# Y1 = xe^(rx) and y2 = e^(rx): how y1 was discovered?

1. May 16, 2012

### gnardog777

hey, can someone explain to me how mathematicians found out that for second order homogenous ODEs with constant coefficents, when for the auxiliary equation, b^2 - 4ac = 0, xe^(rx) is another general solution?

I've seen it proven, but I don't understand why

**EDIT:*** I just realized I may have phrased this in an awkward/weird way. In other words, how did they find out that it works? I mean, I know when you plug it in to the regular equation, it works, but there must be something more to it (at least that's what I imagine, I could very well be wrong though).

Last edited: May 16, 2012
2. May 16, 2012

### HallsofIvy

The derivation comes from using the technique for reducing the order of a differential equation if you know one solution.

I'm sure you know that if you know one solution, say $x_0$, to a polynomial equation, then dividing by $x- x_0$ gives an equation of degree one less to solve for the other roots. Similarly, if $y_0(x)$ is a solution to a linear differential equation, then letting $y(x)= y_0(x)v(x)$ gives a linear differential equation of order one less to solve for other, independent, solutions.

In this case, if r is a double root of the characteristic equation then the characteristic equation must have the form $(x- r)^2= x^2- 2rx+ r^2= 0$ and that means that at least part of the equation must be of the form $y''- 2ry'+ r^2y= 0$. If we let $y(x)= v(x)e^{rx}$ then $y'= v'e^{rx}+ rve^{rx}$ and $y''= v''e^{rx}+ 2rv'e^{rx}+ ve^{rx}$. Putting that into the equation, $y''- 2ry'+ r^2y= v''e^{rx}+ 2rv'e^{rx}+ ve^{rx}- 2rv'e^{rx}- 2r^2ve^{rx}+ r^2ve^{rx}$$= v''e^{rx}= 0$.

The terms involving only v (and not v' and v'') cancel because, not differentiating v, we are treating it like a constant and, of course, since $e^{rx}$ satisfies the equation, so does any constant times $e^{rx}$. And all terms involving v' cancel because r is a double root. So we have $v''e^{rx}= 0$ and, since $e^{rx}$ is never 0, we can divide by it to get $v''= 0$. Integrating that once, $v'= C_1$, a constant, and, integrating again, $v(x)= C_1x+ C_2$. That is $y(x)= v(x)e^{rx}= C_1xe^{rx}+ C_2e^{rx}$.

Of course, the "$C_2e^{rx}$" we already had but now we have discovered that $C_1xe^{rx}$ is also a solution.

Last edited by a moderator: May 17, 2012
3. May 26, 2012

### algebrat

Some other leads, it might come from the very general complex series approach, or it might come from the jordan canonical form, a generalization of diagonalizing matrices. The result from jordan canonical form will be used more generally if your text treats higher order linear ODEs, and systems of ODEs. So your solution you are looking at is the special case in several different and more general methods.