Answers to Selected Problems in Chapter 9



Complete Solutions of Selected Problems in Chapter 9

3. Solve [pic]. The characteristic equation is [pic]. Factoring, we obtain [pic]. Hence, the characteristic roots are [pic] and [pic]. Thus, a general solution is

[pic].

6. Solve [pic]. As [pic], where [pic], the characteristic equation is [pic]. Factoring, we obtain [pic]. Hence, the characteristic roots are [pic]. Thus, a general solution is

[pic].

11. The characteristic equation for [pic] is [pic]. Factoring we have [pic]. Since [pic]is a repeated characteristic root, a general solution is [pic]

13. Solve [pic]. The characteristic equation is [pic]. Factoring, we obtain [pic]. Hence, the characteristic roots are [pic] and [pic]. Thus, a general solution is

[pic].

14. Solve [pic]. As [pic], where [pic], the characteristic equation is [pic]. Completing the square, we obtain

[pic] [pic] [pic] [pic] [pic].

Since the characteristic roots are [pic], a general solution is

[pic].

15. A general solution is [pic].

18. Solve [pic]. The characteristic equation is [pic]. Letting [pic], [pic], and [pic] in the quadratic formula, we obtain the characteristic roots

[pic].

A general solution is [pic].

19. Solve [pic]. The characteristic equation is [pic]. Completing the square, we obtain

[pic] [pic] [pic] [pic] [pic]

Since the characteristic roots are [pic], a general solution is [pic].

21. The characteristic equation for [pic] is [pic] or

[pic].

Thus, [pic] is a repeated real root. A general solution is [pic].

22. The characteristic equation is [pic]. Completing the square or using the quadratic formula, we find the characteristic roots are [pic]. Therefore, a general solution is [pic].

23. The characteristic equation for [pic] is [pic] or

[pic].

Thus, [pic] is a repeated real root. A general solution is [pic].

27. First obtain the general solution of [pic] or [pic]. As the characteristic equation is [pic], the characteristic roots are [pic] and [pic]. Hence, a general solution is

[pic].

From the initial condition [pic], we obtain

[pic].

Differentiating the general solution with respect to t, letting [pic], and using the initial condition [pic], we obtain

[pic], [pic] [pic] [pic].

Then [pic]. Therefore, the solution of the initial value problem is

[pic].

31. First obtain the general solution of [pic] or [pic]. As the characteristic equation is [pic], the characteristic roots are [pic]. Hence, a general solution is

[pic].

From the initial condition [pic], we obtain

[pic] [pic] [pic].

Differentiating the general solution with respect to t, letting [pic], and using the initial condition [pic], we obtain

[pic] [pic] [pic].

Therefore, the solution of the initial value problem is

[pic].

40. The characteristic roots are [pic]. Hence, a general solution is

[pic].

42. The characteristic roots are [pic]. Hence, a general solution is

[pic].

43. The characteristic roots are [pic]. Hence, a general solution is

[pic].

52. Find a particular solution and then a general solution of

[pic].

Solution.   First find a general solution of the associated homogeneous differential equation

[pic].

Since the characteristic equation is [pic], the characteristic roots are [pic] and [pic]. Thus, a general solution of the associated homogeneous equation is

[pic].

The right-hand side of the nonhomogeneous equation suggests a trial particular solution of the form:

[pic].

Now find values for A and B that will actually make this function a solution. The first and second derivatives of [pic] are [pic] and [pic], respectively. Multiply [pic] and its derivatives by the appropriate coefficients of the operator [pic] and add the results as follows:

[pic]

The function [pic] is a particular solution if [pic]; this implies that A and B must be given values so that

[pic]

which has the solution [pic] and [pic]. Hence, a particular solution is

[pic].

Consequently, a general solution is

[pic].

55. Find a particular solution and then a general solution of

[pic].

Solution.   First find a general solution of the associated homogeneous differential equation

[pic].

Since the characteristic equation is [pic], the characteristic roots are [pic] and [pic]. Thus, a general solution of the associated homogeneous equation is

[pic].

According to the Superposition Principle, a particular solution can be expressed as the sum of the particular solutions of the two nonhomogeneous equations

• [pic]

and

• [pic].

By inspection, a particular solution [pic] of [pic] is [pic].

Now we need to find a particular solution of [pic]. Its right-hand side suggests that we try a particular solution of the form

[pic].

However, since [pic], we see that the term [pic] is actually a solution of the associated homogeneous equation. This means that the trial particular solution has the form

[pic].

The first and second derivatives of [pic] are

[pic]

and

[pic],

respectively. Substituting, we obtain

[pic].

