The Method of Undetermined Coefficients (MUC)

The Method of Undetermined Coefficients (MUC) A Lecture in ENGIANA

Introduction To solve a non-homogeneous linear differential equation any(n) + an-1y(n-1) + … + a1y’ + a0y = g(x) we must do two things: 1) Find the complementary function yc and 2) Find any particular solution yp of the non-homogeneous equation.

Introduction • The complementary function yc is the general solution of the associated homogeneous differential equation any(n) + an-1y(n-1) + … + a1y’ + a0y = 0 This was tackled in the last lecture. • We now develop a method for obtaining particular solutions.

Method of Undetermined Coefficients The underlying idea behind this method is a conjecture about the FORM of yp, an educated guess really, that is motivated by the kind of functions that make up the input function g(x).

Limitations The method of undetermined coefficients is limited to linear differential equations in the form of any(n) + an-1y(n-1) + … + a1y’ + a0y = g(x) where (1) the coefficients ai, i = 0, 1, …, n are constants, and

Limitations The method of undetermined coefficients is limited to linear differential equations in the form of any(n) + an-1y(n-1) + … + a1y’ + a0y = g(x) where (2) g(x) is a constant k, a polynomial function, an exponential function ex, a sine or cosine function sinx or cosx, or finite sums and products of these functions.

Applicability of the MUC The function g(x) is a linear combination of functions of the type P(x) = anxn + an-1xn-1 + … + a1x + a0 P(x)ex P(x) exsin P(x) excos where n is a non-negative integer, and  &  are real numbers

Caveat on the MUC The method of undetermined coefficients is not applicable to equations of the form any(n) + an-1y(n-1) + … + a1y’ + a0y = g(x) when g(x) = lnx g(x) = 1/x g(x) tanx g(x) = sin-1x

Working “backwards” • In preparation for the method of undetermined coefficients, we must practice on writing the corresponding homogeneous differential equation of a given function (which happens to be a solution). • In other words, given a solution, find f(m) and, by extension, f(D).

Working “backwards”

Working “backwards” Note: [D – (a + bi)] [D – (a - bi)] = D2 – D(a + bi) – D(a – bi) + (a – bi)(a + bi) = D2 – aD – biD – aD + biD + a2 – b2i2 = D2 – 2aD + a2 – b2i2 = (D – a)2 + b2

Example Obtain in factored form a linear differential equation with real, constant coefficients that is satisfied by the given function. 1) y = 4e2x + 3e-x 2) y = -2x + (1/2)e4x 3) y = 2excos3x 4) y = -2e3xcosx 5) y = xe-xsin2x + 3e-xcos2x 6) y = coskx

Solution to #1 Obtain in factored form a linear differential equation with real, constant coefficients that is satisfied by the given function. 1) y = 4e2x + 3e-x • The presence of e2x means that m = 2 is a root. Hence, (D – 2) is a factor of f(D). • The presence of e-x means that m = -1 is a root. Hence, (D + 1) is a factor of f(D).

Solution to #1 Obtain in factored form a linear differential equation with real, constant coefficients that is satisfied by the given function. 1) y = 4e2x + 3e-x Hence, we have (m – 2) (m + 1) = 0 (D – 2) (D + 1) y = 0

Solution to #2 Obtain in factored form a linear differential equation with real, constant coefficients that is satisfied by the given function. 2) y = -2x + (1/2)e4x • The presence of x means that m = 0 occurs twice (i.e., c1e0x + c2xe0x). So, D2 is a factor of f(D). • The presence of e4x means that m = 4 is a root. So, (D – 4) is a factor of f(D).

Solution to #2 Obtain in factored form a linear differential equation with real, constant coefficients that is satisfied by the given function. 2) y = -2x + (1/2)e4x Hence, we have m2 (m – 4) = 0 D2 (D – 4) y = 0

Solution to #3 Obtain in factored form a linear differential equation with real, constant coefficients that is satisfied by the given function. 3) y = 2excos3x • The presence of excos3x means that m = 1  3i is a root. So, [D – (1 + 3i)] and [D – (1 – 3i)] are factors of f(D).

