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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