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Var. y, linear 11. PDE, 2nd-order, ind. var. t, r, dep. var. N 13. dp / dt ϭ kp, where k is the proportionality constant 15. dT / dt ϭ k A M Ϫ T B , where k is the proportionality constant 17. 594 sec. 8 1 0 1 t Figure 3 Solutions to Problem 3 satisfying y A 0 B ϭ 5, y A 0 B ϭ 8, and y A 0 B ϭ 15 Exercises 2 3. Yes 5. No 7. Yes 9. Yes 11. Yes 13. Yes 19. The left-hand side is always Ն 4. 21. (a) 1 / 3, Ϫ3 (b) 1 Ϯ 26 23. Yes 25. Yes 27. No 29. (b) Yes 31. (a) No (c) y ϭ 2x and y ϭ Ϫ2x 5. (a) p 1 Exercises 3 1.

Dx 1 First-Order Differential Equations and the solution is again elementary: a1 A x B y ϭ y AxB ϭ Ύ b AxB dx ϩ C , 1 c b A x B dx ϩ C d . a1 A x B Ύ One can seldom rewrite a linear differential equation so that it reduces to a form as simple as (2). However, the form (3) can be achieved through multiplication of the original equation (1) by a well-chosen function m A x B . Such a function m A x B is then called an “integrating factor” for equation (1). The easiest way to see this is first to divide the original equation (1) by a1 A x B and put it into standard form (4) dy ؉ P AxBy ؍Q AxB , dx where P A x B ϭ a0 A x B / a1 A x B and Q A x B ϭ b A x B / a1 A x B .

Solutions Not Expressible in Terms of Elementary Functions. As discussed in calculus, certain 2 indefinite integrals (antiderivatives) such as ͐ e x dx cannot be expressed in finite terms using elementary functions. When such an integral is encountered while solving a differential equation, it is often helpful to use definite integration (integrals with variable upper limit). For example, consider the initial value problem ϭϪ 1 1 ϩ . y A x1 B y A2B If we let t be the variable of integration and replace x1 by x and y A 2 B by 1, then we can express the solution to the initial value problem by y A x B ϭ a1 Ϫ Ύ x 2 e t dtb 2 Ϫ1 .