# Thread: DE's and Laplace Transforms

1. ## DE's and Laplace Transforms

Hi all,
Just started an assignment and the first question was on Laplace Transforms. I tried to find a solution to the problem first by finding the general and particular solution . I did this as i haven't had a great deal to do with laplace and was seeing if i could find the solution useing methods i learnt last year (well suposed to have learnt) so i had an idea what the answer should look like as i attempted the question. Can someone please have a look over my work and tell me if my solution is correct (im not very confident with it).
Thanks alot
Elbarto

2. Originally Posted by elbarto
Hi all,
Just started an assignment and the first question was on Laplace Transforms. I tried to find a solution to the problem first by finding the general and particular solution . I did this as i haven't had a great deal to do with laplace and was seeing if i could find the solution useing methods i learnt last year (well suposed to have learnt) so i had an idea what the answer should look like as i attempted the question. Can someone please have a look over my work and tell me if my solution is correct (im not very confident with it).
Thanks alot
Elbarto
Your work looks fine to me

there was no need to try $\displaystyle y = ae^{-4t}$ as a particular solution though, since you already found that was one of the forms to the homogenious solution

3. Ok thanks alot for that Jhevon.
So i am looking for an answer that looks like y(t)=e^t-te^-4t. I will atempt the question useing laplace transforms tonight if you could take another look then? Thanks again for your help, I appreciate it.
Regards Elbarto

4. Jhevon,

Thank You very much for the offer but i will not hold you up. I attempted it roughly with laplace and after a short battle with partial fractions it came out the same supprisingly.

Thanks once again, It makes it so much easier haveing a place that people are willing to look over your working out.

Regards Elbarto

Ps. now onto fourier serries, no doubt you guys will hear from me again.

5. Originally Posted by elbarto
Jhevon,

Thank You very much for the offer but i will not hold you up. I attempted it roughly with laplace and after a short battle with partial fractions it came out the same supprisingly.

Thanks once again, It makes it so much easier haveing a place that people are willing to look over your working out.

Regards Elbarto

Ps. now onto fourier serries, no doubt you guys will hear from me again.
Being bored to death at the moment, I typed up a solution, so even though you found the answer (which I knew you would) I would like to post this anyway, since I took the time to type it.

$\displaystyle y'' + 3y' -4y = 5e^{-4t}$ ................$\displaystyle y(0) = 1 \mbox{, } y'(0) = 0$

(a)

I will write $\displaystyle Y$ to mean $\displaystyle Y(s)$

Let $\displaystyle \mathcal {L} y = Y$

$\displaystyle \Rightarrow \mathcal {L} y' = -y(0) + s \mathcal {L} y = -1 + sY$

$\displaystyle \Rightarrow \mathcal {L} y'' = -y'(0) + s \mathcal {L} y' = -s + s^2 Y$

Also note: $\displaystyle \mathcal {L} 5e^{-4t} = \frac {5}{s + 4}$

$\displaystyle \Rightarrow \mathcal {L} y'' + 3 \mathcal {L} y' - 4 \mathcal {L} y = \frac {5}{s + 4}$

$\displaystyle \Rightarrow -s + s^2 Y - 3 + 3sY - 4Y = \frac {5}{s + 4}$

$\displaystyle \Rightarrow Y(s^2 + 3s - 4) - (s + 3) = \frac {5}{s + 4}$

$\displaystyle \Rightarrow Y(s + 4)(s - 1) = \frac {5}{s + 4} + s + 3$

$\displaystyle \Rightarrow Y = \frac {5}{(s + 4)^2 (s - 1)} + \frac {s + 3}{(s + 4)(s - 1)}$

as desired

(b)

$\displaystyle Y = \frac {5}{(s + 4)^2 (s - 1)} + \frac {s + 3}{(s + 4)(s - 1)}$

$\displaystyle \Rightarrow Y = \frac {5 + (s + 3)(s + 4)}{(s + 4)^2 (s - 1)} = \frac {s^2 + 7s + 17}{(s + 4)^2 (s - 1)}$

Now by the method of Partial Fractions:

$\displaystyle \frac {s^2 + 7s + 17}{(s + 4)^2 (s - 1)} = \frac {A}{s + 4} + \frac {B}{(s + 4)^2} + \frac {C}{s - 1}$

$\displaystyle \Rightarrow s^2 + 7s + 17 = A(s + 4)(s - 1) + B(s - 1) + C(s + 4)^2$

$\displaystyle \Rightarrow s^2 + 7s + 17 = (A + C)s^2 + (3A + B + 8C)s + (16C - 4A - B)$

By equating the coefficients of like powers of $\displaystyle s$, we obtain the system of equations:

$\displaystyle A + C = 1 ........................(1)$
$\displaystyle 3A + B + 8C = 7 .............(2)$
$\displaystyle -4A - B + 16C = 17 .......(3)$

Solving this, we get:

$\displaystyle A = 0$, $\displaystyle B = -1$, $\displaystyle C = 1$

And so we see, $\displaystyle Y = \frac {5}{(s + 4)^2 (s - 1)} + \frac {s + 3}{(s + 4)(s - 1)} = \frac {1}{s - 1} - \frac {1}{(s + 4)^2}$

$\displaystyle \Rightarrow y(t) = e^t - te^{-4t}$ by taking the inverse Laplace Transform...which is what you got before

6. I did the partial fractions in 2 parts, but i just looked at the question again and it says "Use partial fractions to invert this expression and find the solution y(t).". Does this mean that i should have done it the way you did.

my expression for Y(s)=
(-1/5)/(s+4) - 1/(s+4)^2 + (1/5)/(s-1) + (1/5)/(s+4) + (4/5)/(s-1)

answer comes out the same tho i am unsure about the "invert" in the question.

Thanks
Elbarto

7. Originally Posted by elbarto
I did the partial fractions in 2 parts, but i just looked at the question again and it says "Use partial fractions to invert this expression and find the solution y(t).". Does this mean that i should have done it the way you did.

my expression for Y(s)=
(-1/5)/(s+4) - 1/(s+4)^2 + (1/5)/(s-1) + (1/5)/(s+4) + (4/5)/(s-1)

answer comes out the same tho i am unsure about the "invert" in the question.

Thanks
Elbarto
If it works out there's nothing wrong with it. i think that would be more work though. was it? how long did you take to find the partial fractions doing i tthat way compared to the way i did it?

the invert part is just talking about the last step we did. that is, going from the partial fractions back to functions of t

8. Your way was alot shorter. It took me probally about half a page more to come up with the partial fractions because i did it in 2 parts and also used the other method (is it called the coverup method or something like that?). I will try practice the method you have used because it looks like it will save me a bit of time in the future.

Thanks
Elbarto

9. Originally Posted by elbarto
Your way was alot shorter. It took me probally about half a page more to come up with the partial fractions because i did it in 2 parts and also used the other method (is it called the coverup method or something like that?). I will try practice the method you have used because it looks like it will save me a bit of time in the future.

Thanks
Elbarto
well, working with one fraction is bound to be easier than working with two, especally when doing a computationally intensive process like partial fractions. i did, though, skip a few steps that showed how i simplified some stuff. and i didnt show the work for solving the system of equations since i know you would have no problem with that. but whatever makes you comfortable.