Homogeneous Linear Differential Equations with Constant Coefficients

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Homogeneous Linear Differential Equations with Constant Coefficients
A Lecture in ENGIANA

Auxillary Equation Consider a second order equation
ay’’ + by’ + cy = 0 where a, b, and c are constants. If we try to find a solution of the form y = emx, then after substitution of y’ = memx and y’’ = m2emx, the equation becomes am2emx + bmemx + cemx = 0

Auxillary Equation Solving am2emx + bmemx + cemx = 0,
emx(am2 + bm + c) = 0 The quantity in parenthesis, a quadratic equation, is called the auxiliary equation. This means that to find the solution y (see previous slide), we must solve for m.

Auxillary Equation There are three possible cases:
m1  m2; distinct real roots m1 = m2; repeated real roots m1  m2; conjugate complex roots

Case 1: Distinct Real Roots
For this case, we have And hence, Or

Example Find the general solution of (D2 + D – 6) y = 0

Case 2: Real Repeated Roots
Having two real, repeated roots means Now, one solution is

Case 2: Real Repeated Roots
Recall that a2(x)y’’ + a1(x)y’ + a0(x)y = 0 can be written as y” + P(x)y’ + Q(x)y = 0 where P(x) = a1(x)/a2(x) Q(x) = a0(x)/a2(x)

Case 2: Real Repeated Roots
In our case, the coefficients are constants: ay’’ + by’ + cy = 0 Thus, y” + Py’ + Qy = 0 where P = b/a Q = c/a

Case 2: Real Repeated Roots
Recall also that another solution y2 is

Case 2: Real Repeated Roots
Hence,

Case 2: Real Repeated Roots
The general solution is then

Example Find the general solution of y’’ + 8y’ + 16y = 0

Case 3: Conjugate Complex Roots
If m1 and m2 are complex, then we have m1 =  + i m2 =  - i where  and  are real and positive Hence, we can write y = C1e( + i)x + C2e( - i)x

Case 3: Conjugate Complex Roots
However, in practice we prefer to work with real functions instead of complex exponentials. To this end, we use Euler’s formula: ei = cos + isin where  is any real number

Case 3: Conjugate Complex Roots
Thus, we have e ix = cosx + isinx e- ix = cosx - isinx Note that e ix + e- ix = 2cosx & e ix – e- ix = 2isinx

Case 3: Conjugate Complex Roots
Our solution is then y = C1e (+i)x + C2e(-i)x If we let C1 = 1 and C2 = 1: y1 = e (+i)x + e(-i)x y1 = e x(eix + e-ix) y1 = e x(2cosx) y1 = 2e xcosx

Case 3: Conjugate Complex Roots
If we let C1 = 1 and C2 = -1: y2 = e (+i)x - e(-i)x y2 = e x(eix - e-ix) y2 = e x(2isinx) y2 = 2ie xsinx

Case 3: Conjugate Complex Roots
Thus, the solution to y = C1e( + i)x + C2e( - i)x is y = c1y1 + c2y2 y = c1(e xcosx) + c2(e xsinx) or y = e x(c1cosx + c2sinx)

Example Find the general solution of (D2 – 4D + 7) y = 0

Higher-Order (n>2) Equations: Distinct Roots
Consider the case where the auxiliary equation has distinct roots. Say we are given f(D)y = 0. Then one possible solution is emx, f(D)emx = 0, if the auxiliary equation is f(m) = 0

Higher-Order (n>2) Equations: Distinct Real Roots
In other words, if the distinct roots of the auxiliary equation are m1, m2, …, mn, then the corresponding solutions are exp(m1x), exp(m2x), …, exp(mnx). The general solution is

Example Find the general solution of (D3 + 6D2 + 11D + 6) y = 0

Higher-Order (n>2) Equations: Repeated Real Roots
Consider the case where the auxiliary equation has repeated roots. Say we are given f(D)y = 0. If there are several identically repeated roots m1 = m2 = … = mn = b, then this means (D - b)n y = 0

Higher-Order (n>2) Equations: Repeated Roots
If we let y = xkebx [k = 0, 1, 2, …, (n-1)] Then, (D – b)n y = (D – b)n [xkebx] But (D – b)n [xkebx] = ebxDn[xk] = ebx (0) Thus, (D – b)n y = (D – b)n [xkebx] = 0

Higher-Order (n>2) Equations: Repeated Roots
The functions yk = xkebx [e.g., e7x, xe7x, x2e7x, etc.], where k = 0, 1, 2, …, (n – 1) are linearly independent because, aside from the common factor ebx, they contain only the respective powers x0, x1, x2, …, xn-1. The general solution is thus y = c1ebx + c2xebx + … + cnxn-1ebx

Example Find the general solution of (D4 + 6D3 + 9D2) y = 0

Higher-Order (n>2) Equations: Repeated Imaginary Roots
Repeated imaginary roots lead to solutions analogous to those brought in by repeated real roots. For instance, if the conjugate pair m = a  bi occur three times, the corresponding general solution is y = (c1 + c2x + c3x2) eaxcosbx + (c4 + c5x + c6x2) eaxsinbx

Example Find the general solution of (D4 + 18D2 + 81) y = 0

Exercises Find the solution required:
(D2 – 2D – 3)y = 0 y(0)=0; y’(0)=-4 (D3 – 4D)y = y(0)=0; y’(0)=0; y’’(0)=2 (D4 + 2D3 + 10D2)y = 0 (D6 + 9D4 + 24D2 + 16)y = 0 (D3 + 7D2 + 19D + 13)y = 0 y(0)=0; y’(0)=2; y’’(0)=-12 (4D4 + 4D3 – 3D2 – 2D + 1)y = 0 (D4 – 5D2 – 6D – 2)y = 0

Exercises Find the solution required: (D3 + D2 – D – 1)y = 0
y(0)=1; y(2)=0; Find for x = 2 the y value for the particular solution required: (D3 + 2D2)y = 0 y(0)=-3; y’(0)=0; y’’(0)=12