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General Solutions for Homogeneous and Non-Homogeneous Equations

Understand Theorem 4.4 and Theorem 4.5 outlining solutions for linear equations, key factors, proof, and reasoning behind the general solution. Learn how to find fundamental sets and comprehensive solutions to these types of equations.

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General Solutions for Homogeneous and Non-Homogeneous Equations

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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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