# Homework Help: Differential equations - algebraic properties of solutions

1. Mar 6, 2007

Show that the operator A defined by A[y](t) = integ(s(y(s))ds) [integration from t to a] is linear. (Hint: Use integration by parts.)

Attempt at the solution:
A[y](t) = integ(s(y(s))ds) [integ from t to a]

Using integration by parts:
Let u = s
du = ds

let dv = y(s)
v = 1/2 y^2(s)

so,
A[y](t) = s(1/2y^2(s)) - integ(ds 1/2y^2(s)) [i'm not sure how to change the integration limits]
A[y](t) = 1/2s y^2(s) - [1/2 (1/3 y^3(s))] (from ___ to ___)
and just sub in the integrating limits and simplify?
am i doing something wrong.

2. Mar 6, 2007

### cristo

Staff Emeritus
Well, firstly, what is the definition of a linear operator?

Does this mean s times the function y(s)? i.e. the integral $$\int_t^asy(s)ds$$.

y is a function, and so you can't integrate it like this! I would let dv=s ds and u=y(s). But then, I don't know the definition of a linear operator that you are using, and so do not know what you're trying to prove!

3. Mar 6, 2007

### HallsofIvy

What in the world are you doing? No one asked you to actually integrate!

A transformation, T, is linear if and only if T(ay1+ by2)= aT(y1)+ bT(y2).

What can you do with $\int s(ay_1+ by_2)ds$?

4. Mar 7, 2007

i integrated since the hint was to use integration of parts
i have no idea what i am doing

5. Mar 7, 2007

i'm looking at the integral you posted hallsofivy
you can multiply the s in and split it into two integrals???

6. Mar 7, 2007

You can, since linearity is a property of intergals.

7. Mar 7, 2007

can i integrate it then???
or is that not the point.
im not quite sure what i'm doing, ive read the class notes. i dont really understand

8. Mar 10, 2007

anybody know what i can do
this is relaly causing me trouble

9. Mar 10, 2007

### HallsofIvy

First go back and check the problem! If you are really to "Show that the operator A defined by A[y](t) = integ(s(y(s))ds) [integration from t to a] is linear" then do as Cristo suggested originally- check the definition of "linear" and show that it works here. I can not imagine how one would "use integration by parts" to do that.

10. Mar 11, 2007

definition of a linear operator means it must satisfy A[cy] = cA[y] and A[y1+y2]=A[y1] + A[y2]

i would probably know how to do this if it wasn't an integral and i had to show it is linear, but i really do not know where to start.

11. Mar 11, 2007

i think i have an idea how to do it now, without the integration by parts.. but my prof said that we are suppose to do it using that, i have no idea how.

but this is what im doing right now (it seems right to me but correct me if i'm wrong):
A[y] = integ(s(y(s))ds)

A[cy] = integ(s(cy) ds)
A[cy] = c integ(s y(s) ds)
A[cy] = c A[y]
so i satisfied property 1

A[y1 + y2] = integ( s(y1+y2) ds)
A[y1 + y2] = integ( sy1 ds) + integ( sy2 ds)
A[y1 + y2] = A[y1] + A[y2]
so i satisfied property 2

so it is a linear operator?
but since i need integration by parts, i've no idea what to do.