math4b-12 - Math 4B Lecture 12 May 8 2013 Doug Moore The...

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Math 4B Lecture 12 May 8, 2013 Doug Moore
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The method of variation of parameters , also known as varia- tion of constants , was discovered by Euler and Lagrange, and used to understand perturbations of planetary motion from the exact elliptical motion predicted by Kepler’s three laws. It is now used in several contexts. For example, to solve the differential equation d 2 x dt 2 - 8 dx dt + 16 x = 0 , ( D 2 - 8 D + 16)( x ) = 0 , we write down the characteristic equation λ 2 - 8 λ + 16 = 0 , and factor it as ( λ - 4)( λ - 4) = 0 .
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We find that the characteristic equations for d 2 x dt 2 - 8 dx dt + 16 x = 0 , ( D 2 - 8 D + 16)( x ) = 0 , has repeated root , so the method we used before gives only a one-parameter family of solutions x = ce 4 t , where c is a constant , not a two-parameter family of solutions, as is usually the case. We need a two-parameter family to solve initial value problems with arbitrary initial conditions. How do we get another solution? Variation of parameters: Replace c by a variable v ( t ), x ( t ) = v ( t ) e 4 t , substitute into the equation and solve for v.
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In other words, we substitute x = v ( t ) e 4 t into d 2 x dt 2 - 8 dx dt + 16 x = 0 , and solve for v . We find that x = ve 4 t dx dt = v 0 e 4 t + 4 ve 4 t d 2 x dt 2 = v 00 e 4 t + 4 v 0 e 4 t + 4 v 0 e 4 t + 16 ve 4 t , where the prime denotes differentiation with respect to t . Then 0 = d 2 x dt 2 - 8 dx dt + 16 x = v 00 e 4 t + 8 v 0 e 4 t + 16 ve 4 t - 8( v 0 e 4 t + 4 ve 4 t )) + 16 ve 4 t .
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Everything cancels except for v 00 e 4 t = 0 v 00 = 0 v = c 1 t + c 2 . and the general solution to the differential equation d 2 x dt 2 - 8 dx dt + 16 x = 0 is x = c 1 te 4 t + c 2 e 4 t . Repeated root case: When the characteristic equation of a d 2 x dt 2 + b dx dt + cx = 0 reduces to ( λ - λ 1 ) 2 = 0, the general solution is x = c 1 te λ 1 t + c 2 e λ 1 t .
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1. The general solution to the differential equation d 2 x dt 2 - 10 dx dt + 25 x = 0 is a. x = ce 5 t b. x = c 1 e 5 t + c 2 e 5 t c. x = c 1 e 5 t + c 2 te 5 t d. x = c 1 e 5 t + c 2 log | t | e 5 t e. None of these Correct answer: c
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To find the general solution to the Cauchy-Euler equation t 2 d 2 x dt 2 - t dx dt + x = 0 , our method says to try x = t λ , and solve for λ ,
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  • Winter '14
  • Math, Equations, Quadratic equation, Elementary algebra, dt, general solution

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