Reducing u(t)+Int(0->t) to Differential Equation

[yukcream]

How to reduce the equation: Show how the equation u(t)+Int( 0->t) {[e^a(t-t1)]u(t1)}dt1 = k can be reduced to a differential equation and obtain an intial condidtion for the equation.

Remarks: Int(0->t): integral from 0 to t !

 

[arildno]

Evaluate your equation at t=0. What does that say about u(0)?

Differentiate your equation to get your differential equation.

 

[yukcream]

Evaluate your equation at t=0. What does that say about u(0)?

Differentiate your equation to get your differential equation.

Yes I know I have to differential but if the diff. it with respect to t , what will happen to the second term on the left hand side?

 

[lurflurf]

Yes I know I have to differential but if the diff. it with respect to t , what will happen to the second term on the left hand side?

Use the Leibniz Integral Rule.
\frac{d}{dt}\int_{a(t)}^{b(t)}f(x,t)dx=\int_{a(t)}^{b(t)} <br /> \partial_tf(x,t)dx+f(x,t)\partial_tx|_{x=a(t)}^{x=b(t)}
\frac{d}{dt}\int_{a(t)}^{b(t)}f(x,t)dx=\int_{a(t)}^{b(t)} <br /> \frac{\partial}{\partial t}f(x,t)dx+f(x,t){\partial x}{\partial t}\right|_{x=a(t)}^{x=b(t)}
http://mathworld.wolfram.com/LeibnizIntegralRule.html
for the initial condition you should know that
\int_0^0 f(x) dx=0