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March 16th, 2011, 08:07 AM   #11
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Re: Differential equations

ahh for god sake. cheers

Last question today its another Electronics orientated problem, which i hate. but ive attempted

In a LR circuit with applied voltage E=10(1-e^-0.1t) the current i is given by L di/dt + Ri =10(1-e^0.1t).

If the initial current is i0m derive the result :

i=10/R - (100/10R-L) e^-0.1t + [ i0 + 10L/R(10R-L) ] e^-Rt/L.

My working.

L di/dt + Ri = 10 (1-e^-0.1t)

di/dt + (R/L)i = 10(1-e^-0.1t) / L

?(x)=e^?R/L dt = e^Rt/L Multiply by integrating factor

e^Rt/L di/dt + R/L e^Rt/L i = 10(1-e^-0.1t)/L e^Rt/L

d/dt (e^Rt/L i) = 10(1-e^-0.1t)/L e^Rt/L

?d/dt (e^Rt/L i) dt =?10(1-e^-0.1t)/L e^Rt/L dt

e^Rt/L i = ?10-10e^-0.1t/L e^Rt/L dt

e^Rt/L i = 10t +100e^-0.1t + L/R

thats as far as i got as i was digging a deep whole i feared
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March 16th, 2011, 08:45 AM   #12
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Math Focus: Calculus/ODEs
Re: Differential equations

You did well until you integrated the right side of the equation. You did not distribute the integrating factor.

You have the IVP:

,

Put the linear ODE into standard form:



Calculate integrating factor













Now use initial conditions to determine C:





Thus the solution satisfying initial conditions is:

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