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.

a2

ac4bbm

0cbmam

2

2

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

xm2

xm1

21 eyandey

21 yyy

xm2

xm1

21 ececy

Example

Find the general solution of

(D2 + D – 6) y = 0

x2

2x3

1

2

2

ececy

,Hence

2m|3m

0)2m)(3m(

06mm

isequationauxiliarythe

,0y)6DD(From

:Solution

Case 2: Real Repeated Roots

• Having two real, repeated roots means

• Now, one solution isxm

11ey

1

2

2

ma2

bm

a2

ac4bbm

0cbmam

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

dx)x(y

e)x(yy

21

dx)x(P

12

Case 2: Real Repeated Roots

Hence,

)x(ey

dx)e(

eey

dx)e(

eey

dx)x(y

e)x(yy

xm2

2xm

xm2xm

2

2xm

dx)m2(xm

2

21

dx)x(P

12

1

1

11

1

11

Case 2: Real Repeated Roots

The general solution is then

xm2

xm1

21

11 xececy

yyy

Example

Find the general solution of

y’’ + 8y’ + 16y = 0

x42

x41

2

2

xececy

,Hence

)twice(4m

0)4m(

016m8m

isequationauxiliarythe

,0y16'y8''yFrom

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 ix = cosx + isinx

e- ix = cosx - isinx

• Note that

e ix + e- ix = 2cosx &

e ix – e- ix = 2isinx

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

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

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)

Example

Find the general solution of

(D2 – 4D + 7) y = 0

x3sinecx3cosecy

,Hence

i32m

2

122m

)1(2

)7)(1(4)4()4(m

,Then

07m4m

isy)7D4D(ofequationauxiliaryThe

:Solution

x22

x21

2

2

2

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 xm

nxm

2xm

1n21 ec...ececy

Example

Find the general solution of

(D3 + 6D2 + 11D + 6) y = 0

x2

3x3

2x

1

23

ecececy

,Hence

2m|3m|1m

0)2m)(3m)(1m(

,Then

0)6m11m6m(

isequationauxiliaryThe

:Solution

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

x34321

x34

x33

x02

x01

22

22

234

e)xcc(xccy

or

xececxececy

,Hence

)twice(3m

and)twice(0m

0)3m(m

0)9m6m(m

0m9m6m

:Solution

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

x3sin)xcc(x3cos)xcc(y

or

x3sine)xcc(x3cose)xcc(y

,Hence

)twice(i3m

0)9m(

081m18m

:Solution

4321

x043

x021

22

24

Exercises

Find the solution required:

1) (D2 – 2D – 3)y = 0 y(0)=0; y’(0)=-4

2) (D3 – 4D)y = 0 y(0)=0; y’(0)=0; y’’(0)=2

3) (D4 + 2D3 + 10D2)y = 0

4) (D6 + 9D4 + 24D2 + 16)y = 0

5) (D3 + 7D2 + 19D + 13)y = 0 y(0)=0; y’(0)=2;

y’’(0)=-12

6) (4D4 + 4D3 – 3D2 – 2D + 1)y = 0

7) (D4 – 5D2 – 6D – 2)y = 0

Exercises

Find the solution required:

8) (D3 + D2 – D – 1)y = 0

y(0)=1; y(2)=0;

9) 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

0ylimx