1. SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS AP Calculus BC

2. ORDER OF A DIFFERENTIAL EQUATION  The order of a differential equation is the order of the highest derivative.  y ʹʹʹ – 2xy ʹ + y = sin x is a 3 rd order differential equation.

3. SECOND ORDER DIFFERENTIAL EQUATIONS  Second order linear differential equations have the form:  If G(x) = 0, it is a homogeneous differential equation.  If G(x) ≠ 0, it is a non-homogeneous differential equation.

4. USEFUL FACT #1 If y 1 and y 2 are both solutions to a linear homogeneous equation, and c 1 and c 2 are any constants, then the function y(x) = c 1 y 1 (x) + c 2 y 2 (x) is also a solution. (In other words, the linear combination of y 1 and y 2 is also a solution.)

5. PROOF OF USEFUL FACT #1 If y 1 and y 2 are both solutions to the equation, then Now substitute c 1 y 1 (x) + c 2 y 2 (x) for y in the original equation to show that it is a solution:

6. USEFUL FACT #2  y 1 and y 2 are linearly independent if neither is a constant multiple of the other.  y 1 = 5x and y 2 = x  not linearly independent  y 1 = e x and y 2 = xe x  linearly independent

7. USEFUL FACT #3  If y 1 and y 2 are linearly independent solutions to a homogeneous differential equation, and P(x) ≠ 0, then the general solution is given by y(x) = c 1 y 1 (x) + c 2 y 2 (x), where c 1 and c 2 are arbitrary constants.  The general solution to the differential equation is a linear combination of two linearly independent solutions. This means if we know two linearly independent solutions, we know every solution.

8. SOLVING  For now, we will assume that P, Q, and R are constant functions, and we will call them a, b, and c, respectively, which gives us:  We are looking for a function whose second derivative times a constant plus its first derivative times another constant plus another constant times y is equal to 0.  A good candidate is y = e rx (where r is a constant)  So now ar 2 e rx + bre rx + ce rx = 0  Factor out e rx  e rx (ar 2 + br + c) = 0  Therefore, ar 2 + br + c = 0  This is called the characteristic equation (or auxiliary equation) of the differential equation.

9. ROOTS OF THE CHARACTERISTIC EQUATION  This means that y = e rx is a solution to the differential equation.  Find the roots of ar 2 + br + c = 0 by factoring or using the quadratic formula.  There are three cases: 1. b 2 – 4ac > 0  Two real rootsSolution is 2. b 2 – 4ac = 0  One real rootSolution is 3. b 2 – 4ac < 0  Two complex roots  r 1 = α + iβ r 2 = α – iβ General solution is y = e αx (c 1 cos βx + c 2 sin βx)

10. EXAMPLE 1 Solve the equation y ʹʹ + y ʹ – 6y = 0. The auxiliary equation is r 2 + r – 6 = 0 Factor  (r + 3)(r – 2) = 0 r = –3, r = 2 General solution is y = c 1 e –3x + c 2 e 2x You can check this by substituting back into the original equation.

11. EXAMPLE 2 Solve (use quadratic formula for this one) Characteristic equation is 3r 2 + r – 1 = 0 General solution is

12. EXAMPLE 3 Solve the equation 4y ʹʹ + 12y ʹ + 9y = 0 Auxiliary equation is 4r 2 + 12r + 9 = 0 Factor  (2r + 3) 2 = 0 Only root is r = –3/2, so general solution is: y = c 1 e –3/2x + xc 2 e –3/2x

13. EXAMPLE 4 Solve y ʹʹ – 6y ʹ + 13y = 0 Auxiliary equation is r 2 – 6r + 13 = 0 Can’t factor, non-real answer General solution is y = e 3x (c 1 cos 2x + c 2 sin 2x)

14. EXAMPLE 5 – INITIAL VALUE PROBLEM Solve the equation y ʹʹ + y ʹ – 6y = 0 when y(0) = 1 and y ʹ (0) = 0. This is the same as Example 1, so the general solution is y = c 1 e –3x + c 2 e 2x Substitute in values: 1 = c 1 + c 2 0 = –3c 1 + 2c 2 Solve as a system of equations  So the solution is

15. EXAMPLE 6 – ANOTHER INITIAL VALUE PROBLEM Solve the equation y ʹʹ + y = 0 when y(0) = 2 and y ʹ (0) = 3. r 2 + 1 = 0 r = ±i So α = 0 and β = 1 General solution is y = c 1 cos x + c 2 sin x Substitute in values: 2 = c 1 3 = c 2 So solution is y = 2 cos x + 3 sin x.