1. Chapter 9 Differential Equations 9.1 Modeling with Differential Equations *9.2 Direction Fields and Euler’s Method 9.3 Separable Equations *9.4 Exponential Growth and Decay *9.5 The Logistic Equation 9.6 Linear Equations

2. 9.1 Modeling with Differential equations Models of Population Growth Let t =time, P =the number of individuals in the population. Our assumption that the rate of growth of the population is proportional to the population size is written as the equation where k is the proportionality constant.

3. If we let, then. Thus, any exponential function of the form is a solution of Equation 1. Allowing C to vary through all the real numbers, we get the family of solutions. General Differential Equations A differential equation is an equation that contains an unknown function and one or more of its derivatives.

4. The order of a differential equation is the order of the highest derivative (of the unknown function) that appears in the equation. are first-order equations; are equations of the second-order.

5. A function f is called a solution of a differential equation if the equation is satisfied when y=f(x) and its derivatives are substituted into the equation. For example, we know that the general solution of the differential equation is given by where C is an arbitrary constant.

6. Example 1 Show that every member of the family of functions is a solution of the differential equation Solution We differentiate the expression for y: The right side of the differential equation becomes

7. Therefore, for every value of c, the given function is a solution of the differential equation. c is often called parameter. The set of solutions of the second-order differential equation is given by y(t) = -16t 2 + C 1 t + C 2, where C 1 and C 2 are independent arbitrary constants. This Equation (1) has a two-parameter family of solutions. Thus, an nth-order equation has an nth-parameter family of solutions.

8. Generally, the term general solution is often used in place of nth-parameter family of solutions. If specific values are assigned to each of the arbitrary constants in an nth-parameter family of solutions, then the resulting solution is called a particular solution. For example, the function (2) y(t) = -16t 2 + 40t +144 is a particular solution of Equation (1). Specific values for constants are usually determined by imposing additional conditions, called side conditions or initial conditions. The function given in (2) is the solution of (1) which satisfies the side conditions y(0) = 144, y’(0) = 40.

9. 9.3 Separable Equations ; Homogeneous Equations The first-order equation is separable iff it can be put in the form y’ = f(x)g(y) where f(x) and g(y) are continuous. To solve such an equation, we begin by writing as Because f(x) and g(y) are continuous, they have antiderivatives. Thus, we get This is a one-parameter family of solutions of the equation.

10. Example 1 (a) Solve the differential equation (b) Find the solution of this equation that satisfies the initial condition Solution (a) where C is an arbitrary constant.

11. (b) If we put y(0) =2 in the general solution in part (a), we have C= 8/3. Thus, the solution of the initial-value problem is Example 2 Find a solution of the differential equation which satisfies the side condition y(2) = 1. Solution We show first that the equation is separable:

12. is a one-parameter family of solutions which defines y implicitly as a function of x. To find a solution which satisfies the side condition, we set x = 2 and y = 1 in the one-parameter family: 1 + ln1 = 1/2×22 – 2 + C and C = 1. Thus, the particular solution is

13. Homogeneous Equations The first-order differential equation is homogeneous iff it can be put in the form A homogeneous equation can be transformed into a separable equation by setting v = y/x. To see this, write xv = y and differentiate with respect to x: v + xv’ = y’. Substituting y and y’ into equation, gives v + xv’ = f(v), and

14. The above equation is separable. Thus we solve it by solving the transformed equation and then substitute y/x back in for v. Example 1 Find a one-parameter family of solution of the equation xyy’ = 3x 2 + y 2. Solution Note that this equation can be written as Now, set vx = y. Then, v + xv’ = y’ and

15. Substituting y/x back in for v, we have

16. 9.6 Linear Equations A first-order differential equation is linear if it can be written in the form y’ + p(x)y = q(x) where p(x) and q(x) are continuous functions on a given interval. To solve the equation y’ + p(x)y = q(x), multiply both sides by the integrating factor and integrate both sides.

17. Thus, is the general solution of the first-order linear differential equation. Example 1 Solve the differential equation Solution An integrating factor is Multiplying both sides of the differential equation by we get

18. Integrating both sides, we have Example 2 Find the general solution of xy’ – 2y = 2x 2 + x. Solution This equation can be written as follows y’ + (- 2/x)y = 2x + 1.

19. We have and this is the general solution.

20. Chapter 17 Second-Order Differential Equations 17.1 Second-Order Linear Equations 17.2 Nonhomogeneous Linear Equations 17.3 Applications of SODE 17.4 Series Solutions

21. 17.1 Second-Order Linear Equations A second-order linear differential equation has the form (1) where P, Q, R, and G are continuous. (2) homogeneous If for some x, Equation (1) is nonhomogeneous. Theorem If y 1 (x) and y 2 (x) are both solutions of the linear homogeneous equation (2) and c 1 and c 2 are any constants, then the function y(x) = c 1 y 1 (x)+c 2 y 2 (x) is also a solution of Equation (2).

