Thread: 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?

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 , it follows that (def. of product rule)

So it follows that can be written as

Does this make sense?

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

Originally Posted by Chris L T521

Well, if you let , it follows that (def. of product rule)

So it follows that can be written as

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 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 gives the integrating factor, u(x). Multiplying both sides of the equation by that u(x) puts it in the form required.