Q(x)=0 special case self-adjoint eqtns

[ognik]

For the special case $\lambda=0, q(x)=0 $, the self-adjoint (SA) eqtn becomes $\displaystyle\d{}{x} \left[p(x) \d{u(x)}{x}\right]=0$, satisfied by $du/dx=1/p(x) $. Use this to get a 'second' solution of (a) Legendre's eqtn (b) Laguerre's (c) Hermite's. Note, in all 3 cases, $u_1(x)=1$

Not sure why $du/dx=1/p(x) $ automatically produces the 2nd solution?

I don't know where they got $u_1(x)=1$ from for all 3 eqtn types?

(a) With p(x)=$(1-x^2)$, I got $ u(x)=\frac{1}{2} (ln\frac{(1+x)}{(1-x)}- ln(1)) $ integrating between 0 and x, is this INTERVAL correct?

(b) They provide a solution of $ u_2 (x) - u_2(x_0)=\int_{x_0}^{x} \frac{e^{t}}{t} \,dt $. With p(x)=$xe^{-x}$, I got the same integral, but would like to know why they wrote this solution (only this one) as $ u_2 (x) - u_2(x_0)=...$? And why they used $x_0$ instead of 0 for the interval? And why you think they the answer as an integral, when they completed the integration for (a)?

 

[Ackbach]

For the special case $\lambda=0, q(x)=0 $, the self-adjoint (SA) eqtn becomes $\displaystyle\d{}{x} \left[p(x) \d{u(x)}{x}\right]=0$, satisfied by $du/dx=1/p(x) $. Use this to get a 'second' solution of (a) Legendre's eqtn (b) Laguerre's (c) Hermite's. Note, in all 3 cases, $u_1(x)=1$

Not sure why $du/dx=1/p(x) $ automatically produces the 2nd solution?

If you plug in for what $p(x)$ is for each of the three cases, you will get a solution that is not equal to $u_1=1$. So the word "second" is relative to the "first solution" $u_1=1$.

I don't know where they got $u_1(x)=1$ from for all 3 eqtn types?

If you plug in $u_1=1$ into the DE as a candidate solution, its derivative is zero, so when you multiply that by $p$ and differentiate again, you get zero as you should.

(a) With p(x)=$(1-x^2)$, I got $ u(x)=\frac{1}{2} (ln\frac{(1+x)}{(1-x)}- ln(1)) $ integrating between 0 and x, is this INTERVAL correct?

Do you need an interval?

(b) They provide a solution of $ u_2 (x) - u_2(x_0)=\int_{x_0}^{x} \frac{e^{t}}{t} \,dt $. With p(x)=$xe^{-x}$, I got the same integral, but would like to know why they wrote this solution (only this one) as $ u_2 (x) - u_2(x_0)=...$?

To emphasize that it's a displacement? I'm not entirely sure without looking at the context.

And why they used $x_0$ instead of 0 for the interval?

To allow for nonzero initial conditions, presumably.

And why you think they the answer as an integral, when they completed the integration for (a)?

Because that integral isn't do-able, at least not in terms of elementary antiderivatives. You can write it as the "exponential integral", but it has no elementary antiderivative.

 

[ognik]

If you plug in $u_1 =1$ into the DE as a candidate solution, its derivative is zero, so when you multiply that by p and differentiate again, you get zero as you should.

Granted, but in general, how should I know to try $u_1 =1$ if they hadn't told me?

Do you need an interval?

This covers the rest of my questions.
It's a self adjoint eqtn., so I thought it must have boundary conditions? Also an indefinite integral would leave me with an unknown constant in the solution. Finally the book has definite integrals in the other 2 given solutions. (Me, I sometimes find intervals limiting...)

Interestingly the intervals they use don't correspond to the orthogonality intervals.

 

[Ackbach]

Partial reply here:

Granted, but in general, how should I know to try $u_1 =1$ if they hadn't told me?

Intuition, recognizing patterns, experience, imagination. In general, if I see a somewhat complicated-looking DE, or even a not-so-complicated-looking DE, I'm going to see if the trivial solution solves it, which you can usually check by inspection. Then I'm going to see if a constant solves it: again, you can usually check by inspection since all the derivatives will vanish.

 

[ognik]

Partial reply here:
Intuition, recognizing patterns, experience, imagination. In general, if I see a somewhat complicated-looking DE, or even a not-so-complicated-looking DE, I'm going to see if the trivial solution solves it, which you can usually check by inspection. Then I'm going to see if a constant solves it: again, you can usually check by inspection since all the derivatives will vanish.

Yay, nice I hadn't missed something

More interested in understanding the second part ... I think I need intervals as per my attempted justification?