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Extra resources for Ordinary differential equations : methods and applications
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N X xm (iy)n−m . (n − m)! n=0 m=0 59 (10) In the double summation above, for a fixed n which runs from 0 to ∞, we sum over m from 0 to n. 1. 1 Let us now interchange the order of the double summation, that is, we fix m first and sum over n. 1, we find that we have to let m run from 0 to ∞, and for a fixed m, we sum over n from m to ∞. (n − m)! k! m=0 k=0 ∞ ∞ m X X m=0 x m! (iy)k k! k=0 x+iy ⇒e = ex · eiy . 60 (11) With (10) and (11), ex+iy can be evaluated in terms of real exponential, cosine and sine functions, that is, ex+iy = ex · eiy = ex (cos(y) + i sin(y)).
R Thus, we show that y2 y10 − y1 y20 = De− f (x)dx . Theorem 2 If y1 (x) and y2 (x) are two linearly independent solutions of the homogeneous 2nd order linear ODE in (5) then the general solution of (5) is given by y(x) = Ay1 (x) + By2 (x), where A and B are arbitrary constants. 53 From Lemma 1 above, we know that y(x) = Ay1 (x)+By2 (x) is a solution of (5), but is it the general solution? To prove the theorem, we have to show that any solution Y (x) of (5) can be written in the form Ay1 (x)+By2 (x), if y1 (x) and y2 (x) are linearly independent solutions of (5).
Differentiating the general solution, we obtain y 0 = −3Ae−3x + 2Be2x . 61 It follows that y(0) = 1 ⇒ A + B = 1 y0 (0) = 7 ⇒ −3A + 2B = 7. Solving for A and B, we obtain A = −1 and B = 2. Thus, the required particular solution of the ODE is y = −e−3x + 2e2x . 2. Solve the ODE y 00 + 6y 0 + 9y = 0 subject to the conditions y(0) = 1 and y0 (0) = 1. This is a homogeneous 2nd order linear ODE with constant coefficients. Thus, we use y = eλx , y 0 = λeλx and y 00 = λ2 eλx . Substituting into the ODE, we obtain λ2 + 6λ + 9 = 0 ⇒ (λ + 3)2 = 0 ⇒ λ = −3.