Laplace Transform for Solving a First Order Linear IVP

[Temp0]

Homework Statement

Solve the IVP : dy/dt + y = f(t)
y(0) = -5
where f(t) = -1, 0 <= t < 7
-5, t >= 7

y(t) for 0 <= t < 7 = ?
y(t) for t >= 7 = ?

Homework Equations

The Attempt at a Solution

So I have never seen a problem of this type, excuse my silly mistakes if I'm interpreting this question wrong.

At first glance, I assume what the question is asking for is for me to solve what looks like two linear first order equations with the Laplace transform. I start at the first value of f(t), substitute -1 in where f(t) is, and then take the Laplace transform. Giving me:

L(y' + y = -1), which is s * L(S) - y(0) + 1/s^2 = -1/s
I then isolate for L(S), and take the inverse Laplace. Before I go any further, could anyone tell me if I'm coming at this problem correctly? I don't think it's right, since by my logic, I could solve this as if it was a first order linear and get the same result when I use Laplace, but they don't equal. Thanks for any assistance.

 

[Orodruin]

I suggest deriving an expression for the general solution before inserting what f(t) is. Also, if the Laplace transform of y is L(s) in your notation, then you have not transformed the equation correctly.

 

[Ray Vickson]

T

Homework Statement

Solve the IVP : dy/dt + y = f(t)
y(0) = -5
where f(t) = -1, 0 <= t < 7
-5, t >= 7

y(t) for 0 <= t < 7 = ?
y(t) for t >= 7 = ?

Homework Equations

The Attempt at a Solution

So I have never seen a problem of this type, excuse my silly mistakes if I'm interpreting this question wrong.

At first glance, I assume what the question is asking for is for me to solve what looks like two linear first order equations with the Laplace transform. I start at the first value of f(t), substitute -1 in where f(t) is, and then take the Laplace transform. Giving me:

L(y' + y = -1), which is s * L(S) - y(0) + 1/s^2 = -1/s
I then isolate for L(S), and take the inverse Laplace. Before I go any further, could anyone tell me if I'm coming at this problem correctly? I don't think it's right, since by my logic, I could solve this as if it was a first order linear and get the same result when I use Laplace, but they don't equal. Thanks for any assistance.

You cannot do what you tried above: the function L(s) depends on the whole behavior of y(t) over all of t > 0, not just the part from t = 0 to 1. The same is true of F(s) = LT of f(t): you cannot just look at the part from t = 0 to 1.

 

[Temp0]

Thank you for your replies, so I'm restarting on this question and I assume by "general solution", I have to build that step function again.

So I start with f(t) = -1 + u(t-7) - 5u(t-7)

I can take the laplace of that, along with my y' and y, and I get...

L(f(t)) = -1/s - 4e^(7s)/s
L(y') = s Y(s) - y(0)
L(y) = 1/s^2

All my transforms come from the table, so if I read it wrong... Maybe I could be wrong, but am I on the right track now? If I am, any clues to where I go next?

 

[Ray Vickson]

Thank you for your replies, so I'm restarting on this question and I assume by "general solution", I have to build that step function again.

So I start with f(t) = -1 + u(t-7) - 5u(t-7)

I can take the laplace of that, along with my y' and y, and I get...

L(f(t)) = -1/s - 4e^(7s)/s
L(y') = s Y(s) - y(0)
L(y) = 1/s^2

All my transforms come from the table, so if I read it wrong... Maybe I could be wrong, but am I on the right track now? If I am, any clues to where I go next?

(1) Write out the LT of the lhs $y'(t) + y(t)$.
(2) Equate it to the LT of the rhs, which is the LT of $f(t)$.
(3) Solve for the LT of $y(t)$.

Are you sure that the LT of $u(t-7)$ is $e^{7s}/s$?

 

[Orodruin]

L(y) = 1/s^2

I wonder what table you took this from ... If you knew the Laplace transform of y, then you would already know y and there would be no point in making the Laplace transform. By definition, the Laplace transform of y(t) is Y(s).

 

[Temp0]

Woops, I missed a negative sign right? It's $e^{-7s}/s$ right?

Ohh... Wait I see, alright sorry, I'm still learning about Laplace and I didn't realize something I did wrong, thanks, so L(y) is just Y(s)... Let me just change my equation a bit then..

So I isolate my laplace transform of y(t), and I get:

Y(s) = (-1 - 4e^(-7s) - 5s)/ (s(s+1))

 

[Orodruin]

You still need to correct the error I pointed out in my previous post. I would also solve the problem by not assuming anything about $f(t)$, just that it has some Laplace transform that I can denote $F(s)$. You will end up with a product of transforms, which turns into a convolution upon inverse transforming. The given $f(t)$ can then be inserted into the convolution to obtain the solution.

 

[Temp0]

Yup, I edited the post above to reflect that error, and now I'm trying to inverse laplace that equation... Which is difficult.

 

[Temp0]

Well okay, I think I have the answer to something...

I have y(t) = (4e^(7-t) -4)u(t-7) - 4e^(-t) - 1

Now... what do you mean when you say that f(t) can be inserted to give the solution?

 

[Orodruin]

Now... what do you mean when you say that f(t) can be inserted to give the solution?

Well, you did not take that route. I would instead have transformed the differential equation leaving f(t) and assuming it has Laplace transform F(s). This would give you
$$
s Y(s) - y(0) + Y(s) = F(s) \quad \Rightarrow \quad Y(s) = \frac{y(0)}{s+1} + \frac{F(s)}{s+1}.
$$
This is easily inverse transformed if you know what the inverse transform of $1/(s+1)$ is.

