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**Homogeneous Linear Differential Equations with Constant Coefficients**

A Lecture in ENGIANA

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**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

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**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.

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**Auxillary Equation There are three possible cases:**

m1 m2; distinct real roots m1 = m2; repeated real roots m1 m2; conjugate complex roots

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**Case 1: Distinct Real Roots**

For this case, we have And hence, Or

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

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**Case 2: Real Repeated Roots**

Having two real, repeated roots means Now, one solution is

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**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)

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**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

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**Case 2: Real Repeated Roots**

Recall also that another solution y2 is

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**Case 2: Real Repeated Roots**

Hence,

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**Case 2: Real Repeated Roots**

The general solution is then

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

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**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

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**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

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**Case 3: Conjugate Complex Roots**

Thus, we have e ix = cosx + isinx e- ix = cosx - isinx Note that e ix + e- ix = 2cosx & e ix – e- ix = 2isinx

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**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(eix + e-ix) y1 = e x(2cosx) y1 = 2e xcosx

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**Case 3: Conjugate Complex Roots**

If we let C1 = 1 and C2 = -1: y2 = e (+i)x - e(-i)x y2 = e x(eix - e-ix) y2 = e x(2isinx) y2 = 2ie xsinx

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**Case 3: Conjugate Complex Roots**

Thus, the solution to y = C1e( + i)x + C2e( - i)x is y = c1y1 + c2y2 y = c1(e xcosx) + c2(e xsinx) or y = e x(c1cosx + c2sinx)

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

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**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

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**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

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

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**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

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**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

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**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

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

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**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

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

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**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

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**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

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