How Do You Sketch the Graph of i Against t for the Given Differential Equation?

[longball]

Could someone please look over this question

d^2i/dt^2 + 25i = 0

where i(0) =15 and di/dt(0) = 0

sketch the graph of i against t(t>0) over 2 cycles

attempt:

General Solution

i=Acos(5t) + Bsin(5t)

when i(0) = 15 is applied

A = 15

P.S

ip= Acos(5t) + Bcos(5t)

ip'= -5Asin(5t) + 5Bcos(5t)

apply di/dt (0) = 0

therefore 0=5B

so B=0

giving

i-15cos (5t)

where t = 1/5cos t = 1

Am I correct in saying that this calculation has been done correctly and the graph which has to be sketched is a cosine graph peaking and troughing at 15 and -15 with a period of 1

 

[LCKurtz]

Your solution i = 15 cos(5t) is correct. You have a homogeneous equation there, so you don't need to go through the calculation of y_p for a particular solution of a non-homogeneous equation. That's why you got y_p = 0.

But the period of cos(5t) is not 1. Remember the period of cos(bt) is 2pi/b.