Question

Alright, integrodifferential equation attempted to be solved. First time attempting to solve one, posted below in a moment.

Answer 1

\[f(t)=\int\limits_{0}^{t}(t-\tau)f(\tau)d \tau=t\]

Answer 2

General form for Volterra Integra Equation:\[f(t)=g(t)+\int\limits_{0}^{t}f(\tau)h(t-\tau)d \tau\]

Answer 3

@sidsiddhartha (Just dropping you a line, don't need help yet but mentioning the problem so you have an idea of it if you want to help)

Answer 4

Damn, just wrote up a bunch of stuff in LaTeX, web browser problems wiped it, made the identification that h(t-tau)=(t-tau), h(t)=t

Answer 5

\[F(s)=F(s)\frac{1}{s^2}\]
http://thumbpress.com/wp-content/uploads/2013/05/I-Have-No-Idea-What-Im-Doing-1.jpg

Answer 6

lol

Answer 7

Subtract g(t) from the other side? Really not sure what to do, focused *way* too much time on basic Laplace and inverse and heaviside, just now basically learned about these; how should I approach this, solving for f(t)?

Answer 8

\[f(t)=\int\limits_{0}^{t}(t-\tau)f(\tau)d \tau-t\]Right direction?

Answer 9

\[F(s)=F(s)\bigg(\frac{1}{s}\bigg)-\frac{1}{s^2}\]

Answer 10

\[\bigg[1-\frac{1}{s}\bigg]F(s)=-\frac{1}{s^2}\](Are these the right steps? I mean, Iegitimately don't know what I'm doing, I'm just solving for F(s) right now algebraically)

Answer 11

ok wait checking from first

Answer 12

just a little mistake

Answer 13

Interesting, so basically you took the laplace transform both sides and you gona solve F(s) and finally take inverse.

is that the strategry here ?

Answer 14

\[L[\int\limits_{0}^{t}(t-\tau)F(\tau)d \tau]=\frac{ f(s) }{ s^2 }\]
is'nt it?

Answer 15

yes it like a RLC circuit
\[r.i(t)+L \frac{ di(t) }{ dt }+\frac{ 1 }{ C }\int\limits_{0}^{t}i(\tau)d \tau=v(t)\\V(s)=R.I(s)+s.L(s)+\frac{ I(s) }{ Cs }\]
@ganeshie8

Answer 16

Yeah, my bad, that looks right

Answer 17

yeah:)

Answer 18

\[F(s)=\frac{F(s)-1}{s^2}\]

Answer 19

L-C,RLC circuits form intego differential equation