one more undetermined coeffecient

**neven87** · July 2nd 2007, 11:26 AM

the problem is y(quad prime)+2y(double prime) + y = (x-1)^2

I keep tackling this problem and coming up with nothing anyone help me here?

its suppose to come out in the form of y = c1cos(x)+c2 sin(x) +c3 xcos(x)+c4sin(x) +x^2-2x-3

**galactus** · July 2nd 2007, 11:42 AM

From $m^{4}+2m^{2}+1=0$ , we find $m_{1}=m_{3}=i$ and $m_{2}=m_{4}=-i$

Then $y_{c}=C_{1}cos(x)+C_{2}sin(x)+C_{3}xcos(x)+C_{4}xs in(x)$

$y_{p}=Ax^{2}+Bx+C$

Sub into the DE as usual and get:

$A=1, \;\ B=-2, \;\ 4A+C=1$

Then $A=1, \;\ B=-2, \;\ C=-3$

$y_{p}=x^{2}-2x-3$

Therefore,

$y=C_{1}cos(x)+C_{2}sin(x)+C_{3}xcos(x)+C_{4}xsin(x )+x^{2}-2x-3$

Your troubles are probably just coming from the algebra involved after you sub back into the DE?. That's the best place to go astray.

**neven87** · July 2nd 2007, 10:17 PM

Thanks agian man, im just getting back to college and my algerba is hazy at best. I ussally start second guessing some steps i take then i forget what step it was by the time i get there.

Also can you show me the factoring steps on that ? fourth power polynomials is something that is slipping my mind.

**topsquark** · July 3rd 2007, 02:34 AM

Originally Posted by neven87

Thanks agian man, im just getting back to college and my algerba is hazy at best. I ussally start second guessing some steps i take then i forget what step it was by the time i get there.

Also can you show me the factoring steps on that ? fourth power polynomials is something that is slipping my mind.

$m^4 + 2m^2 + 1 = 0$
This is a "biquadratic" equation. Let $a = m^2$ .

Then
$a^2 + 2a + 1 = 0$

$(a + 1)^2 = 0$

$a + 1 = 0$

Thus
$m^2 = -1$

Thus
$m = \pm i$

Both of these roots are doubled since the discriminant of the a equation is 0.

-Dan

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