DE-BOOK Summary & Exercises

Section 5.5 Summary & Exercises

Throughout the chapter, we worked through several examples, each demonstrating how to apply the integrating factor method in different contexts. In each case, the systematic process of identifying the standard form, computing the integrating factor, and applying integration was key to solving the differential equation.

Summary of the Key Ideas.

First-Order Linear Differential Equations
- First-order linear differential equations take the standard form
  $y^{'} + P (x) y = Q (x),$
  where $P (x)$ and $Q (x)$ are functions of $x .$
- The goal is to find the unknown function $y (x) .$
The Product Rule
- The integrating factor method works by rewriting the left side of the standard form as a reversed product rule.
- Unfortunately, most equations are not given in a form where a reversed product rule is possible.
Integrating Factor
- The integrating factor, $μ,$ is the missing function needed for the reverse product rule and is found from
  $μ = e^{\int P (x) d x} .$
- Multiplying the standard form of the equation by $μ$ leads to a new equation where the left-side can be written as
  $\frac{d}{d x} [μ \cdot y] = μ Q (x)$
Integrating Factor Method
- The integrating factor method is the process of solving a first order linear differential equation by turning it into a direct integration problem.

Exercises Exercises

Forward Product Rule.

1.

f (x) = \ln x \cos x

Answer. Answer

f^{'} (x) = (\frac{1}{x}) \cos x + \ln x (- \sin x) = \frac{\cos x}{x} - \ln x \sin x

The Integrating Factor.

2.

y^{'} - 4 y = x

Answer. Answer

Conceptual Questions.

3.

Write the differential equation below in the form

y^{'} + p (x) y = q (x)

and identify

p (x)

and

q (x) .

Also, state the order and whether it is linear or not.

- 4 x y^{'} + 5 y = \cos (x) y - 1

Answer. Answer

$y^{'} + (\frac{\cos x - 5}{4 x}) y = \frac{1}{4 x}$
$p (x) = \frac{\cos x - 5}{4 x}$
$q (x) = \frac{1}{4 x}$
order is 1
it is linear

4.

What classification of DEs be solved by applying an integrating factor?

Answer. Answer

We can use an integrating factor to solve DEs that are linear and first order.

General Solution.

5.

\frac{d y}{d x} - y = e^{3 x}

Answer. Answer

y = \frac{1}{2} e^{3 x} + C e^{x}

6.

\frac{d y}{d x} = \frac{y}{x} + 2 x + 1

, where

x > 0

Answer. Answer

y = 2 x^{2} + x \ln x + C x

7.

\frac{d r}{d θ} + r \tan θ = \sec θ,

where

- \frac{π}{2} \leq θ \leq \frac{π}{2}

Answer. Answer

r = \sin θ + C \cos θ

8.

y \frac{d x}{d y} + 2 x = 5 y^{3}

Answer. Answer

x = y^{3} + \frac{C}{y^{2}}

Initial-Value Problems.

9.

\frac{d y}{d x} - \frac{y}{x} = x e^{x},

where

x > 1

and

y (1) = e - 1

Answer. Answer

10.

e^{t} z^{'} = 1 - 4 e^{t} z, z (0) = \frac{4}{3}

Answer. Answer

Which Method Applies.

You do not need to solve any of the equations.

Can be solved by separating variables, but not using an integrating factor.
Can be solved using an integrating factor, but not by separating variables.
Can be solved both by using integrating factor and by separating variables.
Cannot be solved using either technique.

11.

x^{2} \frac{d y}{d x} + \cos x = y

Answer. Answer

b. Can be solved using an integrating factor, but not by separating variables.

12.

\frac{d x}{d t} + x t = e^{x}

Answer. Answer

d. Cannot be solved using either technique.

13.

(t^{2} + 1) \frac{d y}{d t} = y t - y

Answer. Answer

c. Can be solved both by using integrating factor and by separating variables.

14.

\frac{d y}{d t} - y^{2} t = t

Answer. Answer

a. Can be solved by separating variables, but not using an integrating factor.

15.

3 r = \frac{d r}{d θ} - θ^{3}

Answer. Answer

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