Lecture Notes in Differential Equations - Bruce E. Shapiro

Lecture Notes in Differential Equations - Bruce E. Shapiro Lecture Notes in Differential Equations - Bruce E. Shapiro

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254 LESSON 27. HIGHER ORDER EQUATIONS Example 27.15. Solve y ′′′ + y ′ = tan t using variation of parameters. The characteristic equation is 0 = r 3 + r = r(r + i)(r − i); hence a fundamental set of solutions are y 1 = 1, y 2 = cos t, and y 3 = sin t. From either Abel’s formula or a direct calculation, W (t) = 1, since y P = y 1 ∫ t W = ∣ W 1 (s)f(s) W (s)a 3 (s) ds+y 2 1 cos t sin t 0 − sin t cos t 0 − cos t − sin t ∫ t W 2 (s)f(s) W (s)a 3 (s) ds+y 3 ∣ = 1 (27.207) ∫ W 3 (s)f(s) ds (27.208) W (s)a 3 (s) where a 3 (s) = 1, f(s) = tan s, and 0 cos t sin t W 1 = 0 − sin t cos t ∣ 1 − cos t − sin t ∣ = 1 (27.209) 1 0 sin t W 2 = 0 0 cos t = − cos t (27.210) ∣ 0 1 − sin t ∣ 1 cos t 0 W 3 = 0 − sin t 0 = − sin t (27.211) ∣ 0 − cos t 1 ∣ Therefore ∫ y P = t ∫ ∫ tan sds − cos t cos s tan sds − sin t sin s tan sds (27.212) t t Integrating the first term and substituting for the tangent in the third term, ∫ ∫ sin 2 s y P = − ln |cos t| − cos t sin sds − sin t ds (27.213) cos s t The second integral can not be integrated immediately, and the final integral can be solved by substituting sin 2 s = 1 − cos 2 s ∫ y P =− ln |cos t| + cos 2 1 − cos 2 s t − sin t ds (27.214) cos s Since the first term (the constant) is a solution of the homogeneous equation, we can drop it from the particular solution, giving and a general solution of t y P = ln |cos t| − sin t ln |sec t + tan t| (27.215) y = ln |cos t| − sin t ln |sec t + tan t| + C 1 + C 2 cos t + C 3 sin t. (27.216) t t

Lesson 28 Series Solutions In many cases all we can say about the solution of a(t)y ′′ + b(t)y ′ + c(t)y = f(t) y(t 0 ) = y 0 y ′ (t 0 ) = y 1 ⎫ ⎪⎬ ⎪ ⎭ (28.1) is a statement about whether or not a solution exists. So far, however, we do not have any generally applicable technique to actually find the solution. If a solution does exist, we know it must be twice differentiable (n times differentiable for an nth order equation). If the solution is not just twice, but infinitely, differentiable, we call it an analytic function. According to Taylor’s theorem, any function that is analytic at a point t = t 0 can be expanded in a power series where y(t) = ∞∑ a k (t − t 0 ) k (28.2) k=0 a k = y(k) (t 0 ) (28.3) k! Because of Taylor’s theorem, the term analytic is sometimes used to mean that a function can be expanded in a power series at a point (analytic ⇐⇒ infinitely differentiable ⇐⇒ power series exists). The method of series for solving differential equations solutions looks for analytic solutions to (28.1) by substituting equation (28.2) into the differential equation and solving for the coefficients a 0 , a 1 , ... 255

Lesson 28<br />

Series Solutions<br />

In many cases all we can say about the solution of<br />

a(t)y ′′ + b(t)y ′ + c(t)y = f(t)<br />

y(t 0 ) = y 0<br />

y ′ (t 0 ) = y 1<br />

⎫<br />

⎪⎬<br />

⎪ ⎭<br />

(28.1)<br />

is a statement about whether or not a solution exists. So far, however, we<br />

do not have any generally applicable technique to actually f<strong>in</strong>d the solution.<br />

If a solution does exist, we know it must be twice differentiable (n times<br />

differentiable for an nth order equation).<br />

If the solution is not just twice, but <strong>in</strong>f<strong>in</strong>itely, differentiable, we call it an<br />

analytic function. Accord<strong>in</strong>g to Taylor’s theorem, any function that is<br />

analytic at a po<strong>in</strong>t t = t 0 can be expanded <strong>in</strong> a power series<br />

where<br />

y(t) =<br />

∞∑<br />

a k (t − t 0 ) k (28.2)<br />

k=0<br />

a k = y(k) (t 0 )<br />

(28.3)<br />

k!<br />

Because of Taylor’s theorem, the term analytic is sometimes used to mean<br />

that a function can be expanded <strong>in</strong> a power series at a po<strong>in</strong>t (analytic ⇐⇒<br />

<strong>in</strong>f<strong>in</strong>itely differentiable ⇐⇒ power series exists).<br />

The method of series for solv<strong>in</strong>g differential equations solutions looks<br />

for analytic solutions to (28.1) by substitut<strong>in</strong>g equation (28.2) <strong>in</strong>to the<br />

differential equation and solv<strong>in</strong>g for the coefficients a 0 , a 1 , ...<br />

255

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