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 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: $L\frac{dI}{dt}+RI=10$$1-e^{-0.1t}$$$, $I(0)=I_0$ Put the linear ODE into standard form: $\frac{dI}{dt}+\frac{R}{L}I=\frac{10}{L}$$1-e^{-0.1t}$$$ Calculate integrating factor $\mu(t)=e^{\frac{R}{L}t}$ $e^{\frac{R}{L}t}\frac{dI}{dt}+e^{\frac{R}{L}t}\fra c{R}{L}I=e^{\frac{R}{L}t}\frac{10}{L}$$1-e^{-0.1t}$$$ $\frac{d}{dt}$$e^{\frac{R}{L}t}I$$=\frac{10}{L}$$e^ {\frac{R}{L}t}-e^{\frac{10R-L}{10L}t}$$$ $\int \,d$$e^{\frac{R}{L}t}I$$=\frac{10}{L}\int e^{\frac{R}{L}t}-e^{\frac{10R-L}{10L}t}\,dt$ $e^{\frac{R}{L}t}I=\frac{10}{L}$$\frac{L}{R}e^{\fra c{R}{L}t}-\frac{10L}{10R-L}e^{\frac{10R-L}{10L}t}$$+C$ $e^{\frac{R}{L}t}I=\frac{10}{R}e^{\frac{R}{L}t}-\frac{100}{10R-L}e^{\frac{10R-L}{10L}t}+C$ $I=\frac{10}{R}-\frac{100}{10R-L}e^{-0.1t}+Ce^{-\frac{R}{L}t}$ Now use initial conditions to determine C: $I_0=\frac{10}{R}-\frac{100}{10R-L}+C$ $C=I_0+\frac{100}{10R-L}-\frac{10}{R}=I_0+\frac{100R-10(10R-L)}{R(10R-L)}=I_0+\frac{10L}{R(10R-L)}$ Thus the solution satisfying initial conditions is: $I=\frac{10}{R}-\frac{100}{10R-L}e^{-0.1t}+$$I_0+\frac{10L}{R(10R-L)}$$e^{-\frac{R}{L}t}$

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