How to find r(t) when we are given conditions - ODE

Homework Statement

Consider the following problems

In #2, they start the solution by saying: r(t)=u(t-1)
in #3, they start by saying that r(t)=t-tu(t-1)
I understand how to solve the problem once you get r(t), I just don't understand how they decide what r(t) is going to be.

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phyzguy
I don't understand your question. r(t) is given as part of the problem. It can be whatever the person that designed the problem wants it to be. You are supposed to solve for y(t). Consider a similar algebraic problem:

y + 3 = W

(1) Solve for y when W = 19.
(2) Solve for y when W = 27.

I don't understand your question. r(t) is given as part of the problem. It can be whatever the person that designed the problem wants it to be. You are supposed to solve for y(t). Consider a similar algebraic problem:

y + 3 = W

(1) Solve for y when W = 19.
(2) Solve for y when W = 27.

Here is the solution for #2:

If they would have chose a different r(t) at the very start, they would have gotten an answer (since we take the laplace and use it to solve the problem). My question is how would I know to say r(t)=u(t-1) in this case, where as for #3 they do:

taking r(t)=t-tu(t-1)

Ray Vickson
Homework Helper
Dearly Missed

Homework Statement

Consider the following problems
View attachment 110194
In #2, they start the solution by saying: r(t)=u(t-1)
in #3, they start by saying that r(t)=t-tu(t-1)
I understand how to solve the problem once you get r(t), I just don't understand how they decide what r(t) is going to be.
You do not need to write $r(t)$ in terms of $u(t-1)$; you can just integrate directly, to find
$$({\cal L}\,r)(s) = \int_0^{\infty} e^{-st} r(t) \, dt = \int_1^{\infty} e^{-st} r(t) \, dt,$$
because in both cases we have $r(t) = 0$ when $0 < t < 1$.

epenguin
Homework Helper
Gold Member
Without having looked at your working, it just seems to me unnecessarily complicated just from the length of it.

When you have the = 0 condition you have just got a common or garden homogeneous lde, with an easily factorable differential operator if you want to look at it that way.
When you have the = 1 condition, it is just the same homogeneous lde if you make a new variable, Y say, Y = (y - 1) .
Is that right?
If so then it's usually not such a long calculation so if you're finding it any more complicated than this then if I were you I would do it again simpler and see if its checks with what you have done.

Maybe in problems like this you just have to look and see sometimes the variable in the solution might reach a a point where equation changes.