Partial fraction decomposition with Laplace transformation in ODE

[arhzz]

Homework Statement

Solve the ODE

Relevant Equations

Laplace Transformation,and Partial Fraction Decomposition

Hello!

Im having some trouble with solving ODE's using Laplace transformation,specifically ODE's that require partial fraction decomposition.Now I know how to do partial fraction decomposition,and have done it many times on standard polynoms but here some things just are not clear to me.For example;

$$x'' +25x = cosh(t) $$ with inital values x(0) = 0 and x'(0) = 5.

Now I've transformed it using laplace and pluged in the initial values and this is what I get;

$$ s^2 X(s) - 5 + 25X(s) = \frac{s}{s^2-1} $$

Now this should be correct (so it says in the solution sheet) Now I've tried to isolate X(s) and this is how it went.First I added a 5 to both sides and factored X(s) on the left

$$ X(s) (s^2+25) = \frac{s}{s^2-1} + 5 $$ and now divide with the bracket on the left should give me ;

$$ X(s) = \frac{s}{(s^2-1)(s^2+25} + \frac{5}{s^2+25} $$

Now I think its fairly obvious that the we need partial fraction decomposition here,and that the we have complex zeros. Now here comes the part that I do not undestand.They went ahead and did the following;

$$ X(s) = \frac{s}{(s+1)(s-1)(s^2+25} = \frac{A}{(s+1)} + \frac{B}{(s-1)} + \frac{Cs+D}{s^2+25} $$

I do not understand how they come up with this.I understand where the Cs+D comes from,that is the fact that we have complex roots but where did the 5 go? Why is the second fraction ignored? I would have went with an approach of

$$ X(s) =\frac{s}{(s+1)(s-1)(s^2+25} +\frac{5}{s^2+25} $$ why was the second fraction left out?

Or this example ODE;

$$ x''(t) +4x = e^{2t} $$ with initial values x(0) = 3 and x'(0) = 2 Now after Laplace Transformation and plugging in the initial values I get X(s) to be

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

Again we have complex roots and a simple root of 2.Now again I do not understand their approach here;

$$X(s) = \frac{1}{s-2} * \frac{1}{s^2+4} = \frac{A}{s-2} + \frac{Bs+C}{s^2+4} $$

What happened to the 3s-2?I am not understanding how to go around not having a polynom in the nummerator,finding the roots and the right approach(simple roots complex ect..) is not the problem. And another confusing part about this ODE is,when going ahead and calculation the values of A Bs and C we get them to be A = 1/8 B = -1/8 and C = -1/4

Now plugging in those values in the solutions looks like this;

$$ X(s) = 3 -\frac{1s}{8(s^+4)} + 1 - \frac{2}{8(s^2+4)}+\frac{1}{8(s-2)} $$

Now considering their approach I do not see how they get this.The last fraction makes sense they just plugged in the 1/8 but the first 2 are not clicking with me.Also I don't see that the value of C has been used at all.

Thanks for the help and excuse the long post

 

[vela]

Homework Statement:: Solve the ODE
Relevant Equations:: Laplace Transformation,and Partial Fraction Decomposition

Hello!

Im having some trouble with solving ODE's using Laplace transformation,specifically ODE's that require partial fraction decomposition.Now I know how to do partial fraction decomposition,and have done it many times on standard polynoms but here some things just are not clear to me.For example;

$$x'' +25x = cosh(t) $$ with inital values x(0) = 0 and x'(0) = 5.

Now I've transformed it using laplace and pluged in the initial values and this is what I get;

$$ s^2 X(s) - 5 + 25X(s) = \frac{s}{s^2-1} $$

Now this should be correct (so it says in the solution sheet) Now I've tried to isolate X(s) and this is how it went.First I added a 5 to both sides and factored X(s) on the left

$$ X(s) (s^2+25) = \frac{s}{s^2-1} + 5 $$ and now divide with the bracket on the left should give me ;

$$ X(s) = \frac{s}{(s^2-1)(s^2+25} + \frac{5}{s^2+25} $$

Now I think its fairly obvious that the we need partial fraction decomposition here,and that the we have complex zeros. Now here comes the part that I do not undestand.They went ahead and did the following;

$$ X(s) = \frac{s}{(s+1)(s-1)(s^2+25} = \frac{A}{(s+1)} + \frac{B}{(s-1)} + \frac{Cs+D}{s^2+25} $$

I do not understand how they come up with this.I understand where the Cs+D comes from,that is the fact that we have complex roots but where did the 5 go? Why is the second fraction ignored?

They shouldn't have said that's equal to $X(s)$. It's just the first term. It looks like they're just using partial fractions on the first term as the second term is already identifiable. Either that or they messed up.

 

[arhzz]

They shouldn't have said that's equal to $X(s)$. It's just the first term. It looks like they're just using partial fractions on the first term as the second term is already identifiable. Either that or they messed up.

Well they didnt mess up since the solutions are 100% correct.So it leaves the second option. What are they doing exactly? What is meant with partial fractions?

 

[vela]

How do you know the solutions are correct? In your second example, if there's really a constant term in $X(s)$, the time-domain solution should have a Dirac delta function in it. Is there one in their solution? If there is, their solution is incorrect. If there isn't one, then their expression for $X(s)$ is wrong.

 

[arhzz]

How do you know the solutions are correct? In your second example, if there's really a constant term in $X(s)$, the time-domain solution should have a Dirac delta function in it. Is there one in their solution? If there is, their solution is incorrect. If there isn't one, then their expression for $X(s)$ is wrong.

I've checked with wolfram alpha. And a few other online calculators. But it occurred to me that I made an error in inputing the second equation the X(s) form. I edited my post (sorry its quite late and it was a pretty long latex input)

 

[vela]

I think you're assuming that the work shown must be correct because the answer is correct. I don't think that's a good assumption. If you forget about the solution and just work the problems out on your own, do you arrive at the right answers?

 

[Office_Shredder]

If you just forget about the 5 in that one step, them the form is the same because D changes by 5 is all. As long as you then take that and pick coefficients that match the real function, you'll get the right answer at the end. My guess is that's what happened, they just forgot it in that one line but matched the unknowns to the real function

 

[arhzz]

Well I am not sure, but doing it my way (taking a diffrent approach) does not give me the correct solution. Since in the first example I take the nummerator to be s+5 and that has an impact in how the system of equations looks like at the end. I get that $s^0$ is 5 and they that $s^0$ is 0.

 

[Office_Shredder]

You should just post your solution, and we can tell you where your mistakes is.

 

[arhzz]

Okay so after a week we got the assingments back.I did not get it right,but I understand how to do it now,the lecturer pointed out my mistake and gave me an explanation.

Thanks for the help!