If you know anything about Green's functions, $1/(s+1)$ is the Laplace transform of the Green's function of the problem, and the solution for the same type of equation with any inhomogeneities can be written down directly in terms of it.

 

[Temp0]

Ahh yes, except I've never heard the term Green's function before, but I know that 1/(s+1) has something to do with shifting and the inverse of it is just e^(-t)

Hmm, yes, this would be more convenient to use. I am not seeing how to inverse F(s)/(s+1) though.

 

[Orodruin]

So, let $g(t) = e^{-t}$, which as you say gives $G(s) = 1/(s+1)$. This means that:
$$
Y(s) = y(0) G(s) + F(s) G(s).
$$
Can you take it from here?

 

[Temp0]

Ahh, I see. That makes more sense, I was just confused and forgot that I could just separate the two and inverse them individually.

So I end up with -5e^(-t) + f(t) (e^(-t))

Assuming that the inverse of F(s) is just f(t) again.

 

[Orodruin]

So I end up with -5e^(-t) + f(t) (e^(-t))

Not exactly, the inverse transform of a product of Laplace transforms is not the product of the functions but instead ...

In case of doubt, consult a table of Laplace transforms, this relation usually appears in them.

 

[Temp0]

Wow, I never learned about this, but after looking around for a while I found the convolution theorem that says the inverse transform of G(s) and F(s) is just

(f*g)(t)

 

[Orodruin]

Exactly, so you can use this to write down the general solution for an arbitrary f(t) and then just insert the one you were given and compute the convolution integral, which is fairly straightforward in this case.

 

[Temp0]

Okay so this is just going off what I'm reading off the internet right now...

I know the inverse LT of G(s) is e^(-t)... Then I use an arbitrary f(t) to denote the inverse LT of F(s)

I replace t with a temporary variable v, and determine a definite integral from 0 to t.

$\int^t_0 e^{-v}f(v)\,dv$

However, I don't think I can integrate this without knowing f(v) right?

 

[Orodruin]

You need to have $t-v$ as the argument for one of the functions for it to be a convolution, otherwise ok. When you have corrected that, it is indeed time to insert the $f(t)$ you were given.

If this is your first encounter with Green's functions and you are planning to study more physics (in particular theoretical physics) in the future, Green's functions of different sorts will be central concepts in several fields.

 

[Temp0]

Still kind of confused on where to put the t-v, so am I taking the integral of f(t-v)?

Well, I may come across it in the future, but I'm studying engineering so I might not get into that level of physics.

 

[Orodruin]

Still kind of confused on where to put the t-v, so am I taking the integral of f(t-v)?

Yes, you can either take the argument of $f$ to be $t-v$ or the argument of $g$ to be $t-v$, it does not matter which as they are equivalent. I would personally do it for $g$ (i.e., use $e^{-(t-v)}$) and keep $f(v)$.

Well, I may come across it in the future, but I'm studying engineering so I might not get into that level of physics.

I studied engineering physics for my undergrad and it also appeared in some more applied subjects like automated control. It is still a good thing to keep in the back of your mind.

 

[Ray Vickson]

Homework Statement

Solve the IVP : dy/dt + y = f(t)
y(0) = -5
where f(t) = -1, 0 <= t < 7
-5, t >= 7

y(t) for 0 <= t < 7 = ?
y(t) for t >= 7 = ?

Homework Equations

The Attempt at a Solution

So I have never seen a problem of this type, excuse my silly mistakes if I'm interpreting this question wrong.

At first glance, I assume what the question is asking for is for me to solve what looks like two linear first order equations with the Laplace transform. I start at the first value of f(t), substitute -1 in where f(t) is, and then take the Laplace transform. Giving me:

L(y' + y = -1), which is s * L(S) - y(0) + 1/s^2 = -1/s
I then isolate for L(S), and take the inverse Laplace. Before I go any further, could anyone tell me if I'm coming at this problem correctly? I don't think it's right, since by my logic, I could solve this as if it was a first order linear and get the same result when I use Laplace, but they don't equal. Thanks for any assistance.

If you are not forced to use Laplace transforms on this problem, it would be easier to avoid them and just deal with the DE directly. You have
$$y'(t) + y(t) = -1, \; y(0) = -5, \;\; 0 \leq t < 7\\ y'(t) + y(t) = -5, \; y(7+) = y(7-), \;\; t \geq 7$$
These are both easy to solve. If you set $u(t) = y(t)+1$ you get $u'(t) + u(t) = 0$ for $t \in [0,7)$. If you set $v(t) = y(t)+5$ you get $v'(t) + v(t) = 0$ for $t \in [7,\infty)$. Dealing with the initial values is simple.

 

[Orodruin]

it would be easier to avoid them and just deal with the DE directly.

But, but, but, ... then I would not have been able to make an advertisement for Green's functions ;)

 

[Ray Vickson]

But, but, but, ... then I would not have been able to make an advertisement for Green's functions ;)

I think Green's functions are great, and I much prefer their use to other methods such as "undetermined coefficients". However, maybe they are overkill in this case.

 

[Orodruin]

Why swat a mosquito if you can shoot it with a bazooka? :D


But enough of that, I feel I am drifting off topic and I believe the OP has solved the problem already.

 

[Temp0]

Hmm yes, I solved it somewhat anyway, well, that is to say I got the right answer but I'm going to have to study over the method some more so I can see what happened here. Thanks for all your assistance.

 

[vela]

Well, I may come across it in the future, but I'm studying engineering so I might not get into that level of physics.

You may have heard of the Green's function referred to as the impulse response. It seems like this is your first encounter with Laplace transforms. If so, you'll hear about the impulse response later.