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MATH 374 Lecture 12

MATH 374 Lecture 12. General Solution – Homogeneous and Non-Homogeneous Equations. 4.5: General Solution of a Homogeneous Equation. The next theorem shows that if we know n solutions to an nth order linear, homogeneous equation, we know “all” of the solutions.

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MATH 374 Lecture 12

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  1. MATH 374 Lecture 12 General Solution – Homogeneous and Non-Homogeneous Equations

  2. 4.5: General Solution of a Homogeneous Equation • The next theorem shows that if we know n solutions to an nth order linear, homogeneous equation, we know “all” of the solutions. • One key to this theorem is linear independence! • Theorem 4.4: Let {y1, y2, … , yn } be a linearly independent set of solutions of the homogeneous linear equation b0(x) y(n) + b1(x) y(n-1) + … + bn-1(x) y’ + bn(x) y = 0 (1) for x 2 [a, b]. Suppose further that (1) is normal on [a, b]. If  is any solution of (1), valid on [a, b], then there exists constants ĉ1, ĉ2, … , ĉnsuch that  = ĉ1 y1 + ĉ2 y2 + … + ĉnyn. (2) 2

  3. Proof of Theorem 4.4 • (Case when n = 2, other cases are similar.) • Let y1 and y2 be linearly independent solutions of b0(x) y’’ + b1(x) y’ + b2(x) y = 0 (3) on [a,b] and  be any solution of (3) on [a,b]. By Theorem 4.3, the Wronskian of y1 and y2 is non-zero at some x02 [a,b], i.e. W(x0) = y1(x0) y2’(x0) –y1’(x0) y2(x0)  0. 3

  4. W(x0) = y1(x0) y2’(x0) –y1’(x0) y2(x0)  0. Proof of Theorem 4.4 (continued) • It follows from Theorem 4 in the Coddington Handout that the system c1 y1(x0) + c2 y2 (x0) = (x0) (4a) c1 y1’(x0) + c2 y2’(x0) = ’(x0) (4b) has a unique solution, say c1 = ĉ1 and c2 = ĉ2. Define the function f by: f := ĉ1 y1 + ĉ2 y2. (5) 4

  5. f := ĉ1 y1 + ĉ2 y2 (5) Proof of Theorem 4.4 (continued) • From Theorem 4.1 (Principle of Superposition), it follows that f is a solution of (3) on [a,b], since f is a linear combination of solutions of (3) on [a,b]. From (5), (4a), and (4b), we see that f(x0) = ĉ1 y1(x0) + ĉ2 y2 (x0) = (x0) f’(x0) = c1 y1’(x0) + c2 y2’(x0) = ’(x0). It follows from Theorem 4.2 (Existence and Uniqueness) that  = f = ĉ1 y1 + ĉ2 y2 on [a,b].  • Note: We needed (1) to be normal to apply Theorems 4.2 and 4.3 in this proof! 5

  6. General Solution; Fundamental Set • Definition:The general solution of the nth order linear homogeneous differential equation (1) is y = c1 y1 + c2 y2 + … + cnyn where {y1, y2, … , yn } are linearly independent solutions of (1) and c1, c2, … , cn are arbitrary constants. We call {y1, y2, … , yn } a fundamental set of solutions of (1). 6

  7. 4.6: General Solution of a Non-Homogeneous Equation • Theorem 4.5: Let yc be a solution of the homogeneous linear differential equation b0(x) y(n) + b1(x) y(n-1) + … + bn-1(x) y’ + bn(x) y = 0 (1) on a  x b and let yp be a solution of the related non-homogeneous linear differential equation b0(x) y(n) + b1(x) y(n-1) + … + bn-1(x) y’ + bn(x) y = R(x) (2) on a  x b. Then y = yc + yp is also a solution of (2) on on a  x b. 7

  8. b0(x) y(n) + b1(x) y(n-1) + … + bn-1(x) y’ + bn(x) y = 0 (1) b0(x) y(n) + b1(x) y(n-1) + … + bn-1(x) y’ + bn(x) y = R(x) (2) Proof of Theorem 4.5 • Substituting y = yc + yp into the LHS of (2), we get: b0(x) y(n) + b1(x) y(n-1) + … + bn-1(x) y’ + bn(x) y = (b0(x) yc(n) + b1(x) yc(n-1) + … + bn-1(x) yc’ + bn(x) yc) + (b0(x) yp(n) + b1(x) yp(n-1) + … + bn-1(x) y’p + bn(x) yp) = 0 + R(x) (since yc solves (1) and yp solves (2)) = R(x).  8

  9. b0(x) y(n) + b1(x) y(n-1) + … + bn-1(x) y’ + bn(x) y = 0 (1) b0(x) y(n) + b1(x) y(n-1) + … + bn-1(x) y’ + bn(x) y = R(x) (2) General Solution of a Non-Homogeneous Equation • Theorem 4.6: Let {y1, y2, … , yn } be a linearly independent set of solutions of (1) on a  x  b and yp a particular solution of (2) on a  x  b. Suppose that (2) is normal and Y is any solution of (2) on a  x  b. Then there exists constants ĉ1, ĉ2, … , ĉnsuch that Y = ĉ1 y1 + ĉ2 y2 + … + ĉnyn + yp. 9

  10. b0(x) y(n) + b1(x) y(n-1) + … + bn-1(x) y’ + bn(x) y = R(x) (2) Proof of Theorem 4.6 • Consider the function Y – yp. Substitute Y – yp into LHS of (2): b0(x) (Y – yp)(n) + … + bn-1(x) (Y – yp)’ + bn(x) (Y – yp) = (b0(x) Y(n) + … + bn-1(x) Y’ + bn(x) Y) – (b0(x) yp(n) + … + bn-1(x) y’p + bn(x) yp) = R(x) – R(x) (since Y and yp solve (2)) = 0. By Theorem 4.4, there exist constants ĉ1, ĉ2, … , ĉn such that Y – yp = ĉ1 y1 + ĉ2 y2 + … + ĉnyn.  10

  11. General Solution; Complementary Function • Definition:The general solution of the non-homogeneous linear differential equation (2) is y = c1 y1 + c2 y2 + … + cnyn + yp where {y1, y2, … , yn } are linearly independent solutions of (1), c1, c2, … , cn are arbitrary constants, and yp is any particular solution of (2). We call yc = c1 y1 + c2 y2 + … + cnyn the complementary function of (2). 11

  12. Example 1 • Since yp = -11/12 -1/2 x is a particular solution of y’’’ – 6 y’’ + 11 y’ -6 y = 3x, (3) and it turns out (as we will find later) that the general solution to the homogeneous equation y’’’ – 6 y’’ + 11 y’ -6 y = 0, is yc = c1 ex + c2 e2x + c3 e3x, it follows that the general solution to (3) is: y = yc + yp = c1 ex + c2 e2x + c3 e3x -11/12 - 1/2 x. 12

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