Laplace transforms for a nontrivial solution

[jeannea]

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)}=-$\frac{1}{t}$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
$$\frac{1}{X(s)}\,dx=\frac{s+1}{4}\,ds$$ but that seems wrong

 

[Mark44]

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)}=-$\frac{1}{t}$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.

I see that it is seperable, but I do not know how to go about seperating it.

My best guess would be
$\frac{1}{X(s)}$dx=$\frac{s+1}{4}$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.

 

[jeannea]

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=∫-$\frac{s+1}{4}$ds ?

Thank you

 

[Mark44]

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.

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=∫-$\frac{s+1}{4}$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?

 

[jeannea]

Yes. Thanks

 

[Mark44]

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?

 

[jeannea]

I got X(s)=A$\frac{1}{(s+1)^{4}}$, so x(t)=Bt$^{3}$e$^{-t}$, 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.

 

[Mark44]

I got X(s)=A$\frac{1}{(s+1)^{4}}$, so x(t)=Bt$^{3}$e$^{-t}$, 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.

 

[jeannea]

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

 

[Mark44]

Using L{f(t)}=-$\frac{1}{t}$F'(s), I got
4sX(s)+s(s+1)X'(s)=0.

I got L{tx''}=-$\frac{1}{t}$$\frac{d}{ds}$tL{x''}=-$\frac{d}{ds}$(s$^{2}$X-sx(0)-x'(0))=-(2sX+s$^{2}$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.

 

[vela]

L{f(t)}=-$\frac{1}{t}$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.

 

[LCKurtz]

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.

 

[jeannea]

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