Math 310 Q7 Sol - MATH 310 Fall 2011 Differential Equations...

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MATH 310, Fall 2011: Differential Equations Simon Fraser University Quiz 7: Solutions 1. Solve the differential equation y 00 + 4 y 0 + 5 y = 0 , y (0) = 2 , y 0 (0) = - 1 . Solution: Looking for a solution of the form y = e rt , we find that the characteristic equation is r 2 + 4 r + 5 = 0 , or ( r + 2) 2 + 1 = 0 , which has the pair of complex conjugate roots r = - 2 ± i (we could also find the roots using the quadratic formula: r = - 4 ± 16 - 20 2 = - 2 ± - 1 = - 2 ± i ). Hence two linearly independent (real-valued) solutions of the differential equation are y 1 ( t ) = e - 2 t cos t and y 2 ( t ) = e - 2 t sin t , and the general solution is y ( t ) = c 1 e - 2 t cos t + c 2 e - 2 t sin t. We find the constants c 1 and c 2 by substituting the initial conditions: computing y 0 ( t ) = c 1 ± - 2 e - 2 t cos t - e - 2 t sin t ² + c 2 ± - 2 e - 2 t sin t + e - 2 t cos t ² , and letting t = 0 , we have the two equations y (0) = c 1 = 2 , y 0 (0) = - 2 c 1 + c 2 = - 1 .
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Math 310 Q7 Sol - MATH 310 Fall 2011 Differential Equations...

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