AMS361 Recitation 03/04 Notes 7
1
Wronskian
y1
y2
···
yn
0
0
y10
·
·
·
y
y
n
2
W (y1 , · · · , yn ) = ..
..
..
..
.
.
.
.
(n−1)
(n−1)
(n−1)
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 3.2.9
Question: Check if the following three functions are linearly independent
ex , cos(x), sin(x)
Solution: Calculate the Wronskian of the three functions
x
e
sin(x) x cos(x)
W (y1 , y2 , y3 ) = e −sin(x) cos(x) ex −cos(x) −sin(x) = ex (sin2 (x) + cos2 (x)) − ex (−cos(x)sin(x) + cos(x)sin(x))
+ ex (cos2 (x) + sin2 (x))
= ex ∗ (1) − ex ∗ (0) + ex ∗ (1) = 2ex 6= 0
So, they are linearly independent.
2
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.
1
– 1. Distinct Roots r1 , r2 , ..., rn . The general solution is given by
y(x) = C1 er1 x + C2 er2 x + · · · + Cn ern x .
– 2. m equal roots r1 = r2 = · · · = rm = r. The general solution is given by
y(x) = (C1 + C2 x + · · · + Cm xm−1 )erx .
– 3. 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 3.3.20
Question: Find the general solutions of the differential equation.
y (4) + 2y (3) + 3y 00 + 2y 0 + y = 0
Solution: The characteristic equation is
r4 + 2r3 + 3r2 + 2r + 1 = 0
⇒ (r2 + r + 1)2 = 0
√
√
3
3
1
1
i, r3 = r4 = − −
i
⇒ r1 = r2 = − +
2
2
2
2
This is the combination of case 2 and case 3. The general solution is given by
√
√
3x
3x
x/2
y(x) = e ((c1 + c2 x)cos(
) + (c3 + c4 x)sin(
)).
2
2
2.2
Problem 3.3.27
Question: Find the general solution of the equation. First find a small integral root of the
characteristic equation by inspection; then factor by division.
y (3) + 3y 00 − 4y = 0
Solution: The characteristic equation is
r3 + 3r2 − 4 = 0
⇒ (r − 1)(r + 2)2 = 0
⇒ r1 = 1, r2 = r3 = −2
This is the combination of case 1 and case 2. The general solution is given by
y(x) = c1 ex + (c2 + c3 x)e−2x .
2
2.3
Problem 3.3.34
Question: One solution of the differential equation is given. Find the general solution.
3y (3) − 2y 00 + 12y 0 − 8y = 0, y = e2x/3
Solution: Easy to know r =
2
3
is a root of the characteristic equation
3r3 − 2r2 + 12r − 8 = 0
⇒ (3r − 2)(r2 + 4) = 0
2
⇒ r1 = , r2 = 2i, r3 = −2i
3
This is the combination of case 1 and case 3. The general solution is given by
y(x) = c1 e2x/3 + c2 cos(2x) + c3 sin(2x).
2.4
Problem 3.3.41
Question: Find a linear homogeneous constant-coefficient equation with the given general
solution.
y(x) = Acos(2x) + Bsin(2x) + Ccosh(2x) + Dsinh(2x)
Solution: Observing the formula of the general solution, we know the characteristic equation
has following roots.
r1 = 2i, r2 = −2i, r3 = 2, r4 = −2
⇒ (r2 + 4)(r + 2)(r − 2) = 0
Therefore, the differential equation is
(D2 + 4)(D + 2)(D − 2)y = 0
⇒ y (4) − 16y = 0
3

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Official solutions for Term Test 1