Differential equations - algebraic properties of solutions

In summary, the operator A defined by A[y](t) = integ(s(y(s))ds) [integration from t to a] is linear, satisfying the properties of linearity for linear operators. This can be shown by using the definition of a linear operator and verifying that A[cy] = cA[y] and A[y1 + y2] = A[y1] + A[y2]. However, the use of integration by parts to prove linearity is unclear and may require further clarification from the instructor.
  • #1
braindead101
162
0
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.
 
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  • #2
braindead101 said:
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.)
Well, firstly, what is the definition of a linear operator?

Attempt at the solution:
A[y](t) = integ(s(y(s))ds) [integ from t to a]
Does this mean s times the function y(s)? i.e. the integral [tex]\int_t^asy(s)ds[/tex].

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

let dv = y(s)
v = 1/2 y^2(s)
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
braindead101 said:
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.

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 [itex]\int s(ay_1+ by_2)ds[/itex]?
 
  • #4
i integrated since the hint was to use integration of parts
i have no idea what i am doing
 
  • #5
i'm looking at the integral you posted hallsofivy
you can multiply the s in and split it into two integrals?
 
  • #6
braindead101 said:
i'm looking at the integral you posted hallsofivy
you can multiply the s in and split it into two integrals?

You can, since linearity is a property of intergals.
 
  • #7
can i integrate it then?
or is that not the point.
im not quite sure what I'm doing, I've read the class notes. i don't really understand
 
  • #8
anybody know what i can do
this is relaly causing me trouble
 
  • #9
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
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
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 I am 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.
 

What are differential equations?

Differential equations are mathematical equations that describe how a variable changes over time. They involve the derivatives of the variable with respect to one or more independent variables.

What are algebraic properties of solutions to differential equations?

The algebraic properties of solutions to differential equations refer to the characteristics of the solutions, such as their behavior, stability, and uniqueness, that can be determined by manipulating the equations algebraically.

How are differential equations solved?

Differential equations can be solved using a variety of methods, including separation of variables, substitution, and using integrating factors. The specific method used depends on the type and complexity of the equation.

What is the role of initial conditions in solving differential equations?

The initial conditions, or known values of the variable and its derivative(s) at a specific point, are crucial in solving differential equations. They help determine the constants of integration and provide a starting point for finding the general solution.

Why are differential equations important in science?

Differential equations are used extensively in various fields of science, including physics, chemistry, biology, and engineering, to model and understand natural phenomena. They allow scientists to make predictions and analyze complex systems, making them an essential tool in scientific research and development.

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