# Proving solutions of $y''+p(x)y'+q(x)y=0$ to be linearly independent

When studying Elementary Differential Equations by William, I found trouble understanding Theorem 5.1.5

It says the two solutions are linearly independent iff their Wronskian is never zero, but I think they can still be linearly independent even if the Wronskian is zero for some $$x$$.

In the proof, when $$W(x_0)=0$$, Theorem 5.1.4 is used to show $$W\equiv 0$$.

This theorem is the Abel's identity. It seems flawless, until I saw this answer. So $$p(x)$$ must be continuous on $$(a,b)$$, but it is not as long as $$W(x_0)=0$$, so we shouldn't use Abel's identity. This is because $$y_1''+p(x)y_1'+q(x)y_1,\quad y_2''+p(x)y_2'+q(x)y_2$$

$$y_1''y_2+p(x)y_1'y_2+q(x)y_1y_2,\quad y_2''y_1+p(x)y_2'y_1+q(x)y_2y_1$$

$$p(x) = \frac{y_1''y_2-y_2''y_1}{y_2'y_1-y_1'y_2} = \frac{y_1''y_2-y_2''y_1}{W[y_1,y_2](x)},$$$$p(x)$$ is undefined at $$x_0\in(a,b)$$.

Also, I noticed all this by considering an example: for two solutions $$y_1=1+x$$ and $$y_2=1+x^2$$, $$W(x) = x^2+2x-1$$, on interval $$(0,\infty)$$. So although $$W(\sqrt{2}-1)=0$$, I think the two functions are still linearly independent on the interval, at least according to the definitions here. But using Theorem 5.1.5 they should be linearly dependent, because $$W$$ is zero for a point in the interval?

Now it really confuses me despite thinking about it for hours. Which part did I miss? I am sorry if the question is too dumb, still not accustomed to the linearly independence of functions.

Two functions $$f(x)$$ and $$g(x)$$ are linearly independent if and only if neither is a constant multiple of the other. That's directly from the definition of linear independence: there do not exist constants $$a_1$$ and $$a_2$$, not both $$0$$, such that $$a_1 f(x) + a_2 g(x) = 0$$ for all $$x$$ in the domain of the functions. So, in your example, $$1 + x$$ and $$1 + x^2$$ are obviously linearly independent because their ratio is not constant.

What about Abel's identity? That gives a formula for the Wronskian of two solutions of the differential equation $$y'' + p(x) y' + q(x) y = 0$$ on an interval where $$p(x)$$ and $$q(x)$$ are defined and continuous, and in particular shows that in that interval the Wronskian is either always $$0$$ or never $$0$$. So if you find two linearly independent functions (such as your $$1+x$$ and $$1+x^2$$) whose Wronskian is $$0$$ at some point but not elsewhere, it tells you that something goes wrong with the coefficients of a second-order homogeneous linear differential equation that has these solutions at a point where the Wronskian is $$0$$.

And indeed, in your example you can show that the second-order homogeneous linear differential equation that has $$1+x$$ and $$1+x^2$$ as solutions is

$$y'' - \frac{2(1+x)}{x^2 + 2 x - 1} y' + \frac{2}{x^2 + 2 x - 1} y = 0$$

where the coefficients $$p(x)$$ and $$q(x)$$ blow up at the roots of $$x^2 + 2 x - 1$$, and by no coincidence $$x^2 + 2 x - 1$$ is your Wronskian.

I really don't understand why DE textbooks make such a big deal about the Wronskian for second-order equations as a criterion for linear independence of solutions, when it's so easy to tell whether two functions are linearly independent.

• I agree. So Theorem 5.1.5 is wrong (or replace iff by if)? We actually use it for exams, for continuous $p(t)$ though. So it doesn't change the answer but would change the way it should be justified. Thank you for your clarification and I will ask my teacher tomorrow. Feb 21 at 2:47
• Theorem 5.1.5 is perfectly correct. In your example, the interval $(a,b)$ can't contain any roots of $x^2 + 2 x - 1$. Feb 21 at 15:24
• Ah my bad! I got caught up in the logic, it is correct indeed. Thanks for the help! Feb 22 at 6:13