Download Quiz 11 1 Annihilator 2 Constant

MA 262: Linear Algebra and Differential Equations
Fall 2011, Purdue
Quiz 11
1
Annihilator
Find the annihilator of
f (x) = cos(x) + xex + 1.
Solution: The annihilator of f1 = cos(x) is
D2 + 1.
(1.1)
(D − 1)2 .
(1.2)
D.
(1.3)
The annihilator of f2 = xex is
The annihilator of f3 = 1 is
Now, by (1.1)-(1.3), we can get the annihilator of f = f1 + f2 + f3 is
D(D − 1)2 (D2 + 1).
2
Constant-coefficient nonhomogenous differential equations
Solve the initial value problem
y ′′ + 3y ′ − 4y = 6e2x ,
,
y(0) = 2, y ′ (0) = 3.
(2.1)
Solution: First of all, we can get the auxiliary equation
r2 + 3r − 4 = 0,
⇒
(r + 4)(r − 1) = 0.
(2.2)
The roots of the equation (2.2) is that
r = −4, 1.
Then the general solution to the homogeneous equation y ′′ + 3y ′ − 4y = 0 is
yh (x) = c1 e−4x + c2 ex .
1
(2.3)
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MA 262: Linear Algebra and Differential Equations
Fall 2011, Purdue
Next, we need to find an appropriate trial solution for the nonhomogeneous equation y ′′ + 3y ′ − 4y =
6e in this form
2x
⇒
yp (x) = A0 e2x ,
y ′ = 2A0 e2x , y ′′ = 4A0 e2x ,
⇒
⇒
y ′′ + 3y ′ − 4y = (4A0 + 6A0 − 4A0 )e2x = 6e2x ,
A0 = 1.
It follows that
yp (x) = e2x .
(2.4)
Thus, by Eqs(2.3)-(2.4), the general solution to the nonhomogeneous equation y + 3y − 4y = 6e2x is
′′
y(x)
′
= yh (x) + yp (x),
= c1 e−4x + c2 ex + e2x .
(2.5)
Finally, we need to determine the constants c1 , c2 in Eq.(2.5) by the initial values in (2.1)
y(0) = 2,
,
y ′ (0) = 3.
c1 + c2 + 1 = 2,
⇒
,
−4c1 + c2 + 2 = 3.
c1 = 0,
⇒
,
c2 = 1.
In conclusion, the solution to the problem (2.1) is
y(x) = ex + e2x .
3
Summary
For different kinds of functions, the corresponding annihilators are
1. If f (x) = cos(kx) or f (x) = sin(kx), then the annihilator is
D2 + k2 .
2. If f (x) = xm erx , then the annihilator is
(D − r)m+1 .
3. If f (x) = an xn + an−1 xn−1 · · · a1 x + a0 , which is a polynomial of degree n, then the annihilator is
Dn+1 .
2
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