# Thread: Almost there...

1. ## Almost there...

Hey, I got most of the way through this reduction of order problem, but then my brain gave out right at the end.

Given the equation $y''-4y'+4y=0$ and the solution $y_1=e^{2x}$, write the general solution.

I make the substitution $y_2(x)=u(x)y_1(x)$ and do all the proper work and get here:

$e^{2x}u''=0$ and my brain gives out. I know this is easy, How do I find u?

2. Originally Posted by VonNemo19
Hey, I got most of the way through this reduction of order problem, but then my brain gave out right at the end.

Given the equation $y''-4y'+4y=0$ and the solution $y_1=e^{2x}$, write the general solution.

I make the substitution $y_2(x)=u(x)y_1(x)$ and do all the proper work and get here:

$e^{2x}u''=0$ and my brain gives out. I know this is easy, How do I find u?
Well since $e^{2x} \ne 0,\forall x$ divide both sides by it to get

$u''=0 \iff \frac{d}{dx}u'=0 \iff u'=c \iff u=cx+d$ now choose $c=1,d=0$ to get $u=x$

3. Of course, if you solve the Characteristic equation and realise you get a repeated root $\displaystyle m = 2$, then you immediately know that the solution of the DE is $\displaystyle y = C_1e^{2x} + C_2x\,e^{2x}$...

4. Originally Posted by Prove It
Of course, if you solve the Characteristic equation and realise you get a repeated root $\displaystyle m = 2$, then you immediately know that the solution of the DE is $\displaystyle y = C_1e^{2x} + C_2x\,e^{2x}$...
Right but I think the point of the questions is to see how to get a 2nd linerly inpendant vector when you have a repeated root to the charateristic equation. Without this derivation how would we know what the 2nd vector is? Plus a great gereralization of this is variation of parameters!