1. ## Differential equation theory

This page is about solving a second-order linear differential equation with real and coincident roots. The equation being solved was given earlier but it is almost there halfway down following the word "thus". It is homogeneous, i.e. it equals 0. I have followed the writing down to the line following "So". Can someone explain the rest?

2. ## Re: Differential equation theory

Originally Posted by Stuck Man
I have followed the writing down to the line following "So".
... so, down to but not including, "giving $y = Bxe^{mx}$ as a solution" ...?

If so, the point to get is that

$y = Bxe^{mx}$ (where m happens to be the repeated root of $az^2 + bz + c = 0$ so that $m = \tfrac{-b}{2a}$ and also $c = \tfrac{b^2}{4a})$

has been shown to be a solution to the differential equation because it has been shown to produce the equation. I.e. what has been shown is that setting

$y = Bxe^{mx}$ (where... )

makes

$ay'' + by' + c$

come to zero.

Maybe the 'Now we know' line would be slightly clearer as:

"Now we were assuming that the m in the exponent is equal to $\tfrac{-b}{2a}$ which, with the help of $c = \tfrac{b^2}{4a}$, is going to make both of $am^2 +bm + c$ and $2am + b$ come to zero.

Or maybe not! But you could also dispense with m in the first place and set,

$y = Bxe^{\frac{-b}{2a}x}$

... as here: Pauls Online Notes : Differential Equations - Repeated Roots

Otherwise, if the issue is just the putting together of this solution with the previous one at the top of the page, notice that (A + B) in the original version of the first solution has been replaced by A alone, and B in the second solution has nothing to do with B at the top of the page.

3. ## Re: Differential equation theory

Another way of seeing that is to use the "reduction of order" method. If we have an nth order d.e, and know one solution, u(x), then looking for a solution of the form y(x)= xu(x).

For example, if a second order differential equation has characteristice equation with at double root, that is, of the form $(r- a)^2= r^2- 2ar+ a^2= 0$, then the differential equation is of the form $y''- 2ay'+ a^2y$. Knowing that [tex]e^{ax}[/itex] is a solution ( $(e^{ax})''- 2a(e^{ax})'+ a^2e^{ax}= a^2e^{ax}- 2a^2e^{ax}+ a^2e^{ax}= (a^2- 2a^2+ a^2)e^{ax}= 0$), we try a solution of the form y= $e^{ax}u(x)$. Then $y'= ae^{ax}u+ e^{ax}u'$ and $a^2e^{ax}u+ 2ae^"{ax}u'+ e^{ax}u''$. Putting those into the equation, [tex]a^2e^{ax}u+ 2ae^{ax}u'+ e^{ax}u''- 2a(ae^{ax}u+ e^{ax}u')+ a^2(e^{ax}u)= e^{ax}(u''+ (2a- 2a)u'+ (a^2- 2a^2+ a^2)u)= e^{ax}u''= 0[tex]. The terms involving u only have canceled because $e^{ax}$ is a solution to the differential equation. The terms involving u' only have canceled because a is a double root. Since $e^{ax}$ is never 0, we must have u''= 0. Integrating twice, we have u(x)= C+ Dx so that $y= (C+ Dx)e^{ax}= Ce^{ax}+ Dxe^{ax}$. The " $Ce^{ax}$" we already had but now we know that $xe^{ax}$ is also a solution.