Page 70 - Advanced engineering mathematics

50 CHAPTER 2 Linear Second-Order Equations

SECTION 2.1 PROBLEMS

In each of Problems 1 through 5, verify that y 1 and y 2 are Solve this linear equation to verify the conclusion of
solutions of the homogeneous differential equation, calcu- part (1).
late the Wronskian of these solutions, write the general To prove conclusion (2), show first that, if
solution, and solve the initial value problem. y 2 (x) = ky 1 (x) for all x in I,then W(x) = 0. A sim-
ilar conclusion holds if y 1 (x) = ky 2 (x). Thus, linear
1. y + 36y = 0; y(0) =−5, y (0) = 2 dependence implies vanishing of the Wronskian.

y 1 (x) = sin(6x), y 2 (x) = cos(6x) Conversely, suppose W(x) = 0on I. Suppose
first that y 2 (x) does not vanish on I. Differentiate

2. y − 16y = 0; y(0) = 12, y (0) = 3
4x
y 1 (x) = e , y 2 (x) = e −4x y 1 /y 2 to show that

3. y + 3y + 2y = 0; y(0) =−3, y (0) =−1

y 1 (x) = e −2x , y 2 (x) = e −x 2 d y 1
y =−W(x) = 0
2 dx

4. y − 6y + 13y = 0; y(0) =−1, y (0) = 1 y 2

3x
3x
y 1 (x) = e cos(2x), y 2 (x) = e sin(2x)

5. y − 2y + 2y = 0; y(0) = 6, y (0) = 1 on I. This means that y 1 /y 2 has a zero derivative on

x
x
y 1 (x) = e cos(x), y 2 (x) = e sin(x) I, hence y 1 /y 2 = c,so y 1 = cy 2 and these solutions are
linearly dependent. A technical argument, which we
omit, covers the case that y 2 (x) can vanish at points
In Problems 6 through 10, use the results of Problems 1
of I.
through 5, respectively, and the given particular solution
4
2
3
Y p to write the general solution of the nonhomogeneous 12. Let y 1 (x) = x and y 2 (x) = x . Show that W(x) = x .
equation. Now W(0) = 0, but W(x) = 0if x = 0. Why does this
not violate Theorem 2.3 conclusion (1)?
6. y + 36y = x − 1,Y p (x) = (x − 1)/36

2
13. Show that y 1 (x) = x and y 2 (x) = x are linearly inde-
2
2

7. y − 16y = 4x ;Y p (x) =−x /4 + 1/2
2
pendent solutions of x y − 2xy + 2y = 0on (−1,1)

8. y + 3y + 2y = 15;Y p (x) = 15/2 but that W(0) = 0. Why does this not contradict
x

9. y − 6y + 13y =−e ;Y p (x) =−8e x Theorem 2.3 conclusion (1)?

14. Suppose y 1 and y 2 are solutions of equation (2.2) on
2
2
10. y − 2y + 2y =−5x ;Y p (x) =−5x /2 − 5x − 4

(a,b) and that p and q are continuous. Suppose y 1
11. Here is a sketch of a proof of Theorem 2.2. Fill in the and y 2 both have a relative extremum at some point
details. Denote W(y 1 , y 2 ) = W for convenience. between a and b. Show that y 1 and y 2 are linearly
For conclusion (1), use the fact that y 1 and y 2 are dependent.
solutions of equation (2.2) to write 15. Let ϕ be a solution of y + py + qy = 0onanopen

interval I. Suppose ϕ(x 0 )=0 for some x 0 in this inter-
val. Suppose ϕ(x) is not identically zero. Prove that

y + py + qy 1 = 0

1
1
ϕ (x 0 ) = 0.

y + py + qy 2 = 0.
2
2
16. Let y 1 and y 2 be distinct solutions of equation (2.2)
Multiply the first equation by y 2 and the second on an open interval I.Let x 0 be in I, and suppose
by −y 1 and add. Use the resulting equation to show y 1 (x 0 ) = y 2 (x 0 ) = 0. Prove that y 1 and y 2 are linearly
that
dependent on I. Thus, linearly independent solutions
W + pW = 0. cannot share a common zero.

2.2 The Constant Coefficient Case
We have outlined strategies for solving second-order linear homogeneous and nonhomogeneous
differential equations. In both cases, we must begin with two linearly independent solutions of
a homogeneous equation. This can be a difficult problem. However, when the coefficients are
constants, we can write solutions fairly easily.
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October 14, 2010 14:12 THM/NEIL Page-50 27410_02_ch02_p43-76