Nonhomogeneous Linear Differential Equations

Presentation on theme: "Nonhomogeneous Linear Differential Equations"— Presentation transcript:

1 Nonhomogeneous Linear Differential Equations
AP Calculus BC

2 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.

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

4 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)

5 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.

6 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

7 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 

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

9 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 

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

11 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

12 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

13 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

14 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”.