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
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
From $\displaystyle m^{4}+2m^{2}+1=0$, we find $\displaystyle m_{1}=m_{3}=i$ and $\displaystyle m_{2}=m_{4}=-i$
Then $\displaystyle y_{c}=C_{1}cos(x)+C_{2}sin(x)+C_{3}xcos(x)+C_{4}xs in(x)$
$\displaystyle y_{p}=Ax^{2}+Bx+C$
Sub into the DE as usual and get:
$\displaystyle A=1, \;\ B=-2, \;\ 4A+C=1$
Then $\displaystyle A=1, \;\ B=-2, \;\ C=-3$
$\displaystyle y_{p}=x^{2}-2x-3$
Therefore,
$\displaystyle y=C_{1}cos(x)+C_{2}sin(x)+C_{3}xcos(x)+C_{4}xsin(x )+x^{2}-2x-3$
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.
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.
$\displaystyle m^4 + 2m^2 + 1 = 0$
This is a "biquadratic" equation. Let $\displaystyle a = m^2$.
Then
$\displaystyle a^2 + 2a + 1 = 0$
$\displaystyle (a + 1)^2 = 0$
$\displaystyle a + 1 = 0$
Thus
$\displaystyle m^2 = -1$
Thus
$\displaystyle m = \pm i$
Both of these roots are doubled since the discriminant of the a equation is 0.
-Dan