Solving Differential Equations: the Euler D.E. and Linear D.E.s

[Funzies]

Hello guys, I am new here to the forum and I was wondering if you could help me with some trouble I am having with differential equations. For your information: I am a second year applied physics student from the Netherlands and I loveee everything about physics and mathematics!

The problem concerns the Euler D.E.: x^2*y'' + x*y' + y = f(x). Let's assume for the moment that f(x) = 0, so that the equation is homogenous. I know how to solve it: by substituting y = x^r and calculating the two r's. I also know about the problem af the double root: the solution becomes y(x) = A*x^r1 + B*ln(x)*x^r1.
Well, this last 'solution' concerning the double root was kind of dropped during college: they just told us how to solve it. I am wondering where the ln(x) comes from. I have the same problem with linear D.E.'s with constant coefficients: y'' + y' + y = 0, if you substitute y=exp(kx) you get two k's unless there's a double root and then the solution becomes y(x) = A*exp(k1x) + B*x*exp(k1x). In this case I am wondering where the 'x' comes from. Can someone provide me with a proof for these solutions?
Thanks in advance!

EDIT: by proof I do not mean the proof that they ARE solutions. That's easy checkable: I just fill them in in the D.E.'s. I mean where it comes from, to help me understand it better.

 

[HallsofIvy]

The derivative of xe^{kx} is e^{kx}+ kxe^{kx} and its second derivative is 2e^{kx}+ k^2xe^{kx}, etc. In other words, every derivative of xe^{kx} is a linear combination of e^{kx} and xe^{kx}. The fact that k is a root of the characteristic equation means that [math]e^{kx}[/math] satisifies the differential equation. And putting xe^{kx} into the equation, together with the fact that the equation is linear, allows you to factor "x" out: D(xe^{kx})= xD(e^{kx})= 0 0 because e^{kx} satisfies the equation. And, of course, since the equation is linear, any linear combination of solutions is a solution.

You can extend that to Euler-type equations by the observation that substituting t= ln(x) into an Euler-type equation in the variable x changes it into an equation with constant coefficients with variable t.

 

[Funzies]

Thanks for your replay, but I still have some questions remaining:

And putting xe^{kx} into the equation, together with the fact that the equation is linear, allows you to factor "x" out: D(xe^{kx})= xD(e^{kx})= 0 0 because e^{kx} satisfies the equation. And, of course, since the equation is linear, any linear combination of solutions is a solution.

I do not understand why you can assume: D(xe^{kx})= xD(e^{kx})= 0
Ok, if x*exp(kx) is a solution to the equation, of course x*exp(kx) + exp(kx) is still a solution, but I do not see on what grounds you concluded that x*exp(kx) also satisfies the equation.

You can extend that to Euler-type equations by the observation that substituting t= ln(x) into an Euler-type equation in the variable x changes it into an equation with constant coefficients with variable t.

You are right, by substituting t=ln(x) I get the equation: y''(t) + y'(t) + y(t), but how does that make me conclude that the solution to the homogenous equation is y(x) = x^(r1) + ln(x)*x^(r1)? Do you just solve the D.E. for the variable t, conclude that there's a double root and then solve y(t) = exp(r1*t) + t*exp(r1*t) and then substitue t = ln(x) back? And what about the solution interval, doesn't that get compremised by substituting and ln, which is not defined for x<0?