Nonhomogeneous Linear Differential Equations AP Calculus BC

Nonhomogeneous Differential Equations Recall that second order linear differential equations with constant coefficients have the form: Now we will solve equations where G(x) ≠ 0, which are non-homogeneous differential equations.

Complementary Equation The related homogeneous equation is called the complementary equation, and it is part of the general solution to the nonhomogeneous equation.

General Solution The general solution to a nonhomogenous differential equation is y(x) = yp(x) + yc(x) where yp is a particular solution to the nonhomogenous equation, and yc is the general solution to the complementary equation. Proof for Grant yc(x) = y(x) – yp(x)

Method of Undetermined Coefficients There are two methods for solving nonhomogeneous equations: Method of Undetermined Coefficients Variation of Parameters First, we will learn about the Method of Undetermined Coefficients to solve the equation ayʹʹ + byʹ + cy = G(x) when G(x) is a polynomial. Since G(x) is a polynomial, yp is also a polynomial of the same degree as G. Therefore, we substitute yp = a polynomial (of the same degree as G) into the equation and determine the coefficients.

Example 1 Solve the equation yʹʹ + yʹ – 2y = x2. The auxiliary equation is r2 + r – 2 = 0 Factor  (r – 1)(r + 2) = 0 r = 1, r = –2 Solution of complementary equation is yc = c1ex + c2e–2x Since G(x) = x2, we want a particular solution where yp(x) = Ax2 + Bx + C So ypʹ(x) = 2Ax + B and ypʹʹ(x) = 2A

Example 1 (continued) Substitute these into the differential equation  (2A) + (2Ax + B) – 2(Ax2 + Bx + C) = x2 = – 2Ax2 + (2A – 2B)x + (2A + B – 2C) = 1x2 + 0x + 0 –2A = 1  2A – 2B = 0 

Example 1 (FINAL) Therefore, our particular solution is So our general solution is: y = yc + yp = c1ex + c2e–2x

Example 2 Solve yʹʹ + 4y = e3x When G(x) is of the form ekx, we use yp = Aekx because the derivatives of ekx are constant multiples of ekx and work out nicely. Complementary equation is r2 + 4 = 0  Therefore, yc = c1 cos 2x + c2 sin 2x Solve for yp: y = Ae3x yʹ = 3Ae3x yʹʹ = 9Ae3x Substitute 

Example 2 (continued) 13A = 1 A = 1/13 General solution is

Example 3 Solve the equation yʹʹ + yʹ – 2y = sin x. When G(x) is of the form C sin kx or C cos kx, we use yp = A cos kx + B sin kx Complementary equation is r2 + r – 2 = 0  r = –2, 1  yp = A cos x + B sin x ypʹ = –A sin x + B cos x ypʹʹ = –A cos x – B sin x

Example 3 (continued) Substitute back into original equation: (–A cos x – B sin x) + (–A sin x + B cos x) – 2(A cos x + B sin x) = sin x (–3A + B) cos x + (–A – 3B) sin x = sin x Therefore, –3A + B = 0 and –A – 3B = 1 Solve as a system  General solution is

Review and More Rules for Method of Undetermined Coefficients Form is ayʹʹ + byʹ + cy = G(x) 1. If G(x) is a polynomial, use yp = Axn + Bxn–1 + … + C. 2. If G(x) = Cekx, use yp = Aekx . 3. If G(x) = C sin kx or C cos kx, use yp = A cos kx + B sin kx 4. If G(x) is a product of functions, multiply them for yp. Example: G(x) = x cos 3x  yp = (Ax + B) cos 3x + (Cx + D) sin 3x

Example: G(x) = xex + cos 2x 5. If G(x) is a sum of functions, find separate particular solutions and add them together at the end. Example: G(x) = xex + cos 2x Use yp1 = (Ax + B)ex and yp2 = C cos 2x + D sin 2x Then add  y(x) = yc + yp1 + yp2 6. If yp is a solution to the complementary equation (yc), multiply yp by x or x2, so yc and yp are linearly independent. 7. The particular solutions we try to find using yp (the ones with the A, B, C, etc. in them) are called “trial solutions”.