Hence, [pic]. This implies [pic] and [pic]. Therefore,

[pic].

By the Superposition Principle, a particular solution of [pic] is

[pic].

Therefore, a general solution is

[pic].

57. Find a particular solution and then a general solution of

[pic].

Solution.  Let’s find a general solution of the associated homogeneous differential equation

[pic].

Since the characteristic equation is [pic], the characteristic roots are [pic]. Thus, a general solution of the associated homogeneous equation is

[pic].

By inspection, a particular solution [pic] of [pic] is [pic].

Now we need to find a particular solution [pic] of [pic]. Its right-hand side suggests that we try a particular solution of the form

[pic].

However, we see from above that this is really the form of the general solution of the associated homogeneous equation. Accordingly, the particular solution has the form

[pic].

By the product rule,

[pic]

Thus,

[pic]

From this, we see that [pic] is a particular solution if A and B have values so that

[pic].

This is the case for [pic] and [pic]. So,

[pic].

Therefore, a particular solution of [pic] is

[pic]

and a general solution is

[pic].

63. Find a general solution of [pic], for [pic].

Solution.  First find a general solution of the associated homogeneous differential equation

[pic].

Since the characteristic equation is [pic], the characteristic roots are [pic]. Hence, a general solution of the associated homogeneous equation is

[pic].

Using the Method of Variation of Parameters, we look for a particular solution of the form

[pic],

where the derivatives of [pic] and [pic] satisfy the system of equations

[pic]

Solving for [pic] and [pic] with Cramer’s rule, we obtain

[pic]

and

[pic].

Integrating, we obtain

[pic] and [pic]

since [pic] for [pic]. Hence, a particular solution is

[pic].

Therefore, a general solution is

[pic].

68. To find a general solution of

[pic]

for [pic], let us first find a general solution of the associated homogeneous equation

[pic].

From this we see that the characteristic equation is [pic]. Since its roots are [pic], the general solution of the associated homogenous equation is

[pic].

Now use the Method of Variation of Parameters to look for a particular solution of the form

[pic],

where the derivatives of [pic] and [pic] satisfy the system of equations

[pic]

By Cramer’s Rule,

[pic]

and

[pic]

To obtain [pic]and [pic], integrate [pic]and [pic] with respect to [pic] obtaining

[pic]

and

[pic].

Note that as [pic],

[pic].

Consequently, [pic] Thus, a particular solution is

[pic].

A general solution is

[pic].

72. Find a general solution of [pic].

Solution. The associated homogeneous differential equation is

[pic].

Since the characteristic equation is [pic], the characteristic roots are [pic]. Hence, a general solution of the associated homogeneous equation is

[pic].

If we employ the half-angle formula [pic], we can use the Method of Undetermined Coefficients to find a particular solution of

[pic].

According to the Superposition Principle, a particular solution can be expressed as the sum of the particular solutions of the two nonhomogeneous equations

• [pic]

and

• [pic].

By inspection, a particular solution [pic] of [pic] is [pic].

Now we need to find a particular solution [pic] of [pic] . Its right-hand side suggests that we try a particular solution of the form

[pic].

Differentiating, we obtain,

[pic]

Thus,

[pic]

From this, we see that [pic] is a particular solution if A and B have values so that

[pic]

This is the case for [pic] and [pic]. So,

[pic].

Therefore, a particular solution of [pic] is

[pic]

and a general solution is

[pic].

73. Find a general solution of

[pic]. (N)

on the interval [pic].

Solution. First we find a general solution of the associated homogeneous equation

[pic].

Its characteristic equation is [pic]. Since the roots are [pic], the general solution of the associated homogenous equation is

[pic].

A particular solution of (N) can be found, according to the Superposition Principle, by finding particular solutions of

(A) [pic]

(B) [pic]

and then summing them.

First let’s find a particular solution of (A) using the method of undetermined Coefficients. The right-hand side suggests a trial particular solution of the form

[pic].

To find values for A and B, substitute[pic] and its derivatives in (A) obtaining

[pic].

In order that [pic] be a solution, values for A and B must be found so that

[pic],

holds for all [pic]. Matching coefficients, we obtain

[pic]

Thus, [pic] Consequently, a particular solution of (A) is

[pic].

By Problem 68, a particular solution of (B) is

[pic].

Thus, a particular solution [pic] of (N) is

[pic]

and a general solution of (N) is

[pic].

Check with Maple:

> ode1 := diff(x(t), t$2) + x(t) = 4 - t^2 +tan(t);

[pic]

> dsolve(ode1, x(t));

[pic]

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