+1313 X(t)=Ce^3t; x(0)=5.
Please assist with this equation. Is the C a constant. How is x(0)=5 applied. This is integration?
+33616 Have a look at http://www.physics.umd.edu/hep/drew/IntegralTable.pdf, for example, to see a list of standard integrals.
+33616 Either this is
1. x(t) = Ce3t where C is a constant and you are expected to find C from the fact that x at time 0 is 5. If so, then x(0) = C*e0 = C, so C = 5
or
2. this is meant to be x`(t) = Ce3t (note the `) or dx(t)/dt = Ce3t , where C is a constant, with initial condition x at time 0 being 5, and you are expected to integrate it to find x as a function of t. If so, then
$$$$\int dx = \int Ce^{3t}dt$$
$$$$x(t)=\frac{C}{3}e^{3t}+k$$
where k is another constant, found from x(0) = 5
x(0) = (C/3) + k = 5 so k = 5 - (C/3) and
$$x(t)=5+\frac{C}{3}(e^{3t}-1)$$
+1313 Thanks. Doesnt n+1 happen in taking the ontgration in the later part of your reply Alan?
+33616 Doesnt n+1 happen in taking the integration...
No. You are thinking of
$$\int t^ndt=\frac{t^{n+1}}{n+1}$$
But
$$\int e^{kt}dt = \frac{e^{kt}}{k}$$
+33616 Have a look at http://www.physics.umd.edu/hep/drew/IntegralTable.pdf, for example, to see a list of standard integrals.
