4.2 - 4.6

auxiliary equation/ characteristic equation

ar^2 + br + c = 0

homogeneous form of equation

ay" + by' + cy = 0

(Repeated Root) If the auxiliary equation ar^2 + br + c = 0 has a repeated root r, then

both y1(t)= e^(rt) and y2(t)=te^(rt) are solutions to ar^2 + br + c = 0, and y(t) = c1e^(rt) + c2te^(rt) is a general solution

the method of undetermined coefficients applies only to

nonhomogeneities that are polynomials, exponentials, sines or cosines, or products of these functions

Theorem 1 (Existence and Uniqueness: Homogeneous Case) For any real numbers a(!=0), b, c, t0, Y0, and Y1,

there exists a unique solution to the initial value problem ay" + by' + cy = 0; y(t0)= Y0, y'(t0)= Y1. The solution is valid for all t in (-inf, inf)

(Complex Conjugate Roots) If the auxiliary equation has complex conjugate roots a +- iB, then

two linearly independent solutions to ay" + by' + cy = 0 are e^(at)cos(Bt) and e^(at)sin(Bt), and a general solution is y(t) = c1e^(at)cos(Bt) + c2e^(at)sin(Bt), where c1 and c2 are arbitrary constants

(Proof) If the auxiliary equation ar^2 + br + c = 0 has a repeated root r, then both y1(t)= e^(rt) and y2(t)=te^(rt) are solutions to ar^2 + br + c = 0, and y(t) = c1e^(rt) + c2te^(rt) is a general solution

(1) If r is repeated, this means that (-b+-sqrt(b^2-4ac))/2a) has only one solution -> b^2-4ac = 0 => r = -b/(2a) => 2ar + b = 0 (2) Let's try y(t) = te^(rt) y'(t)= e^(rt) + rte^(rt); y"(t)= re^(rt) + re^(rt) + r^2te^(rt) = 2re^(rt) + r^2te^(rt) a[2re^(rt) + r^2te^(rt)] + b[e^(rt) + rte^(rt)]+c[te^(rt) ]= 0 [2ar + b]e^(rt) + [ar^2 + br + c]te^(rt) = 0 = zero ----- = zero since r is a root

(Proof) Let z(t) = u(t) + iv(t) be a solution to equation ay" + by' + cy = 0, where a, b, and c are real numbers. Then, the real part u(t) and the imaginary part v(t) are real-valued solutions of ay" + by' + cy = 0.

(1) az" + bz' + cz = 0 (2) a(u" + iv") + b(u' + iv') + c(u + iv) = 0 (au" + bu' + cu) + i(av" + bv' +cv) = 0 (3) But a complex number is zero if and only if its real and imaginary parts are both zeros. (4) => (au" + bu' + cu) = 0; (av" + bv' +cv) = 0 so we know that e^(a+iB)t = e^(at)cos(Bt) + i e^(at)sin(Bt) real --------- imaginary is a solution, and therefore, e^(at)cos(Bt) is a solution; e^(at)sin(Bt) is a solution you show that e^(a-iB)t would lead to the same pair of solutions

variation of parameter

(1) consider the nonhomogeneous linear second-order equation ay" + by' + cy = f(t) * (2) let y1(t) and y2(t) be two linearly independent solutions for the corresponding homogeneous equation ay" + by' + cy = 0 ** (3) yh(t) = c1y1(t) + c2y2(t) solves ** (4) to find a solution to * (5) let's look for solution of yp(t) = v1(t)y1(t) + v2(t)y(t) unknown v1(t) and v2(t)

(method of variation of parameter) to determine a particular solution to ay" + by' + cy = f(t)

(1) find two linearly independent solutions y1(t) and y2(t) to the corresponding homogeneous equation and take yp(t) = v1(t)y1(t) + v2(t)y2(t) (2) determine v1(t) and v2(t) by solving the system in v1'y1 + v2'y2 = 0 for v1'(t) and v2'(t) and integrating (3) substitute v1(t) and v2(t) into the expression for yp(t) to obtain a particular solution

method of undetermined coefficients

(1) guess a solution of the form yp(t) = Amt^m + ... + A1t + A0 (2) using the guess to help us build the solution --- a simple procedure for determining a particular solution when the equation has constant coefficients and the nonhomogeneous term is of a special type.