Solution to #3 Obtain in factored form a linear differential equation with real, constant coefficients that is satisfied by the given function. 3) y = 2excos3x • Thus, we have [m – (1 + 3i)] [m – (1 – 3i)] = 0 [D – (1 + 3i)] [D – (1 – 3i)] = 0 [(D – 1)2 + 9] y = 0 (D2 – 2D + 10) y = 0

Exercises List the roots of the auxiliary equation for a homogeneous linear differential equation with real, constant coefficients that has the given function as a particular solution. 1) y = 3xe2x 2) y = e-xcos4x 3) y = x(e2x + 4) 4) y = xex 5) y = 4cos2x

Exercises List the roots of the auxiliary equation for a homogeneous linear differential equation with real, constant coefficients that has the given function as a particular solution. 6) y = xcos2x 7) y = xcos2x – 3sin2x 8) y = (1/4)(3sinx – sin3x) 9) y = x2 – x + e-x(x + cosx) 10) y = x2sinx + xcosx

Method of Undetermined Coefficients • Let f(D) be a polynomial and a differential operator such that f(D) y = R(x) • Let the roots of the auxiliary equation f(m) = 0 be m = m1, m2, …, mn. • The general solution of f(D) y = R(x) is y = yc + yp

Method of Undetermined Coefficients • yc can be obtained from the values of m in f(m) = 0. • Now, suppose that R(x) is itself a particular solution of some homogeneous linear differential equation with constant coefficients g(D) R = 0 with auxiliary equation roots m’ = m’1, m’2, m’3, …, m’k

Note that g(D) “annihilates” R(x); leading to the right side of the equation becoming zero. Method of Undetermined Coefficients • Recall that the values of m’ can be obtained by inspection from R(x). • From the values of m’, g(D) can be constructed. • Now, multiply g(D) to both sides of f(D)y = R(x): f(D) y = R(x) g(D) f(D) y = g(D) R(x) g(D) f(D) y = 0

Method of Undetermined Coefficients • The differential equation g(D) f(D) y = 0 has as the roots of its auxiliary equation the values of m and m’. • Hence, the solution of g(D) f(D) y = 0 is of the form y = yc + yq • ycis the solution obtained from the m values. • yq is the solution obtained from the m’ values.

Method of Undetermined Coefficients • But any particular solution of f(D) y = R(x) must also be a solution of g(D) f(D) y = 0. • In other words, from f(D) ( yc + yp) = R(x) and g(D) f(D) (yc + yq) = 0, we can get yp from yq by determining the correct numerical values of the coefficients of yq to make it equivalent to yp.

Example 1 Obtain the general solution of (D2 + D)y = – cosx

Solution Step 1: Get the roots of the auxiliary equation. (m2 + m) = 0 m (m + 1) = 0 m = 0 & m = -1 Step 2: Get the complementary function yc. yc = c1e0x + c2e-x or yc = c1 + c2e-x

Solution Step 3: Get g(D) from R(x). R(x) = -cosx g(D) = D2 + 1 g(D) = [D – (0 + i)] [D – (0 – i)] Step 4: Get the FORM of ypfrom g(D). yp = Ae0xcosx + Be0xsinx or yp = Acosx + Bsinx m’ = i

Solution Step 5: Substitute yp into the given differential equation f(D)y = R(x) and simplify. yp = Acosx + Bsinx yp’ = – Asinx + Bcosx yp’’ = – Acosx – Bsinx (D2 + D)y = – cosx (D2 + D)yp = – cosx

Solution Step 5: Substitute yp into the given differential equation f(D)y = R(x) and simplify. (D2 + D)yp = – cosx D2yp + Dyp = – cosx (– Acosx – Bsinx) + (– Asinx + Bcosx) = – cosx

Solution Step 6: Determine (solve for) the coefficients. (– Acosx – Bsinx) + (– Asinx + Bcosx) = – cosx – Acosx + Bcosx = – cosx – A + B = – 1 – Bsinx – Asinx = 0 – B – A = 0

