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

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SUMMARY

The discussion centers on solving the initial value problem represented by the equation R(dQ/dt) + (1/C)Q = E_0 e^-t with the initial condition Q(0) = 0. The user correctly rearranged the equation and applied the integrating factor method, ultimately deriving the general solution for Q. However, a suggestion was made to express the second term using a negative exponent instead of a fraction to enhance clarity. The final correct solution is Q = (E_0/R)e^-t - (E_0/(R(1/CR-1)))e^(-t/CR).

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fufufu
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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
 
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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.

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

ehild
 

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