Laplace transforms for a nontrivial solution

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Homework Help Overview

The discussion revolves around solving a differential equation using Laplace transforms, specifically focusing on finding a nontrivial solution with the initial condition x(0)=0 for the equation tx''+(t-2)x'+x=0.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Laplace transforms to the differential equation and explore the separation of variables. There are questions about the correctness of the transformation steps and the integration process. Some participants express uncertainty about the implications of initial conditions on the uniqueness of the solution.

Discussion Status

The conversation includes attempts to clarify the steps involved in applying Laplace transforms and separating variables. Some participants provide guidance on the integration process and raise questions about the assumptions made regarding initial conditions. There is an acknowledgment of potential confusion in the application of the Laplace transform formula.

Contextual Notes

Participants note the lack of explicit initial conditions beyond x(0)=0, which raises questions about the uniqueness of the solution to the second-order differential equation. There is also mention of a singular point at t=0, which may affect the application of existence and uniqueness theorems.

jeannea
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Homework Statement



Transform the given DE to find a nontrivial solution such that x(0)=0.

tx''+(t-2)x'+x=0

Homework Equations


The Attempt at a Solution



Using L{f(t)}=-[itex]\frac{1}{t}[/itex]F'(s), I got

4sX(s)+s(s+1)X'(s)=0.

I see that it is separable, but I do not know how to go about separating it.

My best guess would be
[tex]\frac{1}{X(s)}\,dx=\frac{s+1}{4}\,ds[/tex] but that seems wrong
 
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jeannea said:

Homework Statement


Transform the given DE to find a nontrivial solution such that x(0)=0.

tx''+(t-2)x'+x=0


Homework Equations




The Attempt at a Solution


Using L{f(t)}=-[itex]\frac{1}{t}[/itex]F'(s), I got
4sX(s)+s(s+1)X'(s)=0.
I am assuming that the above is correct - I didn't check your work.
jeannea said:
I see that it is seperable, but I do not know how to go about seperating it.

My best guess would be
[itex]\frac{1}{X(s)}[/itex]dx=[itex]\frac{s+1}{4}[/itex]ds
but that seems wrong

For one thing, it's not separated.

Starting with 4sX(s)+s(s+1)X'(s)=0, and abbreviating a bit, we get
s(s + 1)X' = -4sX
==> X'/X = -4s/(s(s + 1))

The equation is now separated.
What you should probably do now is use partial fractions decomposition to break up the fraction on the right into the sum of two fractions.
 
Oh, I see. I knew I was doing the separation wrong. Wouldn't the s cancel so I wouldn't need to do partial fractions? Also, for the next step I'm not sure what to integrate with respect to on each side. The notation is throwing me off.
Should I do:
∫XdX=∫-[itex]\frac{s+1}{4}[/itex]ds ?

Thank you
 
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jeannea said:
Oh, I see. I knew I was doing the separation wrong. Wouldn't the s cancel so I wouldn't need to do partial fractions?
Good point - yes, the s factors cancel. I didn't notice that.
jeannea said:
Also, for the next step I'm not sure what to integrate with respect to on each side. The notation is throwing me off.
Should I do:
∫XdX=∫-[itex]\frac{s+1}{4}[/itex]ds ?
No.

X'/X = -4/(s + 1)
==> (1/X) * dX/ds = -4/(s + 1)
==> dX/X = -4/(s + 1) * ds
Can you take it from here?
 
Yes. Thanks
 
When you get your solution, by all means check that it satisfies tx'' + (t - 2)x' + x = 0, and x(0) = 0.
Was there another initial condition that you forgot to write?
 
I got X(s)=A[itex]\frac{1}{(s+1)^{4}}[/itex], so x(t)=Bt[itex]^{3}[/itex]e[itex]^{-t}[/itex], which turned out to be correct in the back of my book. When I plugged in the derivatives of x(t) it worked out. I wasn't given any other initial conditions.
 
jeannea said:
I got X(s)=A[itex]\frac{1}{(s+1)^{4}}[/itex], so x(t)=Bt[itex]^{3}[/itex]e[itex]^{-t}[/itex], which turned out to be correct in the back of my book. When I plugged in the derivatives of x(t) it worked out. I wasn't given any other initial conditions.
Then how did you go from tx'' + (t -2)x' + x = 0 to 4sX(s)+s(s+1)X'(s)=0? In particular, what did you get for L(tx'')? This will involve both x(0) -- which you show -- and x'(0) -- which you didn't show.
 
I got L{tx''}=-[itex]\frac{1}{t}[/itex][itex]\frac{d}{ds}[/itex]tL{x''}=-[itex]\frac{d}{ds}[/itex](s[itex]^{2}[/itex]X-sx(0)-x'(0))=-(2sX+s[itex]^{2}[/itex]X'). I treated x(0) and x'(0) as constants to take the derivative with respect to s
 
  • #10
jeannea said:
Using L{f(t)}=-[itex]\frac{1}{t}[/itex]F'(s), I got
4sX(s)+s(s+1)X'(s)=0.

jeannea said:
I got L{tx''}=-[itex]\frac{1}{t}[/itex][itex]\frac{d}{ds}[/itex]tL{x''}=-[itex]\frac{d}{ds}[/itex](s[itex]^{2}[/itex]X-sx(0)-x'(0))=-(2sX+s[itex]^{2}[/itex]X'). I treated x(0) and x'(0) as constants to take the derivative with respect to s

Using the formula you wrote in post #1, it seems to me that L{tx''} would be -1/t * X'''(s).

Maybe I'm missing something.

In any case, it still bothers me that you're able to get a unique solution to a 2nd order diff. equation with only one initial condition.

For example, if y'' + y = 0, then all we can say is that the solution is the family of curves y = Acos(t) + Bsin(t), where A and B are arbitrary constants.
 
  • #11
jeannea said:
L{f(t)}=-[itex]\frac{1}{t}[/itex]F'(s)
This can't be correct. The Laplace transform of f(t) should yield a function only of ##s##; ##t## is integrated out. So ##t## shouldn't appear on the righthand side.
 
  • #12
Of course, the formula she actually used is ##\mathcal L(tf(t))=-F'(s)## although it is poorly written in her writeup. Notice that the solution ##x(t) = Bt^3e^{-t}## satisfies ##x'(0)=0## although that wasn't given or required, and the solution isn't unique. But we do have a singular point at ##t=0## so the existence and uniqueness theorems we like so much don't necessarily apply.
 
  • #13
Sorry for the confusion, I should have written it differently. I used L{tf(t)}=-F'(s). I had the other one written down as part of the inverse L to find f(t), which was not as relevant here
 

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