University:
Whitman College
Course:
MATH-244 | Differential Equations
Academic year:

2023
Views:

352

Pages:

2
Author:

Greyson Washington

Partial Fraction Decomposition Partial Fraction Decomposition is used when we have a fraction, P (x)/Q(x), where P, Q are polynomials, and the degree of P is less than the degree of Q1 Assume Q is fully factored. We have 4 cases that we will consider. Case I : Q has distinct linear factors, Q(x) = (a1 x + b1 )(a2 x + b2 ) . . . (ak x + bk ) Then: P (x) A1 A2 Ak = + + ... + Q(x) a1 x + b1 a2 x + b2 ak x + bk Case II : Q has some repeated linear factors. Let a1 x + b1 be repeated r times. Then, instead of the single term A1 /(a1 x + b1 ), we have: B1 B2 Br + + ... + 2 a1 x + b1 (a1 x + b1 ) (a1 x + b1 )r Case III : Q has some irreducible quadratic factors, not repeated. Let ax2 + bx + c be an irreducible quadratic factor for Q. Then the decomposition will have the term: Ax + B ax2 + bx + c Case IV : Q has some irreducible quadratic factors, some repeated. Suppose that ax2 + bx + c is a repeated quadratic factor (repeated r times). Then, instead of the single expression in Case III, we will have: A2 x + B2 Ar x + Br A1 x + B1 + + ... + 2 2 2 ax + bx + c (ax + bx + c) (ax2 + bx + c)r Worked Examples and Exercises 1. Worked Example: x2 + 2x − 1 2x3 + 3x2 − 2x Solution: First, factor the denominator: x(2x − 1)(x + 2). We have distinct linear factors. Now we try to solve the following for A, B, and C: x2 + 2x − 1 A B C A(2x − 1)(x + 2) + Bx(x + 2) + Cx(2x − 1) = + + = x(2x − 1)(x + 2) x 2x − 1 x + 2 x(2x − 1)(x + 2) Now, simplify the numerator, and we can see that we are solving: x2 + 2x − 1 = (2A + B + 2C)x2 + (3A + 2B − C)x − 2A which is really three equations, where we set the coefficients of x2 equal, coefficients of x equal, and constants equal: 1 = 2A + B + 2C, 2 = 3A + 2B − C, −1 = −2A Solving, we get: A = 1/2, B = 1/5 and C = −1/10. 1 If the degree of P > Q, then perform long division first. 1 2. Worked Example: x2 − 2 x(x2 + 2) Expanding: x2 − 2 A Bx + C = + 2 x(x2 + 2) x x +2 Getting a common denominator, x2 − 2 = A(x2 + 2) + (Bx + C)x = (A + B)x2 + Cx + 2A so that: 1 = A + B, 0 = C, 2A = −2 so: A = −1, B = 2 and C = 0: −1 2x x2 − 2 = + 2 2 x(x + 2) x x +2 The solutions to the following questions are on my web site: 3. 2x + 3 (x + 1)2 4. 4x2 − 7x − 12 x(x + 2)(x − 3) 5. x2 + 3 x3 + 2x 6. x2 − 2x − 1 (x − 1)2 (x2 + 1) 7. 2x3 − x2 + 3x − 1 (x2 + 1)(x2 + 2) 8. x4 + 1 x(x2 + 1)2 2

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