AMS361 Recitation 03/04 Notes 6
1
Wronskian
y1
y2 · · · yn
0
0
0
y1
y2 · · · yn
W (y1 , · · · , yn ) = ..
..
..
..
.
.
.
.
(n) (n)
(n)
y
y2
· · · yn
1
n×n
For n functions y1 , y2 , · · · , yn , if the Wronskian W (y1 , · · · , yn ) 6= 0, then they are linearly independent. Conversely, if y1 , y2 , · · · , yn are linearly dependent, then the Wronskian
W (y1 , · · · , yn ) = 0.
1.1
Problem 3.1.20
Question: Determine if f (x) and g(x) are linearly independent.
f (x) = π, g(x) = cos2 (x) + sin2 (x)
Solution: Easy to know f 0 (x) = 0 and g 0 (x) = 0, because both of them are constant. So
the Wronskian is
f (x) g(x) W (f (x), g(x)) = 0
f (x) g 0 (x) = f (x)g 0 (x) − f 0 (x)g(x) = 0
Therefore, they are not linearly independent.
1.2
Problem 3.2.6
Question: Check if the following three functions are linearly independent
ex , cosh(x), sinh(x)
Solution: Calculate the Wronskian of the three functions
x
e cosh(x) sinh(x) x
W (y1 , y2 , y3 ) = e sinh(x) cosh(x) ex cosh(x) sinh(x) = ex (sinh2 (x) − cosh2 (x)) − ex (cosh(x)sinh(x) − cosh(x)sinh(x))
+ ex (cosh2 (x) − sinh2 (x))
= ex ∗ (−1) − ex ∗ (0) + ex ∗ (1) = 0
So, they are not linearly independent.
1
2
Second Order, Constant Coefficient DEs
General formula
ay 00 + by 0 + cy = 0
Characteristic equation
ar2 + br + c = 0
There are three cases for the roots of the characteristic equation.
• 1. b2 − 4ac > 0 There are two distinct roots r1 and r2 . The general solution is given
by
y(x) = C1 er1 x + C2 er2 x .
• 2. b2 − 4ac = 0 There are two equal roots r1 = r2 = r. The general solution is given
by
y(x) = (C1 + C2 x)erx .
• 3. b2 − 4ac < 0 There are two complex roots r1 = α + iβ and r2 = α − iβ. The general
solution is given by
y(x) = eαx (c1 cos(βx) + c2 sin(βx)).
2.1
Problem 3.1.34
Question: Find the general solution of the following DE
y 00 + 2y 0 − 15y = 0
Solution: The characteristic equation is
r2 + 2r − 15 = 0
⇒r1 = 3, r2 = −5
⇒y = C1 e3x + C2 e−5x
2.2
Problem 3.1.40
Question: Find the general solution of the following DE
9y 00 − 12y 0 + 4y = 0
Solution: The characteristic equation is
9r2 − 12r + 4 = 0
2
⇒r1 = r2 = r =
3
2
⇒y = (C1 + C2 x)e 3 x
2
2.3
Problem 3.1.11
Question: Find the general solution of the following DE
y 00 − 2y 0 + 2y = 0
Solution: The characteristic equation is
r2 − 2r + 2 = 0
⇒r1 = 1 + i, r2 = 1 − i
⇒y = ex (C1 cos(x) + C2 sin(x))
3
Higher Order, Constant Coefficient DEs
The general idea is similar to the second order DE.
• step 1: write down the characteristic equation.
• step 2: solve the characteristic equation for r.
• step 3: find the general solution according to the three different cases.
3.1
Extra Problem 1
Question: Find the general solution of the following DE
3y 000 + 2y 00 = 0
Solution: The characteristic equation is
3r3 + 2r2 = 0
⇒r1 = r2 = 0, r3 = −
2
3
− 23 x
⇒y = (C1 + C2 x)e0x + C3
3.2
Extra Problem 2
Question: Find the general solution of the following DE
y (4) = y 000 + y 00 + y 0 + 2y
Solution: The characteristic equation is
r4 = r3 + r2 + r + 2
⇒(r + 1)(r − 2)(r2 + 1) = 0
⇒r1 = −1, r2 = 2, r3 = i, r4 = −i
⇒y = C1 e−x + C2 e2x + e0x (C3 cos(x) + C4 sin(x))
3
3.3
Extra Problem 3
Question: Find the general solution of the following DE
y 4 + 8y 00 + 16y = 0
Solution: The characteristic equation is
r4 + 8r2 + 16 = 0
⇒(r2 + 4)2 = 0
⇒r1 = r2 = 2i, r3 = r4 = −2i
⇒y = (C1 + C2 x)cos(2x) + (C3 + C4 x)sin(2x)
4