Math 54. Selected Solutions for Week 11
Section 4.6 (Page 435)
7.
Find a general solution to the differential equation
y 00 + 4y 0 + 4y = e−2t ln t
using the method of variation of parameters.
The characteristic polynomial is r2 + 4r + 4 = (r + 2)2 , so a fundamental solution
set is y1 = e−2t , y2 = te−2t . One therefore needs to solve the system
e−2t v10 + te−2t v20 = 0 ;
−2e−2t v10 + (1 − 2t)e−2t v20 = e−2t ln t .
Adding two times the first equation to the second gives
e−2t v20 = e−2t ln t ,
so v20 = ln t , and (from the first equation) v10 = −t ln t . Integrating gives
v1 = −
t2
t2
ln t +
2
4
and
v2 = t ln t − t .
Therefore
t2
t2
e−2t + (t ln t − t)te−2t
yp = − ln t +
2
4
ln t 3 2 −2t
=
−
t e
.
2
4
The general solution is therefore
ln t 3 2 −2t
y=
−
t e
+ c1 e−2t + c2 te−2t .
2
4
Section 6.1 (Page 482)
2.
Determine the largest interval (a, b) for which Theorem 1 guarantees the existence of
a unique solution on (a, b) to the given initial value problem:
√
y(π) = 0 , y 0 (π) = 11 , y 00 (π) = 3 .
y 000 − xy = sin x ;
Since
√
x is only defined for x ≥ 0 , the largest possible interval is (0, ∞) .
1
2
4.
Determine the largest interval (a, b) for which Theorem 1 guarantees the existence of
a unique solution on (a, b) to the given initial value problem:
x(x + 1)y 000 − 3xy 0 + y = 0 ;
y(−1/2) = 1 ,
y 0 (−1/2) = y 00 (−1/2) = 0 .
We have to divide the equation by x(x + 1) to get the coefficient of y 000 to equal
1 (which the theorem requires). This function is zero at x = 0 and x = −1 (and
nowhere else), so the largest interval is (−1, 0) .
10. Determine whether the given functions are linearly dependent or linearly independent
on the specified interval. Justify your decision.
{sin x, cos x, tan x} on (−π/2, π/2) .
They are linearly independent. To show this, we use the method of Example 3.
Suppose that
c1 sin x + c2 cos x + c3 tan x = 0
for all x ∈ (−π/2, π/2) . Plugging in x = 0 gives c2 = 0 , leaving c1 sin x+c3 tan x = 0 .
Plugging in two other values gives
1/2
√
1/ 2
are linearly independent.
Since the matrix
12.
c3
c1
+√ =0
(x = π/3) ;
2
3
c
√1 + c3 = 0
(x = π/4) .
2
√ 1/ 3
is invertible, this forces c1 = c3 = 0 , so the functions
1
Determine whether the given functions are linearly dependent or linearly independent
on the specified interval. Justify your decision.
{cos 2x, cos2 x, sin2 x} on (−∞, ∞) .
We have cos 2x = cos2 x − sin2 x on (−∞, ∞) , so the functions are linearly dependent.
34.
Constructing Differential Equations. Given three functions f1 (x) , f2 (x) , f3 (x) that
are each three times differentiable and whose Wronskian is never zero on (a, b) , show
that the equation
y f1 (x) f2 (x) f3 (x)
0
0
0
0 f1 (x) f2 (x) f3 (x) y 00
=0
f1 (x) f200 (x) f300 (x) y 00 000
f1 (x) f2000 (x) f3000 (x) y 000
3
is a third-order linear differential equation for which {f1 , f2 , f3 } is a fundamental solution set. What is the coefficient of y 000 in this equation?
The function y = f1 satisfies the differential equation because the first and fourth
columns of the matrix are equal, so the determinant is zero. Similarly, y = f2 and
y = f3 are also solutions. This equation is a linear differential equation because you
can expand about the last column to get an expression
C44 y 000 + C34 y 00 + C24 y 0 + C14 y = 0 ,
where the cofactors Ci4 are functions of x not involving y . The coefficient of y 000 is
just the Wronskian of f1 , f2 , f3 , and we are given that it is never zero, so the equation
must be of third order (the term C44 y 000 does not disappear), and we can divide by the
Wronskian and apply Theorem 3 to find that f1 , f2 , and f3 form a fundamental set
of solutions of the differential equation.
Section 6.2 (Page 488)
14.
Find a general solution for the differential equation
y (4) + 2y 000 + 10y 00 + 18y 0 + 9y = 0
with x as the independent variable. [Hint: y(x) = sin 3x is a solution.]
The auxiliary polynomial is r4 + 2r3 + 10r2 + 18r + 9 . We are given that sin 3x is
a solution, which suggests that ±3i are roots. In fact, dividing by r2 + 9 works out,
and we have the factorization (r2 + 9)(r2 + 2r + 1) = (r2 + 9)(r + 1)2 . Therefore the
general solution is y = c1 sin 3x + c2 cos 3x + c3 e−x + c4 xe−x .
Section 9.1 (Page 503)
4.
Express the system of differential equations in matrix notation:
x01 = x1 − x2 + x3 − x4 ,
x02 = x1 + x4 ,
√
x03 = πx1 − x3 ,
x04 = 0 .
0 
x1
1
x
 2
 1
  = √
x3
π
x4
0

−1
0
0
0
1
0
−1
0
 
−1
x1
1   x2 
 .
0
x3
0
x4
4
12.
Express the given system of higher-order differential equations as a matrix system in
normal form:
x00 + 3x0 − y 0 + 2y = 0 ,
y 00 + x0 + 3y 0 + y = 0 .
Letting x1 = x , x2 = x0 , x3 = y , and x4 = y 0 gives
0 
x1
0
x
 2
0
  =
x3
0
x4
0

1
−3
0
−1
0
−2
0
−1
 
0
x1
1   x2 
  .
1
x3
−3
x4