MIT18_03S10_c04

MIT18_03S10_c04 - 18.03 Class 4, Feb 10, 2010 First order...

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18.03 Class 4 , Feb 10, 2010 First order linear equations: integrating factors [1] First order homogeneous linear equations [2] Newtonian cooling [3] Integrating factor (IF) [4] Particular solution, transient, initial condition [5] General formula for IF Definition: A "linear ODE" is one that can be put in the "standard form" _______________________________ | | | r(t)x' + p(t)x = q(t) | x = x(t) |_______________________________| r(t), p(t) are the "coefficients" [I may have called q(t) also a coefficient also on Monday; this is not correct, fix it if I did.] The left hand side represents the "system," and the right hand side arises from an "input signal." A solution x(t) is a "system response" or "output signal." We can always divide through by r(t), to get an equation of the Reduced standard form: _______________________________ | | | x' + p(t)x = q(t) | x = x(t) (*) |_______________________________| The equation is "homogEneous" if q is the "null signal," q(t) = 0 . This corresponds to letting the system evolve in isolation: In the bank example, no deposits and no withdrawals. In the RC example, the power source is not providing any voltage increase. The homogeneous linear equation x' + p(t) x = 0 (*)_h is separable. Here's the solution, in general on the left, with an example (with p(t) = 2t ) on the right: x' + p(t)x = 0 x' + 2tx = 0 Separate: dx/x = - p(t) dt dx/x = - 2t dt Integrate: ln|x| = - int p(t) dt + c ln|x| = - t^2 + c Exponentiate: |x| = e^c e^{ - int p(t) dt } |x| = e^c e^{-t^2} Eliminate the absolute value and reintroduce the lost solution:
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x = C e^{- int p(t) dt} x = C e^{-t^2} In the example, we chose a particular anti-derivative of k , namely kt. That is what I really have in mind to do in general. The constant of integration is taken care of by the constant C . So the general solution to
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MIT18_03S10_c04 - 18.03 Class 4, Feb 10, 2010 First order...

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