# Thread: [SOLVED] Non-homogenous linear ode particular solution help.

1. ## [SOLVED] Non-homogenous linear ode particular solution help.

Hey all.

I am having a lot of trouble finding particular solutions.

I have three problems:

A) $y'' + y' = 6t^2$

B) $y'' - 4y = sin(2t)$

C) $y'' + 4y' + 5y = 5t + e^-t$

I am trying to find the particular solutions using the method of undetermined coefficients.

For A), I found the general solution to be
$Ce^-t + D$

and the particular solution to be

$(2t^4)/(2+t)$

For B), I found the general solution to be
$C + De^(4*t)$

and the particular solution to be

$(-1/8)sin(2t)$

For C), I found the general solution to be
$Ce^(-2*t+i*t) + De^(-2*t-i*t)$ (everything in the parenthesis after e is what e is raised to.

and the particular solution to be

$t + (1/2)e^(-t)$

None of these check out. I am flustered and frustrated with this and don't know what else to do. PLEASE HELP!!!

Thank you

2. First, you need to solve their homogeneous equivalent so you can come up with the complement function.

For a, $m^2+m=0$ is the homogeneous equivalent.

$m(m+1)=0$

$y_{c}=C_{1}+C_{2}e^{-t}$

Now what annihilates $6t^{2}$ and that is $D^{3}$

$m^{3}*m*(m+1)$

$y_{p}=tC_{3}+t^{2}C_{4}+t^{3}C_{5}$

$y'_{p}=C_{3}+2tC_{4}+3t^{2}C_{5}$

$y''_{p}=2C_{4}+6tC_{5}$

$2C_{4}+6tC_{5}+C_{3}+2tC_{4}+3t^{2}C_{5}=6t^{2}$

Now combing and replace Cs with A,Bs.... and solve for the coefficients.

$2B+6tC+A+2tB+3t^{2}C=6t^{2}$

$A+2B+t(2B+6C)+3t^{2}C$

$\left[ \begin{array}{c c c c}
1 & 2 & 0 & 0 \\
0 & 2 & 6 & 0 \\
0 & 0 & 3 & 6 \end{array} \right]$

Reduced row echelon of our augmented matrix will solve for A B and C
$\left[ \begin{array}{c c c c}
1 & 0 & 0 & 12 \\
0 & 1 & 0 & -6 \\
0 & 0 & 1 & 2 \end{array} \right]$

$A=C_{3}=12;
B=C_{4}=-6;
C=C_{5}=2$

Answer is: $y=C_{1}+C_{2}e^{-t}+12t+-6t^{2}+2t^{3}$

3. Originally Posted by dwsmith
First, you need to solve their homogeneous equivalent so you can come up with the complement function.

For a, $m^2+m=0$ is the homogeneous equivalent.

$m(m+1)=0$

$y_{c}=C_{1}+C_{2}e^{t}$
Why is it $y_{c}=C_{1}+C_{2}e^{t}$ the general solution to the homogenous, and not
$y_{c}=C_{1}+C_{2}e^{-t}$?

I thought you solved for m (0 and -1 in this case), got two distinct answers a and b and you could plug into
$y_{c}=C_{1}e^{at}+C_{2}e^{bt}$

Have I missed something, or was it a typo?

Also, if not too much trouble, could you elaborate on how you found $
y_{p}=tC_{3}+t^{2}C_{4}+t^{3}C_{5}
$
would annihilate the 6t^2? I sat and tried a bunch of different things before I gave up and thought it to be hopeless. What is your strategy/technique to attack finding these particular solutions?

4. You are correct it is supposed to be -1.

5. Originally Posted by dwsmith
You are correct it is supposed to be -1.
Radical...okay, so I was on the right track (I realize I butchered the general solution to the homogenous in C pretty badly. Another go with the CORRECT formula will probably yield better results).

Thank you very much.

I don't know if you saw the edit to my reply in time to see my last minute question about strategy/technique, so I shall ask again:

What is your strategy in finding these particular solutions? I tried a bunch of stuff that kept NOT working that I thought I was doing it completely wrong or had the wrong idea or something.

6. I edited my original solution because I wasn't paying attention. I didn't finish it but all you need to do is group the like coefficients and factor them out.

7. $D^{3}$ annihilates square values. What we do to one side we need to do to the other. Since I am using m as my factors, we need to multiple the left which was m(m+1) by m^3.

Now we have 3 repeated real roots 0. The other 2 are solutions belong to yc.

8. I finished the solution for you.

9. Originally Posted by dwsmith
$D^{3}$ annihilates square values. What we do to one side we need to do to the other. Since I am using m as my factors, we need to multiple the left which was m(m+1) by m^3.

Now we have 3 repeated real roots 0. The other 2 are solutions belong to yc.
Ok. I was able to get that the particular solution for a is

$12t - 6t^2 + 2t^3$ which checks out. Thank you kindly.

But I'm having trouble seeing how you went from $D^3$
$m^3*m*(m+1)$ to $At + Bt^2 + Ct^3$

Well...I hadn't seen that you had finished it for me until I had posted this. Heh. Thank you.

I don't see what the ms did to get to the generic particular equation...

10. For B, I have taken steps back.

General to the homogenous:

$r^2 - 4r = 0$
r=0, 4

so... $Yc = C_1 + C_2e^{4t}$

I took
$Yp = Asin(2t)$
$Y'p = 2Acos(2t)$
$Y''p = -4Asin(2t)$

Substitution:

( $-4Asin(2t)$) - 4 ( $Asin(2t)$) = $-8Asin(2t)$

then, I did (more substitution)

$-8Asin(2t)$ = sin(2t)

and found $A = -1/8$

so, $Yp = -(1/8)sin(2t)$

But when I try to plug the general solution

$y = C_1 + C_2e^{4t} - (sin(2t)/8)$

It doesn't work. Where did I go wrong?

11. Found the error. Wrong general solution...needed

$r^2 - 4 = 0$

$r^2 - 4r = 0$

Yp was correct.

Correct general equation

$y = C_1e^{2t} + C_2e^{-2t} - (sin(2t)/8)$

and everything works.

12. Yup B is correct.