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Section 2.1 Introduction: Second-Order Linear Equations

Section 2.1 Introduction: Second-Order Linear Equations. Second Order Differential Equations: F ( x , y , y ′, y ″) = 0 2 nd Order Linear Differential Equations: A ( x ) y ″ + B ( x ) y ′ + C ( x ) y = f ( x )

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Section 2.1 Introduction: Second-Order Linear Equations

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  1. Section 2.1 Introduction: Second-Order Linear Equations

  2. Second Order Differential Equations: F(x, y, y′, y″) = 0 2nd Order Linear Differential Equations: A(x) y″ + B(x) y′ + C(x) y = f (x) often these will be in the form y″ + P(x) y′ + Q(x) y = f (x) nth Order Linear Differential Equations

  3. Homogeneous Linear Differential Equations: A(x) y″ + B(x) y′ + C(x) y = 0 Solutions to homogeneous linear Differential Equations form vector spaces!

  4. Theorem Let y1 and y2 be solutions to y″ + P(x) y′ + Q(x) y = 0. Then functions of the form f (x) = c1y1 + c2y2 (with c1 and c2 being any constants) are also solutions.

  5. General Solutions Initial Conditions

  6. Theorem Suppose the P(x) and Q(x) are continuous on an interval J containing xo. Then the initial value problem y″ + P(x) y′ + Q(x) y = 0, y′(xo) = a , y(xo) = b has a unique solution on J.

  7. Ex. 1 Solve y″ + y = 0,

  8. Suppose we've found two linearly independent solutions, y1 and y2, to the initial value problem y″ + P(x) y′ + Q(x) y = 0, y′(xo) = a, y(xo) = b. How do we find the general solution to this differential equation?

  9. Recall: Two functions f (x) and g(x) are said to be linearly independent if one function cannot be written as a multiple of the other (or equivalently, f (x) and g(x) are linearly independent if the equation c1 f (x) + c2 g(x) = 0 has only a solution of c1 = c2 = 0).

  10. Ex. 2 Show that y1 = sin(x) and y2 = 3sin(x) are both solutions to y″ + y = 0, but we cannot use these to find our particular solution.

  11. Definition: The Wronskian of f (x) and g(x) is defined as:

  12. Ex. 3 Calculate the Wronskian of f (x) = sin(x) and g(x) = x2.

  13. Theorem: Suppose y1 and y2 are solutions of y″ + P(x) y′ + Q(x) y = 0 on an interval J in which P(x) and Q(x) are continuous. (a) If y1 and y2 are linearly dependent then W(f, g) = 0 for all x on the interval J. (b) If y1 and y2 are linearly independent then W(f, g) ≠ 0 for any x on the interval J.

  14. Ex. 4 (a) What does this theorem imply regarding the solutions y1 = sin(x) and y2 = cos(x) to the differential equation y″ + y = 0?

  15. Ex. 4 (b) What does this theorem imply regarding the solutions y1 = sin(x) and y2 = 3sin(x) to the differential equation y″ + y = 0?

  16. Theorem: Suppose y1 and y2 are linearly independent solutions of y″ + P(x) y′ + Q(x) y = 0 where P(x) and Q(x) are continuous on an interval J. If g(x) is any solution to this differential equation then g(x) = c1y1 + c2y2 for some choices of constants c1 and c2.

  17. Big Important Question: How do we determine two linearly independent solutions, y1 and y2? We shall examine this question for the differential equation ay″ + by′ + cy = 0 (where a, b, and c are constants).

  18. Claim: y = erx is a solution (for some constant r) to ay″ + by′ + cy = 0. We just have to figure out what value of r will work.

  19. Ex. 5 Solve y″ – 5y′ + 6y = 0, y(0) = 1, y′(0) = –1.

  20. Theorem: If the roots of the characteristic equation ar2 + br + c = 0 are r1 and r2 with r1 ≠ r2 then y1 = _________ and y2 = _________ are two linearly independent solutions to the differential equation ay″ + by′ + cy = 0. Thus, y = _________________ is the general solution to ay″ + by′ + cy = 0.