Solution

Example 2 Obtain the general solution of (D2 – 4)y = e2x + 2

Solution Step 1: Get the roots of the auxiliary equation. (m2 – 4) = 0 (m – 2) (m + 2) = 0 m = 2 & m = – 2 Step 2: Get the complementary function yc. yc = c1e2x + c2e-2x

Solution Step 3: Get g(D) from R(x). R(x) = 2 + e2x g(D) = D (D – 2) Step 4: Get the FORM of ypfrom g(D). yp = Ae0x + Bxe2x yp = A + Bxe2x Note! The root “2” appeared twice: once in yc and once in yp. Hence, we write Bxe2x. It is incorrect to use Be2x. m’1 = 0 & m’2 = 2

Solution Step 5: Substitute yp into the given differential equation f(D)y = R(x) and simplify. yp = A + Bxe2x yp’ = B(e2x + 2xe2x) = Be2x + 2Bxe2x yp’’ = 2Be2x + 2B(e2x + 2xe2x) = 4Be2x + 4Bxe2x

Solution Step 5: Substitute yp into the given differential equation f(D)y = R(x) and simplify. (D2 – 4)yp = e2x + 2 D2yp – 4yp = e2x + 2 (4Be2x + 4Bxe2x) – 4 (A + Bxe2x) = e2x + 2 4Be2x – 4A = e2x + 2

Solution Step 6: Determine (solve for) the coefficients. 4Be2x – 4A = e2x + 2 4B = 1  B = ¼ – 4A = 2  A = – ½

Solution

Example 3 Obtain the general solution of y’’ – 4y’ + 3y = 20cosx

Solution Step 1: Get the roots of the auxiliary equation. (m2 – 4m + 3) = 0 (m – 3) (m – 1) = 0 m = 3 & m = 1 Step 2: Get the complementary function yc. yc = c1e3x + c2ex

Solution Step 3: Get g(D) from R(x). R(x) = 20cosx g(D) = D2 + 1 g(D) = (D – i)(D + i) Step 4: Get the FORM of ypfrom g(D). yp = Ae0xcosx + Be0xsinx yp = Acosx + Bsinx m’ = i

Solution Step 5: Substitute yp into the given differential equation f(D)y = R(x) and simplify. yp = Acosx + Bsinx yp’ = A(– sinx) + Bcosx yp’= – Asinx + Bcosx yp’’ = – Acosx – Bsinx

Solution Step 5: Substitute yp into the given differential equation f(D)y = R(x) and simplify. (D2 – 4D + 3)yp = 20cosx D2yp – 4Dyp + 3yp = 20cosx (– Acosx – Bsinx ) – 4(– Asinx + Bcosx) + 3(Acosx + Bsinx) = 20cosx (2Acosx – 4Bcosx) + (2Bsinx + 4Asinx) = 20cosx

Solution Step 6: Determine (solve for) the coefficients.

Example & Exercises Obtain the general solution: • (D2 + D)y = – cosx (solved already) • (D2 + 3D + 2)y = 12x2 • (D2 + 9)y = 5ex – 162x • y’’ – 3y’ – 4y = 30ex • (D2 – 4)y = e2x + 2 (solved already) • y’’ – 4y’ + 3y = 20cosx (solved already) • y’’ + 2y’ + y = 7 + 75sin2x

Example & Exercises Obtain the general solution: • (D2 + 1)y = cosx • (D2 – 1)y = e-x(2sinx + 4cosx) • (D3 – D)y = x • (D3 + D2 – 4D – 4)y = 3e-x – 4x – 6 • (D4 – 1)y = e-x • (D2 + 1)y = 12cos2x • y’’ – 3y’ – 4y = 16x – 50cos2x

Example & Exercises Obtain the general solution: • y’’ + 4y’ + 3y = 15e2x + e-x • y’’ – y’ – 2y = 6x + 6e-x • (D3 – 3D2 + 4)y = 6 + 80cos2x • (D3 + D2 – 4D – 4)y = 8x + 8 + 6e-x

The Method of Undetermined Coefficients (MUC)