Math 54. Selected Solutions for Week 11 Section 4.6 (Page 435)

Math 54. Selected Solutions for Week 11

Section 4.6 (Page 435)

7. Find a general solution to the differential equation y + 4y + 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-2tv1 + te-2tv2 = 0 ; -2e-2tv1 + (1 - 2t)e-2tv2 = e-2t ln t .

Adding two times the first equation to the second gives

e-2tv2 = e-2t ln t ,

so v2 = ln t , and (from the first equation) v1 = -t ln t . Integrating gives

t2

t2

v1

=- 2

ln t +

4

and v2 = t ln t - t .

Therefore

yp = =

t2

t2

- ln t +

e-2t + (t ln t - t)te-2t

2

4

ln t 3 -

t2e-2t .

24

The general solution is therefore

y=

ln t 3 -

24

t2e-2t + c1e-2t + c2te-2t .

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 - xy = sin x ; y() = 0 , y () = 11 , y () = 3 .

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 - 3xy + y = 0 ; y(-1/2) = 1 , y (-1/2) = y (-1/2) = 0 .

We have to divide the equation by x(x + 1) to get the coefficient of y 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 (-/2, /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 (-/2, /2) . Plugging in x = 0 gives c2 = 0 , leaving c1 sin x+c3 tan x = 0 . Plugging in two other values gives

c1 + c3 = 0 23

c1 2

+

c3

=

0

(x = /3) ; (x = /4) .

Since the matrix

1/2 1/ 2

1/ 3 1

is invertible, this forces c1 = c3 = 0 , so the functions

are linearly independent.

12. 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

f1(x) f2(x) f3(x) y

f1(x) f1 (x)

f2(x) f2 (x)

f3(x) f3 (x)

y y

=0

f1 (x) f2 (x) f3 (x) y

3

is a third-order linear differential equation for which {f1, f2, f3} is a fundamental solution set. What is the coefficient of y 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

C44y + C34y + C24y + C14y = 0 ,

where the cofactors Ci4 are functions of x not involving y . The coefficient of y 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 C44y 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 + 10y + 18y + 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 + c3e-x + c4xe-x .

Section 9.1 (Page 503)

4. Express the system of differential equations in matrix notation:

x1 = x1 - x2 + x3 - x4 ,

x2

=

x1

+

x4

,

x3 = x1 - x3 ,

x4 = 0 .

x1 1 -1 1 -1 x1

x2 x3

=

1

0 0

0 -1

1 0

x2 x3

.

x4

0 0 0 0 x4

4

12. Express the given system of higher-order differential equations as a matrix system in normal form:

x + 3x - y + 2y = 0 , y + x + 3y + y = 0 .

Letting x1 = x , x2 = x , x3 = y , and x4 = y gives

x1 0 1 0 0 x1

x2 x3

=

0 0

-3 0

-2 0

1 1

x2 x3

.

x4

0 -1 -1 -3 x4

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