let's start around

1) yp(t) = v1(t)y1(t) + v2(t)y2(t) 2) y'p(t) = (v1'y1 + v2'y2) + (v1y1' + v2y2') 3) let's try requiring that v1'y1 + v2'y2 = 0 4) then, y'p(t) = v1y1' + v2y2' 5) so, y"p(t) = v1'y1' + v1y1" + v2'y2' + v2y2" 6) f = ay" + by' + cy = a[v1'y1' + v1y1" + v2'y2' + v2y2"] + b[v1y1' + v2y2'] + c[v1y1 + v2y2] = a[v1'y1' + v2'y2'] + v1[ay1" + by1' + cy1] + v2[ay2" + by2' + cy2] =a[v1'y1' + v2'y2'] 7) f/a = v1'y1' + v2'y2 8)~~

variation of parameter

a more general method of finding a particular solution to ay" + by' + cy = f

(Proof) Let y1 be a solution to the differential equation ay" + by' + cy = f1(t), and y2 be a solution to ay" + by' + cy = f2(t). Then for any constants k1 and k2, the function k1y1 + k2y2 is a solution to the differential equation ay" + by' + cy = k1f1(t) + k2f2(t)

a(k1y1 + k2y2)" + b(k1y1 + k2y2)' + c(k1y1 + k2y2) = k1( ay1" + by1' + cy1) + k2(ay2" + by2' + cy2) = k1f1(t) + k2f2(t)

nonhomogeneity

ay" + by' + cy = f(t)

(Distinct Real Roots) If the auxiliary equation ar^2 + br + c = 0 has distinct real roots r1 and r2, then

both y1(t)= e^(r1t) and y2(t)=e^(r2t) are solutions to ay" + by' + cy = 0 and y(t)= c1e^(r1t) + c2e^(r2t) is a general solution

Theorem 3 (Superposition Principle) Let y1 be a solution to the differential equation ay" + by' + cy = f1(t), and y2 be a solution to ay" + by' + cy = f2(t). Then

for any constants k1 and k2, the function k1y1 + k2y2 is a solution to the differential equation ay" + by' + cy = k1f1(t) + k2f2(t)

Def 1 (Linear Independence of Two Functions) A pair of functions y1(t) and y2(t) is said to be linearly independent on the interval I if and only if

neither of them is a constant multiple of the other on all of I.

y1 and y2 are linearly dependent on I if

one of them is a constant multiple of the other on all of I.

Lemma 2 (Real Solutions Derived from Complex Solutions) Let z(t) = u(t) + iv(t) be a solution to equation ay" + by' + cy = 0, where a, b, and c are real numbers. Then,

the real part u(t) and the imaginary part v(t) are real-valued solutions of ay" + by' + cy = 0.

Theorem 2 (Representation of Solutions to Initial Value Problem) If y1(t) and y2(t) are any two solutions to the differential equation ay" + by' + cy = 0 that are linearly independent on (-inf, inf), then

unique constants c1 and c2 can always be found so that c1y1(t) + c2y2(t) satisfies the initial value problem ay" + by' + cy = 0; y(t0)= Y0, y'(t0)= Y1 on (-inf, inf).

to find a particular solution to the differential equation ay" + by' + cy = Ct^me^(at)cos Beta t or Ct^me^(at)sin Beta t for Beta != 0,

use the form yp(t) = t^s(Amt^m + ... + A1t + A0)e^(at)cos (Beta) t + t^s(Bmt^m + ... + B1t + B0)e^(at)sin Beta t, with (iv) s =0 if a + i Beta is not a root of the associated auxiliary equation; (v) s =1 if a + i Beta is a root of the associated auxiliary equation

(method of undetermined coefficients) to find a particular solution to the differential equation ay" + by' + cy = Ct^me^(rt), where m is a nonnegative integer,

use the form yp(t) = t^s(Amt^m + ... + A1t + A0)e^(rt), with (i) s =0 if r is not a root of the associated auxiliary equation; (ii) s =1 if r is a simple root of the associated auxiliary equation; (iii) s =2 if r is a double root of the associated auxiliary equation;

a general solution to the nonhomogeneous equation ay" + by' + cy = f(t)

y = yp + c1y1 + c2y2

(proof)

y = yp + c1y1 + c2y2 is a general solution to the nonhomogeneous equation ay" + by' + cy = f(t) since any solution yg(t) can be expressed in this form. we simply pick c1 and c2 so that yp + c1y1 + c2y2 matches the value and the derivative of yg at any point; by uniqueness, yp + c1y1 + c2y2 and yg have to be the same function

if y1 and y2 are linearly independent solutions to ay" + by' + cy = 0 on (-inf, inf), then

y(t)= c1y1(t) + c2y2(t) is a general solution