1. ## Second order DEs and Self Adjoint Form

How do you go about proving that all second order DEs

a(x)y" + b(x)y' + c(x)y = 0

can be written in self adjoint form

[p(x)y']' + q(x)y = 0?

2. Originally Posted by kiwijoey
How do you go about proving that all second order DEs

a(x)y" + b(x)y' + c(x)y = 0

can be written in self adjoint form

[p(x)y']' + q(x)y = 0?
Well, if you let $b(x)=a^{\prime}(x)$, it follows that $a(x)y^{\prime\prime}+a^{\prime}(x)y^{\prime}=\left[a(x)y^{\prime}\right]^{\prime}$ (def. of product rule)

So it follows that $a(x)y^{\prime\prime}+b(x)y^{\prime}+c(x)y=0$ can be written as $\left[a(x)y^{\prime}\right]^{\prime}+c(x)y=0\sim\left[p(x)y^{\prime}\right]^{\prime}+q(x)y=0$

Does this make sense?

3. Yeah it does, thanks mate, I knew it wouldn't be very complicated.

4. Originally Posted by Chris L T521
Well, if you let $b(x)=a^{\prime}(x)$, it follows that $a(x)y^{\prime\prime}+a^{\prime}(x)y^{\prime}=\left[a(x)y^{\prime}\right]^{\prime}$ (def. of product rule)

So it follows that $a(x)y^{\prime\prime}+b(x)y^{\prime}+c(x)y=0$ can be written as $\left[a(x)y^{\prime}\right]^{\prime}+c(x)y=0\sim\left[p(x)y^{\prime}\right]^{\prime}+q(x)y=0$

Does this make sense?
That does not show what was asked. The original questions was to show that any second order differential equation could be put in that form. Chris L has only showed that you can write an equation of the form $a(x)y"+ a'(x)y'+ c(x)y= 0$ in that form. What if b(x) is not equal to a'(x)?

Just as with first order equations, try to find an "integrating factor", u(x). Look at (a(x)u(x)y')'= a(x)u(x)y"+ (a'(x)u(x)+ a(x)u'(x))y' and compare that with a(x)u(x)y"+ b(x)u(x)y'. In order that those be the same we must have a'(x)u(x)+ a(x)u'(x)= b(x)u(x) which is the same as the first order, separable, equation a(x)u'(x)= (b(x)- a'(x))u(x). Integrating $\frac{du}{u}= \frac{b(x)- a'(x)}{a(x)}$ gives the integrating factor, u(x). Multiplying both sides of the equation by that u(x) puts it in the form required.