- #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.
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.