Initial Value Problem (done but i think its wrong please check workthanks)

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
fufufu
Messages
14
Reaction score
0

Homework Statement



R(dQ/dt) + (1/C)Q = E_0 e^-t ...Q(0) = 0 and E_0 = a constant


Homework Equations





The Attempt at a Solution




first i rearranged to give:
Q' + (1/CR)Q = (E_0e^-t)/R

next i multiplied all by integrating factor of: u(t) = e^integ:(1/CR) = e^(t/CR)

(e^(t/CR) Q)' = (E_0e^-t)/R (e^(t/CR))

e^(t/CR) Q = integ: (E_0e^-t)/R (e^(t/CR))

now integrating right side to give...
e^(t/CR) Q = (E_0/R)e^-t) (e^(t/CR) / (1/CR-1) + C_1

now rearrange for gen solution:

Q = (E_0/R)e^-t) / (1/CR-1) + C_1/e^(t/CR)

then i applied initial conditions to get C_1. The initial condion is: Q(0) = 0

C = - E_0/R / (1/CR-1)

so solution is:

Q = (E_0/R)e^-t) / (1/CR-1) - E_0/R / (1/CR-1) /e^(t/CR)


is this correct? It doesn't match the solution on exam but not sure if its just because i can rearrange it another way..thanks
 
Physics news on Phys.org
fufufu said:
so solution is:

Q = (E_0/R)e^-t) / (1/CR-1) - E_0/R / (1/CR-1) /e^(t/CR)

It is correct, but use negative exponent in the second term instead of a fraction to avoid confusion.

[tex]Q=\frac{E_0}{R}e^{-t}-\frac{E_0}{R(\frac{1}{CR}-1)} e^{-\frac{t}{CR}}[/tex]

ehild