# Correct answer a 3 the general solution to the

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3. The general solution to the differential equation 6 d 2 x dt 2 - 5 dx dt + x = 0 is a. x = c 1 e 2 t + c 2 e 3 t b. x = c 1 e - 2 t + c 2 e - 3 t c. x = c 1 e t/ 2 + c 2 e t/ 3 d. x = c 1 e - t/ 2 + c 2 e - t/ 3 e. None of these Correct answer: c
Suppose that we know that the general solution to the differential equation d 2 x dt 2 - 3 dx dt + 2 x = 0 is x = c 1 x 1 ( t ) + c 2 x 2 ( t ) = c 1 e t + c 2 e 2 t , and we want to solve the initial value problem d 2 x dt 2 - 3 dx dt + 2 x = 0 , x (1) = 3 , dx dt (1) = 4 . We then need to solve the system c 1 x 1 (1) + c 2 x 2 (1) = 3 , c 1 x 0 1 (1) + c 2 x 2 (1) = 4 . We can solve this linear system for the unknowns c 1 and c 2 by Cramer’s rule.

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The solution is c 1 = 3 x 2 (1) 4 x 0 2 (1) x 1 (1) x 2 (1) x 0 1 (1) x 0 2 (1) , c 2 = x 1 (1) 3 x 0 1 (1) 4 x 1 (1) x 2 (1) x 0 1 (1) x 0 2 (1) The denominator in these expressions is the so-called Wronskian W ( x 1 , x 2 )( t ) = x 1 ( t ) x 2 ( t ) x 0 1 ( t ) x 0 2 ( t ) , evaluated at t = 1. Since x 1 ( t ) = e t and x 2 ( t ) = e 2 t , W ( x 1 , x 2 )( t ) = e t e 2 t e t 2 e 2 t = e 3 t .
Using this expression for the Wronskian, we find that c 1 = 3 e 2 4 2 e 2 e 3 = 2 e , c 2 = e 3 e 4 e 3 = 1 e 2 , and the solution to the initial value problem d 2 x dt 2 - 3 dx dt + 2 x = 0 , x (1) = 3 , dx dt (1) = 4 . is x = 2 e t e + e 2 t e 2 .

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4. Suppose that we know that the general solution to the dif- ferential equation d 2 x dt 2 - 2 dx dt - 15 x = 0 is x = c 1 x 1 ( t ) + c 2 x 2 ( t ) = c 1 e 5 t + c 2 e - 3 t . Then the Wronskian is W ( x 1 , x 2 ) = Correct answer: c

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