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Unformatted text preview: 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 differential 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 differential equation above: y (t) = c1 e−2t + c2 e−3t for some constants c1 and c2 . We must find which constants satisfy the initial conditions. We have the derivatives ￿ ￿ 2= c1 + c2 y (t) = c1 e−2t + c2 e−3t =⇒ ￿ −2t −3t 3 = (−2) c1 + (−3) c2 y (t) = (−2) c1 e + (−3) c2 e Upon multiplying the first equation by 3 then by 2, we have the following systems of equations: 3 c1 −2 c1 c1 + 3 c2 + −3 c2 =6 =3 =9 2 c1 −2 c1 + 2 c2 + −3 c2 −c2 =4 =3 =7 The constants are c1 = 9 and c2 = −7. Hence the solution to the initial value problem is y (t) = 9 e−2t − 7 e−3t . Non-Example. Now say that we wish to find the general solution to the differential equation y ￿￿ + 2 y ￿ + y = 0. Again, we consider the characteristic equation: r2 + 2 r + 1 = 0. This polynomial factors as r2 + 2 r + 1 = (r + 1) 2 =⇒ r1 = r2 = −1. In this case, the two roots are not distinct. Hence the trick we have introduced does not give the general solution to the differential equation. We will introduce in a future lecture a trick which will solve such equations – even when the roots of the characteristic equation are not distinct. Existence and Uniqueness Theorems. We are motivated by two questions: #1. Does one solution exist? #2. When do we have all solutions? We attempt to answer these questions by stating the “Existence and Uniqueness Theorem” for second order equations. To this end, we recall the corresponding result for first order differential equations. Say we have the initial value problem a(t) y ￿￿ + b(t) y ￿ + c(t) y = 0; y (t0 ) = y0 , If we have an interval I = (α, β ) such that i. f (t) is continuous on I , ii. b(t) and c(t) are continuous on I , iii. a(t) is continuous yet a(t) ￿= 0 on I , and iv. t0 ∈ I ; then the initial value problem has a unique solution y = y (t). 1 ￿ y ￿ (t0 ) = y0 ...
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue University-West Lafayette.

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