# Thread: Help with nonhomogeneous second order Diff. Eq.

1. ## Help with nonhomogeneous second order Diff. Eq.

Hello,

I am currently working on solving nonhomogeneous differential equations using the method of undetermined coefficients. Im doing well solving problems, but I have come across one that has puzzled me, possibly due to its simplicity.

The initial value problem is:
$\displaystyle y'' + y' - 2y = 2t, ~ ~ y(0) = 0, ~ ~ y'(0) = 1$

I have found the complementary general solution,
$\displaystyle y(t) = c_1e^t + c_2e^{-2t}$
but before I can solve for the constants, I must account for the nonhomogeneous term.

I make my first assumption that the term is of the form:
$\displaystyle Y(t) = At$
therefore
$\displaystyle Y'(t) = A,$
$\displaystyle Y''(t) = 0$
$\displaystyle A - 2At = 2t$
Obviously this is not a solution so I try the only other thing I know to try.
$\displaystyle Y(t) = At^2$
$\displaystyle \therefore Y'(t) = 2At,$
$\displaystyle \ Y''(t) = 2A$
so
$\displaystyle 2A + 2At - 2At^2 = 2t$
This isnt going to work either and now Im out of ideas. Any help is greatly appreciated.

2. Originally Posted by sixstringartist
Hello,

I am currently working on solving nonhomogeneous differential equations using the method of undetermined coefficients. Im doing well solving problems, but I have come across one that has puzzled me, possibly due to its simplicity.

The initial value problem is:
$\displaystyle y'' + y' - 2y = 2t, ~ ~ y(0) = 0, ~ ~ y'(0) = 1$

I have found the complementary general solution,
$\displaystyle y(t) = c_1e^t + c_2e^{-2t}$
but before I can solve for the constants, I must account for the nonhomogeneous term.

I make my first assumption that the term is of the form:
$\displaystyle Y(t) = At$
therefore
$\displaystyle Y'(t) = A,$
$\displaystyle Y''(t) = 0$
$\displaystyle A - 2At = 2t$
Obviously this is not a solution so I try the only other thing I know to try.
$\displaystyle Y(t) = At^2$
$\displaystyle \therefore Y'(t) = 2At,$
$\displaystyle \ Y''(t) = 2A$
so
$\displaystyle 2A + 2At - 2At^2 = 2t$
This isnt going to work either and now Im out of ideas. Any help is greatly appreciated.
Try: $\displaystyle Y(t)=At+B$

RonL

3. $\displaystyle y'' + y' -2y = 2t, ~~ y(0) = 0, ~~ y'(0) = 1$
$\displaystyle Y(t) = At + B$
$\displaystyle Y'(t) = A$
$\displaystyle Y''(t) = 0$
$\displaystyle A - 2At - 2B = 2t$
$\displaystyle -2A = 2, ~~ A-2B = 0$
$\displaystyle A = -1, ~~ B = \frac {-1}{2}$

And that is the correct coefficients. Thank you very much.