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# lecture_14 (dragged) - MA 36600 LECTURE NOTES WEDNESDAY...

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MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 18 Second Order Differential Equations Example. Consider the initial value problem y + 5 y + 6 y = 0; where y (0) = 2 , y (0) = 3 . To solve this equation, we proceed in two steps: First, we find the general solution to the di ff erential equation. Second, we find the particular solution to the initial value problem by taking in account the initial conditions. We found the general solution to this di ff erential equation above: y ( t ) = c 1 e 2 t + c 2 e 3 t for some constants c 1 and c 2 . We must find which constants satisfy the initial conditions. We have the derivatives y ( t ) = c 1 e 2 t + c 2 e 3 t y ( t ) = ( 2) c 1 e 2 t + ( 3) c 2 e 3 t = 2 = c 1 + c 2 3 = ( 2) c 1 + ( 3) c 2 Upon multiplying the first equation by 3 then by 2, we have the following systems of equations: 3 c 1 + 3 c 2 = 6 2 c 1 + 3 c 2 = 3 c 1 = 9 2 c 1 + 2 c 2 = 4 2 c 1 + 3 c 2 = 3 c 2 = 7 The constants are c 1 = 9 and c 2 = 7. Hence the solution to the initial value problem is y ( t ) = 9 e 2 t 7 e
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