Download Quiz 13 1 Reduction of order

MA 262: Linear Algebra and Differential Equations
Fall 2011, Purdue
Quiz 13
1
Reduction of order
Use the method of reduction of order to determine a second linearly independent solution and write the
general solution.
We know that y1 (x) = x2 is a solution to this differential equation:
x2 y ′′ − 3xy ′ + 4y = 0,
x > 0.
(1.1)
Solution: We let
y(x) = x2 u(x),
(1.2)
where u(x) is to be determined. Differentiating y(x) twice yeilds
y ′ = 2xu + x2 u′ ,
(1.3)
y ′′ = x2 u′′ + 4xu′ + 2u.
(1.4)
Substituting into Eq.(1.1) and collecting terms, we obtain the following differential equation for u(x):
xu′′ + u′ = 0,
(1.5)
xw′ + w = 0,
(1.6)
or equivalently,
′
where w = u .
By Eq.(1.6), we can get
dw
+ w = 0,
dx
dw
dx
=− ,
w
x
1
w = c1 ,
x
1
′
u = c1 ,
x
u = c1 ln x + c2 ,
x
⇒
⇒
⇒
⇒
where c1 and c2 are constants.
Thus, the two linearly independent solutions are y1 = x2 , y2 = x2 ln x and the general solution of
Eq.(1.1) is
y(x) = c1 x2 ln x + c2 x2 .
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MA 262: Linear Algebra and Differential Equations
2
Fall 2011, Purdue
Linear system of differential equations
Find the general solution of the system
x′ = Ax,
where
A=
3
2
(2.1)
2
.
3
(2.2)
Solution:
Method I: Annihilator method.
From the system (2.1) we know that
⇒
x′1 = 3x1 + 2x2 ,
x′2 = 2x1 + 3x2
(2.3)
(D − 3)x1 − 2x2 = 0,
−2x1 + (D − 3)x2 = 0,
(2.4)
Multiplying by 2 on both sides of the first equation in (2.4) while multiplying (D − 3) on the second
one in (2.4), we can get
2(D − 3)x1 − 4x2 = 0,
−2(D − 3)x1 + (D − 3)2 x2 = 0,
⇒
⇒
((D − 3)2 − 4)x2 = 0,
(D2 − 6D + 5)x2 = 0,
⇒
(D − 1)(D − 5)x2 = 0.
(2.5)
The general solution to (2.5) is
x2 = c1 et + c2 e5t .
(2.6)
Now by the second equation in (2.3), we can find the general solution of x1 :
2x1 = x′2 − 3x2 ,
⇒
⇒
2x1 = (c1 et + c2 e5t )′ − 3(c1 et + c2 e5t ),
2x1 = −2c1 et + 2c2 e5t ,
⇒
x1 = −c1 et + c2 e5t .
(2.7)
Finally, the general solution of the system (2.1) is
x
x
= 1 ,
x2
−1
1
= c1 et
+ c2 e5t
.
1
1
2
(2.8)
Copy right reserved by Yingwei Wang
MA 262: Linear Algebra and Differential Equations
Fall 2011, Purdue
Method II: Eigenvalues and eigenvectors.
We need to find the eigenvalues and eigenvectors of matrix A defined by (2.2).
⇒
det(λI − A) = λ2 − 6λ + 5 = 0,
λ1 = 1, λ2 = 5.
It is easy to check that for λ1 = 1, the corresponding eigenvalue is v1 = (−1, 1)T while for λ2 = 5,
the corresponding eigenvalue is v2 = (1, 1)T .
Thus, the general solution of the system (2.1) is
x
x
= 1 ,
x2
= c1 et v1 + c2 e5t v2 ,
t −1
5t 1
.
= c1 e
+ c2 e
1
1
(2.9)
Remark 2.1. We can conclude that it is a convenient way to get the general solution (or fundamental
solution) by computing the eigenvalues and eigenvectors of the coefficient matrix.
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