second_hom

second_hom - EXAMPLES FOR SOLVING INITIAL VALUE PROBLEMS ax...

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EXAMPLES FOR SOLVING INITIAL VALUE PROBLEMS a ¨ x + b ˙ x + cx = 0 x ( t 0 ) = x 0 , ˙ x ( t 0 ) = ˙ x 0 . Ich kam unerwartet auf meine L¨ osung und hatte vorher keine Ahnung, daß die L¨ osung einer algebraischen Gleichung in dieser Sache so n¨utlich sein onnte. I came to my solution unexpectedly having had, beforehand, no idea that the solution of an algebraic equation could be so useful in this case. Leonhard Euler In order to solve the second order linear initial value problem in the case of constant coeffients, we always follow the same steps to first find exponential solutions. 1. Write down the characteristic equation 2 + + c = 0 . 2. Find the roots of the characteristic equation. The nature of the roots is determined by the behavior of the discriminant D ( a, b, c ) := b 2 - 4 ac . (a) If D ( a, b, c ) > 0, then the roots λ 1 and λ 2 are real and λ 1 = λ 2 . We then form the general solution x ( t ) = c 1 e λ 1 t + c 2 e λ 2 t . (b) If D ( a, b, c ) = 0 then the roots of the characteristic equation are real and equal. Call this root λ . Then the two independent solutions are e λt , and t e λt , and we form the general solution: x ( t ) =

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