepsilon x}-1}{\varepsilon} \right) e^{rx} . $$ Now, if we take the limit, $(e^{\varepsilon x}-1)/\varepsilon \to x$, because this is the limit in the definition of the derivative with respect to $t$ of $e^{xt}$ at $t=0$.

You may find it instructive to draw graphs of the solution for a fixed $a$ and $b$ and various small $\varepsilon$, to see the behaviour.

As for when the multiply-by-$x$ rule works for linear differential equations: let $$ y'' + py' + qy = 0 \tag{1} $$ be solved by $y=u$ and $y=xu$. Then we have $$ u'' + pu' + qu = 0 \qquad xu'' + 2u' + p(xu'+u) + qxu = 0 , $$ and subtracting, we find that $$ 0 = 2u' + pu , \tag{2} $$ which is certainly necessary, but not sufficient, in the case of constant coefficients we considered above. Can we say more? Differentiating, we find $$ 0 = 2u'' + p'u + pu' = 0 , $$ or using the differential equation, $$ 0 = -2(pu'+qu) + p'u + pu' \implies 0 = -pu' - 2qu + p'u = p^2 u/2 - 2q u + p' u . $$ Assuming that $u \neq 0$, we therefore must have $$ 2p' + p^2 - 4q = 0 . \tag{3} $$ Again, this is true for the constant coefficients case since in that case $p'=0$ and we are left with the discriminant of the characteristic equation. Indeed, this shows that in the case of constant coefficients, this is sufficient as well as necessary (although we knew that).

Indeed, it is sufficient in the general case, providing we assume $p$ and $q$ are sufficiently well-behaved: if $(3)$ is satisfied, $ u(x) = \exp{(\frac{1}{2}\int p(x) \, dx)}$ satisfies the differential equation $(1)$ (plugging it into $(1)$ gives $(3)$). It is easy to check that $ 2u' + pu = 0 $, and then plugging $xu(x)$ into the operator in $(1)$ gives $$ xu'' + 2u' + p(xu'+u) + qxu = x(u''+pu'+qu) + (2u'+pu) = 0 , $$ as the $u$ we defined satisfies $(1)$ and $(2)$.