Differential Equations Ordinary and Partial Differential Equations Math Forum

 March 16th, 2011, 08:07 AM #11 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 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  March 16th, 2011, 08:45 AM #12 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 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: Tags differential, equations ,
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