345_pdfsam_math 54 differential equation solutions odd

345_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: CHAPTER 6: Theory of Higher Order Linear Differential Equations EXERCISES 6.1: Basic Theory of Linear Differential Equations, page 324 1. Putting the equation in standard form, y− we find that p1 (x) ≡ 0, 3 p2 (x) = − , x p3 (x) = ex , x and q (x) = x2 − 1 . x 3 ex x2 − 1 y + y= , x x x Functions p2 (x), p3 (x), and q (x) have only one point of discontinuity, x = 0, while p1 (x) is continuous everywhere. Therefore, all these functions are continuous on (−∞, 0) and (0, ∞). Since the initial point, x0 = 2, belongs to (−∞, 0), Theorem 1 guarantees the existence of a unique solution to the given initial value problem on (−∞, 0). 3. For this problem, p1 (x) = −1, p2 (x) = √ x − 1, and g (x) = tan x. Note that p1 (x) is contin- uous everywhere, p2 (x) is continuous for x ≥ 1, and g (x) is continuous everywhere except at odd multiples of π/2. Therefore, these three functions are continuous simultaneously on the intervals 1, π , 2 π 3π , 22 , 3π 5π , 22 ,... . Because 5, the initial point, is in the interval (3π/2, 5π/2), Theorem 1 guarantees that we have a unique solution to the initial value problem on this interval. √ 5. Dividing the equation by x x + 1, we obtain 1 1 y +√ y = 0. y−√ x x+1 x+1 341 ...
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