Homogenoues Higher Order DE
Homogenoues Higher Order DE
Homogenoues Higher Order DE
Equations
with Constant Coefficients
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.
2
am bm c 0
2
b b 4ac
m
2a
Auxillary Equation
(D2 + D – 6) y = 0
Solution :
2
From ( D D 6)y 0,
the auxiliary equation is
2
m m6 0
(m 3)(m 2) 0
m 3 | m 2
Hence,
3x 2x
y c1e c 2e
Case 2: Real Repeated Roots
Case 2: Real Repeated Roots
• Having two real, repeated roots means
2
am bm c 0
2
b b 4ac
m
2a
b
m m1
2a
m1x
• Now, one solution is y1 e
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
P ( x )dx
e
y 2 y 1 ( x) 2
dx
y1 ( x)
Case 2: Real Repeated Roots
P ( x )dx
Hence, e
y 2 y 1 ( x) 2
dx
y1 ( x)
( 2m1 )dx
e
m1 x
y2 e m1 x 2
dx
(e )
2 m1 x
e
(em x )2 dx
m1 x
y2 e
1
m1 x
y2 e (x)
Case 2: Real Repeated Roots
The general solution is then
y y1 y 2
m1 x m1 x
y c1e c 2 xe
Example
Find the general solution of
(D2 – 4D + 7) y = 0
Solution :
2
The auxiliary equation of ( D 4D 7)y is
m 2 4m 7 0
Then,
( 4) ( 4) 2 4(1)(7)
m
2(1)
12
m 2
2
m 2 3i
Hence,
y c1e 2x cos 3x c 2e 2 x sin 3x
Higher-Order (n>2) Equations:
Distinct Roots
Consider the case where the auxiliary
equation has distinct roots.