Homogeneous ODE with variable coefficient

[Machete_B]

Hello everyone!

I'm trying to solve 2 questions for my assignment on homogeneous ODEs

I can solve ODE's with variation of parameters and with the method of undetermined coefficients, but these 2 methods seem useless when the coefficients are not constant :

(x^2)y" - 2xy' -54y = 0

and

(x^2)y" - 3xy' + 4y = 0


I'm not asking anyone to solve it for me, but just to tell me what method I have to use, and I'll find on my book or on the net.

P.S : I know that you can use power series method, but I think there is another one since we are learning the power series method right now, with a brand new homework just on power series!

Thanks guys

 

[HallsofIvy]

Those are "Euler type" or "equi-potential" equation. You can look for a solution of the form "y= x^r" just as, with equations with constant coefficients, you try "y= e^rx". Of course, in either situation, there are other kinds of solutions.

That works because making the substitution u= ln(x) converts a Euler-type equation to an equation with constant coefficients. For example, letting u= ln x, then dy/dx= (dy/du)(du/dx)= (1/x)dy/du. d²y/d^x= (d/dx)((1/x) dy/du)= (-1/x²)dy/dx+ (1/x)((1/x)d²y/du²)= (1/x)²d²y/du²- (1/x²)dy/du.

So x²y"- 3xy'+ 4y= 0 becomes x²(1/x²)d²/du²- (1/x²)dy/dx)- 3x((1/x)dy/du)+ 4y= d²y/du²- 4 dy/du+ 4y= 0. Solve that differential equation for y as a function of u and then replace u by ln(x).

 

[Machete_B]

Thank you very much :D

I'm trying it right now

 

[Machete_B]

damn it!

all terms cancel out when I got with

y(u) = Ae^2u+ Bue^2u

I'm going to try again tomorrow.

thanks again!

 

[HallsofIvy]

damn it!

all terms cancel out when I got with

y(u) = Ae^2u+ Bue^2u

I'm going to try again tomorrow.

thanks again!

What?? Yes, of course, "all terms cancel out"! They are supposed to- they make the equation equal to 0! That proves that y(u) satisfies y"- 4y'+4y= 0.

And with u= ln x, y(x)= Ax²+ B x² ln(x). Put that into the original equation. "All terms cancel out" is what you want.

 

[Machete_B]

haha lol thanks :P

amd another question : we always use u = ln (x) for Euler's type ODE ?

 

[HallsofIvy]

Yes, that gives a one to one relation between the set of all Euler-type equations and the set of all equations with constant coefficients. A solution of one automatically gives a solution of the other.