22. Proof Since y 1 and y 2 are solutions of Equation (2), we have Therefore, Thus, y(x) = c 1 y 1 (x)+c 2 y 2 (x) is a solution of Equation (2).

23. linearly independent solutions y 1 and y 2 : y 1 /y 2 ≠constant For instance, the functions f(x) = x 2 and g(x) = 5x 2 are linearly dependent, but f(x) = e x and g(x) = xe x are linearly independent. Theorem If y 1 and y 2 are linearly independent solutions of equation (2), and P(x) is never 0, then the general solution is given by y(x) = c 1 y 1 (x)+c 2 y 2 (x) where c 1 and c 2 are any constants.

24. If the differential equation has the form (3) where a( ≠0), b, and c are constants. If y = e rx, then y’ = re rx and y’’ = r 2 e rx. Substitution into the differential equation gives ar 2 e rx + bre rx + ce rx = 0 and, since e rx ≠0, ar 2 + br + c = 0. This show that the function y = e rx satisfies the differential equation iff ar 2 + br + c = 0. (characteristic equation)

25. By the quadratic equation formula, the roots of the characteristic equation are Case 1. If b 2 > 4ac, the characteristic equation has two distinct real roots: Both are linearly independent solutions of the differential equation (3).

26. Therefore, the general solution of is Case 2. If b 2 = 4ac, the characteristic equation has only one root: r 1 = -b/2a. In this case, we can see that are solutions of the differential equation (3). Thus, the general solution of is

27. Case 3. If a 2 < 4b, the characteristic equation has two complex conjugate roots: Setting we can write r 1 = α+ iβ, r 2 = α- iβ. The functions y 1 = e αx cosβx, y 2 = e αx sinβx are real-valued solutions. Thus, the general solution of is y = C 1 e αx cosβx + C 2 e αx sinβx.

28. Example 1 Find the general solution of the equation y ’’ + 2y ’ – 15y = 0. Then find the particular solution that satisfies the side conditions y(0) = 0, y ’ (0) = - 1. Solution The characteristic equation is the quadratic r 2 + 2r – 15 = 0. There are two real roots: - 5 and 3. The general solution takes the form y = C 1 e -5x +C 2 e 3x. Differentiating the general solution, we have y ’ = -5C 1 e -5x + 3C 2 e 3x.

29. The conditions y(0) = 0, y ’ (0) = - 1 are satisfied iff C 1 + C 2 = 0 and -5C 1 + 3C 2 = -1. Solving these two equations simultaneously, we find that C 1 = 1/8, C 2 = -1/8. The solution that satisfies the prescribed side conditions is the function

30. Example 2 Find the general solution of the equation y ’’ + 4y ’ + 4y = 0. Solution The characteristic equation is the quadratic r 2 + 4r + 4 = 0. The number – 2 is the only root. The general solution can be written y = C 1 e -2x + C 2 xe -2x.

31. Example 3 Find the general solution of the equation y ’’ + y ’ + 3y = 0. Solution The characteristic equation is r 2 + r + 3 =0. The quadratic formula shows that there are two complex roots: The general solution takes the form

32. 17.1 Nonhomogeneous Linear Equations (1) (second-order nonhomogeneous linear differential equation with constant coefficients) (2) is called the complementary equation. Theorem The general solution of the nonhomogeneous differential equation (1) can be written as y(x) = y p (x)+y c (x) where y p is a particular solution of Equation (1) and y c is the general solution of Equation (2).

33. THE SUPERPOSITION PRINCIPLE Given the equation (*) ay ’’ + by ’ + cy = G 1 (x) + G 2 (x). If y 1 is a solution of ay ’’ + by ’ + cy = G 1 (x) and y 2 is a solution of ay ’’ + by ’ + cy = G 2 (x), then y * = y 1 + y 2 is a solution of (*). The Method of Undetermined Coefficients The method of undetermined coefficients works for equations (1) ay ’’ + by ’ + cy = G(x), where the function G(x) has a special form.

34. A Particular Solution of ay ’’ + by ’ + cy = G(x) Note: If y p satisfies the equation ay ’’ + by ’ + cy = 0, try xy p ; if xy p also satisfies the reduced equation, then x 2 y p will give a particular solution.

35. Example 1 Find a particular solution of each of the following differential equations: (a) y ’’ + 2y ’ +5y = 10e -2x. (b) y ’’ + 2y ’ +y = 10cos3x + 6sin3x. Solution (a) We assume that the equation has a solution of the form y p = Ae -2x. Then (y p ) ’ = -2Ae -2x and (y p ) ’’ = 4Ae -2x. Substituting y p and its derivatives into the equation, we have 4Ae -2x + 2(-2Ae -2x ) + 5Ae -2x = 10e -2x

36. 5Ae -2x = 10e -2x. Therefore, 5A = 10 and A = 2. Thus, a particular solution of the equation (a) is y p = 2e -2x. We can verify that the general solution of the equation (a) is y = C 1 e -x cos2x + C 2 e -x sin2x + 2e -2x. (b) We assume that the equation has a solution of the form y p = Acos3x + Bsin3x. Then

37. (y p ) ’ = -3Asin3x + 3Bcos3x and (y p ) ’’ = -9Acos3x -9Bsin3x. Substituting y p and its derivatives into the equation, we have (-8A + 6B)cos3x + (-6A -8B)sin3x = 10cos3x + 6sin3x This equation will be satisfied for all x iff -8A + 6B = 10, -6A - 8B = 6. The solution of this pair of equations is A = -29/25, B =3/25.