  21. What if there is a repeated root of the characteristic equation? Ex. 6 Solve y″ – 8y′ + 16y = 0, y(0) = 5, y′(0) = 21.

  22. Theorem: If r1 is a repeated root of the characteristic equation ar2 + br + c = 0 then y1 = _________ and y2 = _________ are two linearly independent solutions to the differential equation ay″ + by′ + cy = 0. Thus, y = _________________ is the general solution to ay″ + by′ + cy = 0.

  23. Ex. 6 Solve y″ – 8y′ + 16y = 0, y(0) = 5, y′(0) = 21.

  24. Why does this theorem work? Let's prove that it produces the correct general solution for this past problem. (Introduction to the linear operator D.)

  25. Why does this theorem work? Let's prove that it produces the correct general solution for this past problem. (Introduction to the linear operator D.)

  26. Why does this theorem work? Let's prove that it produces the correct general solution for this past problem. (Introduction to the linear operator D.)

  27. Why does y″ + y = 0 have solutions of y1 = sin(x) and y2 = cos(x)???? Our previous theorems couldn't have possibly generated these solutions.

  28. Ex. 7 Use our past two theorems to solve y″ + y = 0.

  29. Recall that ez =

  30. Recall that ez = cos(z) =

  31. Recall that ez = cos(z) = sin(z) =

  32. Recall that ez = cos(z) = sin(z) =

  33. Ex. 8 Solve y″ – 6y′ + 13y = 0

  34. Theorem: If the roots of the characteristic equation ar2 + br + c = 0 are A ± Bi then y1 = eAxsin(Bx) and y2 = eAxcos(Bx) are two linearly independent solutions to the differential equation ay″ + by′ + y = 0. Thus, y= c1 eAxsin(Bx) + c2eAxcos(Bx) is the general solution to ay″ + by′ + y = 0.

  35. Section 2.2 General Solutions of Linear Equations

  36. Higher order linear differential equations: y(n) + pn–1(x) y(n–1) + pn–2(x) y(n–2) + · · · · · + p2(x) y″ + p1(x) y′ + p0(x) y = f (x)

  37. If y1, y2, . . . , yn are linearly independent solutions to the homogeneous diff. eq. y(n) + pn–1(x) y(n–1) + pn–2(x) y(n–2) + · · · · · + p1(x) y′ + p0(x) y = 0, then the general solution to y(n) + pn–1(x) y(n–1) + · · · · · + p1(x) y′ + p0(x) y = 0 is y = c1y1 + c2y2 + · · · + cnyn.

  38. To prove this we needed the following definitions and theorems: Theorem: If y1, y2, . . . , yn are solutions to y(n) + pn–1(x) y(n–1) + · · · · · + p1(x) y′ + p0(x) y = 0 then so is y = c1y1 + c2y2 + · · · + cnyn Theorem: Suppose y1, y2, .., yn are solutions to y(n) + pn–1(x) y(n–1) + ··· + p1(x) y′ + p0(x) y = 0 on an interval J where the pi(x) are continuous. (a) If y1, y2, . . . , yn are linearly dependent then W(y1, y2, . . . , yn) = 0 at each point on the interval J. (b) If y1, y2, . . . , yn are linearly independent then W(y1, y2, . . . , yn) ≠ 0 at each point on the interval J. Theorem: If y1, y2, . . . , yn are solutions to y(n) + pn–1(x) y(n–1) + · · · · · + p1(x) y′ + p0(x) y = 0, then so is y = c1y1 + c2y2 + · · · + cnyn for any choice of constants c1, c2, . . . , cn. Theorem: If y1, y2, . . . , yn are linearly independent solutions to y(n–1) + pn–2(x) y(n–2) + · · · · · + p1(x) y′ + p0(x) y = 0, then general solution is y = c1y1 + c2y2 + · · · + cnyn.

