reduction of order...

**slapmaxwell1** · February 28th, 2011, 12:09 AM

the indicated y1 is a solution of the given differential equation. find a second solution y2(x).

so my problem is: y'' -4y' + 4y = 0 , y1 = e^(2x)

if y=u(x)y1(x) then y= u(x)e^(2x).

after finding y'' and y' i substituted back into the original equation and got,

u''e^(2x) = 0. and this is where i am stuck. I am not sure what to do.

i know that u'' must be 0.

my book goes on to say that, u= c1x + c2? and that c1= 1 and c2=0 and that
y2= xe^(2x)?

a little help please....

thanks in advance.

**Prove It** · February 28th, 2011, 12:11 AM

If $\displaystyle \frac{d^2u}{dx^2} = 0$ , integrate once to get $\displaystyle \frac{du}{dx}$ , integrate again to get $\displaystyle u$ .

**slapmaxwell1** · February 28th, 2011, 12:17 AM

ok so with that being said u'=c1 and u would be c1x + c2

ok i see that now thanks...ok but how did the book know to make c1 = 1 and c2 = 0?

**Prove It** · February 28th, 2011, 01:02 AM

Were you given any initial conditions or boundary conditions?

**slapmaxwell1** · February 28th, 2011, 01:20 AM

no i wasnt

**Prove It** · February 28th, 2011, 01:55 AM

In that case you will need to substitute $\displaystyle y = e^{2x} + (c_1x + c_2)e^{2x}$ into your DE, simplify, then equate the coefficients of like powers of $\displaystyle x$ ...

**Danny** · February 28th, 2011, 05:19 AM

What you're looking for is the second independent solution. If you keep your two constants as they are you have

$y = \left(c_1 + c_2x\right)e^{2x} = c_1 e^{2x} + c_2 x e^{2x}$

The first is your first solution ( $c_1 = 1, c_2 = 0$ ) and the second the second independent solution ( $c_1 = 0, c_2 = 1$ )

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