38. Thus, a particular solution of the equation (b) is y p = -29/25cos3x +3/25sin3x. We can verify that the general solution of the equation (b) is y = C 1 e -x + C 2 xe -x -29/25cos3x +3/25sin3x Example 2 Find the general solution of y ’’ – 5y ’ +6y = 4e 2x. Solution The complementary equation y ’’ – 5y ’ +6y = 0 has characteristic equation

39. r 2 – 5r + 6 = (r – 2)(r – 3) = 0 Thus, the general solution of the complementary equation is given by y c = C 1 e 2x + C 2 e 3x Noting that y = Ae 2x satisfies the equation y ’’ – 5y ’ +6y = 0, we should try y p = Axe 2x instead of y = Ae 2x. The derivatives of y p are (y p ) ’ = Ae 2x + 2Axe 2x, (y p ) ’’ = 4Ae 2x + 4Axe 2x.

40. Substituting into the differential equation, we have 4Ae 2x + 4Axe 2x – 5(Ae 2x + 2Axe 2x ) + 6Axe 2x = 4e 2x -Ae 2x = 4e 2x and A = -4. Thus, y p = C 1 e 2x + C 2 e 3x – 4xe 2x is the general solution of y ’’ – 5y ’ +6y = 4e 2x. In a similar manner, we can verify that the form of a particular solution of y ’’ – 4y ’ +4y = e 2x is y p = Ax 2 e 2x since e 2x and xe 2x are the

41. solutions of the complementary equation y ’’ – 4y ’ +4y = 0. Example 3 Solve y ’’ – 4y = xe x +cos2x. Solution The complementary equation y ’’ – 4y = 0 has characteristic equation r 2 – 4 = 0. Thus, the general solution of the complementary equation is given by y c = C 1 e 2x + C 2 e -2x

42. For the equation y ’’ – 4y = xe x we try y p1 (x)=(Ax+B)e x Then (y p1 ) ’ = (Ax+A+B)e x, (y p1 ) ’’ = (Ax+2A+B)e x, so substitution in the equation gives (Ax+2A+B)e x – 4 (Ax+A+B)e x = xe x and A = – 1/3, B = – 2/9, so y p1 (x)=( – 1/3x – 2/9)e x For the equation y ’’ – 4y = cos2x, we try y p2 (x)=Ccos2x+Dsin2x Substitution gives

43. – 4Ccos2x – 4Dsin2x – 4(Ccos2x+Dsin2x)=cos2x or – 8Ccos2x – 8Dsin2x =cos2x Therefore, C = – 1/8, D = 0, and y p2 (x)= – 1/8 cos2x By the superposition principle, the general solution is y = y c + y p1 + y p2 = C 1 e 2x + C 2 e -2x – (1/3x + 2/9)e x – 1/8 cos2x

44. The Method of Variation of Parameters Suppose we have solved the equation ay ’’ + by ’ + cy = 0 and written the general solution as (2) y(x)=c 1 y 1 (x)+ c 2 y 2 (x) We look for a particular solution of the equation ay ’’ + by ’ + cy = G(x) of the form (3) y p (x)=u 1 (x)y 1 (x)+ u 2 (x)y 2 (x) (This method is called variation of parameters.)

45. Since u 1 and u 2 are arbitrary function, we can impose two conditions on them u 1 ’ y 1 + u 2 ’ y 2 =0 and a(u 1 ’ y 1 ’ + u 2 ’ y 2 ’ )=G After solving this system of two equations in the unknown functions u 1 ’ and u 2 ’ we may be able to find u 1 and u 2 and then the particular solution is given by Equation (3).

46. Example 4 Solve y ’’ + y = tanx, 0<x< π/2. Solution The general solution of the complementary equation y ’’ + y = 0 is c 1 sinx+c 2 cosx. Let y p (x)= u 1 (x) sinx+u 2 (x)cosx We have u 1 ’ sinx+u 2 ’ cosx=0 u 1 ’ cosx -u 2 ’ sinx=tanx Solving the equations, we get u 1 ’ =sinx, u 2 ’ =cosx - secx

47. (We seek a particular solution, so we don ’ t need a constant of integration here.) Therefore, y p (x)=-cosxsinx+[sinx-ln(secx+tanx)]cosx =-cosxln(secx+tanx) and the general solution is y(x)=c 1 sinx+c 2 cosx-cosxln(secx+tanx)