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

**kpizle** · April 7th 2010, 06:10 AM

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

**dwsmith** · April 7th 2010, 06:54 PM

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

**kpizle** · April 7th 2010, 07:21 PM

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?

**dwsmith** · April 7th 2010, 07:23 PM

You are correct it is supposed to be -1.

**kpizle** · April 7th 2010, 07:29 PM

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.

Thanks again for your reply. It was very helpful!

**dwsmith** · April 7th 2010, 07:32 PM

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.

**dwsmith** · April 7th 2010, 07:34 PM

$D^{3}$ 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.

**dwsmith** · April 7th 2010, 08:09 PM

I finished the solution for you.

**kpizle** · April 7th 2010, 08:17 PM

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...

**kpizle** · April 7th 2010, 08:31 PM

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?

**kpizle** · April 8th 2010, 12:59 AM

Found the error. Wrong general solution...needed

$r^2 - 4 = 0$

instead of

$r^2 - 4r = 0$

Yp was correct.

Correct general equation

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

and everything works.

**dwsmith** · April 8th 2010, 06:13 AM

Yup B is correct.

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