How to prove this LDE can be simplified to with only real coefficients via subsitution.

Question

Question:

Prove that all the Second Order LDEs that can be written in the form of $$\frac{d}{dx}\left[P(x)\frac{dy}{dx}\right] + Q(x)y = 0$$ with $P(x),Q(x)$ continous and always greater than $0$ and statisfy $$\frac{1}{Q}\frac{d}{dx}\sqrt{PQ} = const.$$ can be simplified to LDEs with only real coefficients by performing a subsituion $x=\phi(t)$

My attempt first was to simplify the above equations with the unkonwn function $\phi$ and hope to find out what $\phi$ is eventually, but it turns out there are too many unkonwns and its impossible to figure it out.

So next I was attempting specific subsitutions, i.e. $t=\int{Q(x)dx}$ which kind of eliminate the $\frac{1}{Q}$ in the front of the second equation. By this subsitution they simplified to

$$\frac{d}{dt}\sqrt{PQ} =cosnt. $$ and $$\frac{d}{dt}\left[P(x)Q(x)\frac{dy}{dt}\right] + y = 0$$ Thoungh it looks cleaner, I still could not proceed.

Answer 1

So during recent days I was still thinking about my question, and eventually I proved it. Nevertheless I am going to share with you guys about my solution .

——TL,DR——

Use the subsitution $$ t = \int {\sqrt{\frac{Q(x)}{P(x)}}dx }$$

in both equation, then the question will be solved automatically.

The following content is why this subsitution, which is not a mathematical proof , but I still thought it is much more interesting.

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My guess that the subsitution $\phi(t)$ might not be dervied directly from the question itself is correct. Basically the question only says that the coefficients are real with no more constraints, this leaves direct subsitution method full of unknowns and at the end we obtain no more infomation about $\phi(t)$.

Instead, which can be already tell from the question, I started to "guess" some specific subsitutuion. However it is not an random guess, which will take forever. I start to use some example LDEs to become my basics of my "guesses"

The first LDE is $$\frac{d}{dx}\left(\sqrt{1-x^2}\frac{dy}{dx}\right)+\frac{1}{\sqrt{1-x^2}}y = 0$$ which is an example from the same book which I should also shared, the simpicity of this example is that $PQ=1$, and I thought this would be a perfect way to start the "guesses"

Apparently the subsitution is $t = sin(x)$, so I immedaitely gussess $ t= \int{Q dx } $, and it holds as long as PQ = constant. So the approch is good but need another step.

What really solved the problems is the Eluer's formula $$x^2y''+pxy'+qy=0$$ or write in clear forms $$\frac{d}{dx}\left(x^2\frac{dy}{dx}\right)+y=0$$ with the subsitution $x = e^t$, when I write it down as $t = ln(x)$, immediately realise it also has something to do with P(x), (because in this case Q(x) is 1! ), and the reason that P(x) is not appearing in the former subsituition is because P(x)Q(x) is just a constant so it canceled.

ln(x) reminds me $\int{\frac{1}{x}dx}$, Thus I guess again that $$t=\int{\frac{Q(x)}{\sqrt{P(x)Q(x)}}}dx$$ or $$t=\int{\sqrt{\frac{Q(x)}{P(x)}}}dx$$ which turns out to be the answer.

After that it is just chain rule stuff, one can tell that the equations turns to $$\frac{1}{\sqrt{PQ}}\frac{d}{dt}\left[\sqrt{PQ}\frac{dy}{dt}\right]+y=0$$ $$\frac{1}{\sqrt{PQ}}\frac{d}{dt}\left[\sqrt{PQ}\right]=c_1$$ applying product rule on the first equation, one yeild $$\frac{d^2y}{dt^2}+c_1*\frac{dy}{dt}+y=0$$ Q.E.D.