  39. Ex. 1 Suppose that the following functions are all particular solutions to y(3) + 3y″ + 4y′ + 12y = 0. We would like to have a set of three linearly independent solutions. If this can be done then which three can we use? If it cannot be done then show that it cannot. y1 = sin(2x) y2 = cos(2x) y3 = 5 sin(2x) + 3 cos(2x) y4 = sin(x) cos(x) y5 = cos2(x) – sin2(x) y6 = e–3x

  40. Ex. 2 Use the definition of linear dependence to show that the following functions are linearly dependent. f (x) = x2 + 3x, g(x) = 7x2, h(x) = 31x2 + 6x.

  41. The Wronskian of the n functions f1(x), f2(x), . . . , fn(x):

  42. Theorem: Suppose y1, y2, ..., yn are solutions of y(n) + pn–1(x) y(n–1) + ··· + p1(x) y′ + p0(x) y = 0 on an interval J in which pn–1(x), pn–2(x), · · · p1(x), p0(x) are continuous. (a) If y1, y2, . . . , yn are linearly dependent then W(y1, y2, . . . , yn) = 0 for all x on the interval J. (b) If y1, y2, . . . , yn are linearly independent then W(y1, y2, . . . , yn) ≠ 0 for all x on the interval J.

  43. Ex. 3 (a) y1 = sin(2x), y2 = cos(2x), y3 = e–3x are solutions to y(3) + 3y″ + 4y′ + 12y = 0. Use the Wronskian to determine whether or not they are linearly independent.

  44. Ex. 3 (b) y1 = sin(2x), y2 = cos(2x), y3 = e–3x are solutions to y(3) + 3y″ + 4y′ + 12y = 0. Use the Wronskian to determine whether or not they are linearly independent.

  45. Ex. 3 (c) y1 = x, y2 = xln(x), y3 = x2 are solutions to x3y(3) – x2y″ + 2xy′ – 2y = 0. Use the Wronskian to determine whether or not they are linearly independent.

  46. Suppose we can find three linearly independent solutions to y(3) + 3 y″ + 4y′ + 12y = 0. Are we guaranteed a solution to the initial valued problem with given initial conditions of y(0) = 1, y′(0) = 2, y″(0) = 3? Why??

  47. Theorem: Suppose pn–1(x), pn–2(x), · · · p1(x), p0(x) are continuous on an interval J which contains xo. If we can find n linearly independent solutions to the diff. eq. y(n) + pn–1(x) y(n–1) + · · · · · + p1(x) y′ + p0(x) y = 0 then we are guaranteed a solution to the initial value problem of y(n) + pn–1(x) y(n–1) + · · · · · + p1(x) y′ + p0(x) y = 0 y(x0) = b0 y′(x0) = b1 y″(x0) = b2 : : y(n)(x0) = bn

  48. Theorem: If y1, y2, . . . , yn are linearly independent solutions to y(n) + pn–1(x) y(n–1) + · · · + p1(x) y′ + p0(x) y = 0, then general solution is y = c1y1 + c2y2 + · · · + cnyn.

  49. Nonhomogeneous linear Differential Equations: y(n) + pn–1(x) y(n–1) + · · · · · + p1(x) y′ + p0(x) y = f (x) We solve a nonhomogeneous differential equation by first examining the corresponding homogeneous differential equation. Review:

  50. Theorem: Suppose that yc = c1y1 + c2y2 + · · · + cnyn is the general solution to the homogeneous differential equation y(n) + pn–1(x) y(n–1) + ··· + p1(x) y′ + p0(x) y = 0. Further suppose that yp is any particular solution to the nonhomogeneous differential equation y(n) + pn–1(x) y(n–1) + · · · · · + p1(x) y′ + p0(x) y = f (x). Then the general solution to the nonhomogeneous differential equation y(n) + pn–1(x) y(n–1) + · · · · · + p1(x) y′ + p0(x) y = f (x) is: y = yp + yc = yp + c1y1 + c2y2 + · · · + cnyn

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