# [SOLVED] Can't find particular solution of nonhomogenous linear ode

• Apr 8th 2010, 01:08 AM
kpizle
[SOLVED] Can't find particular solution of nonhomogenous linear ode
My problem:

$\displaystyle y'' + 4y' + 5y = 5t + e^{-t}$

I found the general solution of the homogenous equation

$\displaystyle r^2 + 4r + 5 = 0$

to be

$\displaystyle Y_c = C_1e^{-2t}cos(t) + C_2e^{-2t}sin(t)$

since

$\displaystyle r = -2 + i, -2 - i$ (complex)

Now, I have tried all of the following as particular solutions and all have come up 'no solution'

$\displaystyle Be^{-t}$
$\displaystyle Bte^{-t}$
$\displaystyle A + Be^{-t}$
$\displaystyle A + Bte^{-t}$
$\displaystyle At + Be^{-t}$
$\displaystyle At + Bte^{-t}$
$\displaystyle At^2 + Be^{-t}$
$\displaystyle At^2 + Bte^{-t}$
$\displaystyle At^2 + Bt^2e^{-t}$
$\displaystyle At^3 + Be^{-t}$
$\displaystyle At^3 + Bte^{-t}$
$\displaystyle At^3 + Bte^{-t}$
$\displaystyle At^3 + Ct^2 + Dt + Be^{-t}$

All of them come up with conflicting values (A will equal 0 in one spot and need to be a value in another).

Can anyone tell what I am doing wrong?

thanks
• Apr 8th 2010, 01:29 AM
Prove It

Now for the particular solution:

$\displaystyle y_p = A + Bt + De^{-t}$

$\displaystyle y_p' = B - De^{-t}$

$\displaystyle y_p'' = De^{-t}$.

Therefore

$\displaystyle y'' + 4y' + 5y = De^{-t} + 4(B - De^{-t}) + 5(A + Bt + De^{-t})$

$\displaystyle = De^{-t} + 4B - 4De^{-t} + 5A + 5Bt + 5De^{-t}$

$\displaystyle = 5A + 4B + 5Bt + 2De^{-t}$.

Since this is equal to $\displaystyle 5t + e^{-t}$ that means

$\displaystyle 5A + 4B = 0, 5B = 5, 2D = 1$

Solving simultaneously gives

$\displaystyle A = -\frac{4}{5}, B = 1, D = \frac{1}{2}$.

So $\displaystyle y_p = -\frac{4}{5} + t + \frac{1}{2}e^{-t}$.

So finally, the general solution is

$\displaystyle y_g = y_c + y_p$

$\displaystyle = C_1e^{-2t}\cos{t} + C_2e^{-2t}\sin{t} - \frac{4}{5} + t + \frac{1}{2}e^{-t}$.
• Apr 8th 2010, 01:35 AM
kpizle
Quote:

Originally Posted by Prove It

Now for the particular solution:

$\displaystyle y_p = A + Bt + De^{-t}$

$\displaystyle y_p' = B - De^{-t}$

$\displaystyle y_p'' = De^{-t}$.

I was so close! I need to be a little more careful with my guesses and more methodical in my choices.

Using the method of undetermined coefficients was presented to me in a guess & check manner. Is there a better way? This is too tedious and the longer it takes to find a solution, the more my confidence in my work subsides.

• Apr 8th 2010, 01:42 AM
Prove It
The undetermined coefficients methods is exactly that - guess and check.

If you have a polynomial, you would expect your particular solution to have a polynomial of the same degree. Of course, that means you need coefficients for each power and a constant.

If you have a sinusoidal function, you would expect your particular solution to be a sum of a sine and cosine function (with the same multiple angle).

If you have an exponential function, you would expect your particular solution to have an exponential function of the same power.

If you have any sum of the above types of functions, your particular solution will have a sum of these types of functions.

If you have a function that already appears in your complementary solution, you will need to make a guess of that function multiplied by the independent variable.

I suggest you have a read of http://www.mathhelpforum.com/math-he...-tutorial.html
as it explains these in more detail, and also gives other methods of solving second order DEs (Annihilator Method and Variation of Parameters).