Page 94 - Advanced engineering mathematics

74 CHAPTER 2 Linear Second-Order Equations

Then
y(x) = c 1 e −ln(x) cos(3ln(x)) + c 2 e −ln(x) sin(3ln(x))
1
= (c 1 cos(3ln(x)) + c 2 sin(3ln(x))).
x
As usual, we solve an initial value problem by finding the general solution of the differential
equation and then using the initial conditions to determine the constants.

EXAMPLE 2.24
Solve
2

x y − 5xy + 10y = 0; y(1) = 4, y (1) =−6.

The Euler equation transforms to Y − 6y + 10Y = 0 with the general solution

3t
3t
Y(t) = c 1 e cos(t) + c 2 e sin(t)
for x > 0. Then
3
y(x) = x (c 1 cos(ln(x)) + c 2 sin(ln(x))).
Then
y(1) = 4 = c 1 .
Thus far,
3
3
y(x) = 4x cos(ln(x)) + c 2 x sin(ln(x)).
Compute
2
2
y (x) =12x cos(ln(x)) − 4x sin(ln(x))

2
2
+ 3c 2 x sin(ln(x)) + c 2 x cos(ln(x)).
Then

y (1) = 12 + c 2 =−6,
so c 2 =−18. The solution of the initial value problem is
y(x) = 4x cos(ln(x)) − 18x sin(ln(x)).
3
3

SECTION 2.5 PROBLEMS

2
In each of Problems 1 through 10, find the general solution. 8. x y − 5xy + 58y = 0

2
9. x y + 25xy + 144y = 0

2
1. x y + 2xy − 6y = 0

2

10. x y − 11xy + 35y = 0
2

2. x y + 3xy + y = 0

2

3. x y + xy + 4y = 0 In each of Problems 11 through 16, solve the initial value
problem.
2
4. x y + xy − 4y = 0

2
2

5. x y + xy − 16y = 0 11. x y + 5xy − 21y = 0; y(2) = 1, y (2) = 0

2
2

6. x y + 3xy + 10y = 0 12. x y − xy = 0; y(2) = 5, y (2) = 8

2
2

7. x y + 6xy + 6y = 0 13. x y − 3xy + 4y = 0; y(1) = 4, y (1) = 5

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October 14, 2010 14:12 THM/NEIL Page-74 27410_02_ch02_p43-76