# Non-Homogeneous DQ's

• July 8th 2008, 12:40 PM
ccdelia7
Non-Homogeneous DQ's
Hello, I am having some trouble with these second and third order non-homogeneous dq's.

I can identify a Y compliment, but I freeze up when I have to distinguish the Y particular.

ex 1) Solve:

y'' - y' = 11x + 1

Ycompliment -

(D^2 - D)y = 0
D(D-1)y = 0

Roots: D= 0, 1

Yc = c1 *e^0x + c2 *e^x = 1 + c2 *e^x

Now this is where I get lost... I can ID Yc but not Yp. Please help! Thanks!
• July 8th 2008, 01:40 PM
bobak
For the particular integral try $y = ax^2 + bx$

Bobak
• July 8th 2008, 03:00 PM
ccdelia7
Thanks for the help!

One question, how do you arrive at this solution? If you could explain further, the process, I would greatly appreciate it!

I understand that it's more of a guessing thing, but how should I make the best guess?

Thanks again, Chris
• July 8th 2008, 03:33 PM
bobak
Quote:

Originally Posted by ccdelia7
Thanks for the help!

One question, how do you arrive at this solution? If you could explain further, the process, I would greatly appreciate it!

I understand that it's more of a guessing thing, but how should I make the best guess?

Thanks again, Chris

It is not guessing at all. From the differential equation it should be obvious that the particular integral is a polynomial, in which case all you need to determine is the order of the polynomial, Can you explain why such a polynomail for this differential equation must have a zero ceofficient on any terms higher than $x^2$?

Bobak
• July 8th 2008, 05:06 PM
topsquark
Quote:

Originally Posted by bobak
For the particular integral try $y = ax^2 + bx$

Bobak

Actually you can get away with $y = ax^2 + bx + c$

-Dan
• July 8th 2008, 05:10 PM
bobak
Quote:

Originally Posted by topsquark
Actually you can get away with $y = ax^2 + bx + c$

-Dan

No point there is a constant term in the complementary solution.

Bobak