##
Section 5.4 Additional Examples

In this section, we will look at more advanced examples of the integrating factor method. These examples will demonstrate how the technique applies to a wider variety of differential equations, including those with more complex functions or integration techniques. Each example will follow the

four-step process outlined in the previous section.

##
Example 22.

Solve the differential equation:

\begin{equation*}
x^2 y' - y = 1, \quad x > 0\text{.}
\end{equation*}

## Solution. Solution

This is a first-order linear differential equation, so we can apply the integrating factor method.

\begin{gather*}
y' + \us{P}{\boxed{-\frac{1}{x^2}}}\ y = \frac{1}{x^2}
\end{gather*}

\begin{gather*}
\mu(x) = e^{\large \int -\sfrac{1}{x^2} dx} = e^{1/x}
\end{gather*}

\begin{align*}
e^{1/x}y' - \frac{e^{1/x}}{x^2} y =\amp\ \frac{e^{1/x}}{x^2} \\
\frac{d}{dx}\left[e^{1/x} y\right] =\amp\ \frac{e^{1/x}}{x^2}
\end{align*}

\begin{align*}
\int \frac{d}{dx}\left[e^{1/x} y\right] dx =\amp\ \int \frac{e^{1/x}}{x^2} dx \quad \left(u = \sfrac{1}{x}\right) \\
e^{1/x} y =\amp\ \int -e^u du \\
e^{1/x} y =\amp\ -e^u + c \\
e^{1/x} y =\amp\ -e^{1/x} + c
\end{align*}

Now, we can solve for \(y\text{:}\)

\begin{gather*}
y = -1 + ce^{-1/x}
\end{gather*}

##
Example 23.

Solve the equation:

\begin{equation*}
t^2 y' + 2ty = t^3 \ln t \text{.}
\end{equation*}

## Solution. Solution

This is a linear equation, and we will solve it using the integrating factor method.

\begin{gather*}
y' + \frac{2}{t} y = t \ln t
\end{gather*}

\begin{gather*}
\mu(t) = e^{\large \int (\sfrac{2}{t})\, dt} = e^{2 \ln t} = t^2
\end{gather*}

\begin{align*}
t^2 y' + 2t y =\amp\ t^3 \ln t \\
\frac{d}{dt} \left[t^2 y\right] =\amp\ t^3 \ln t
\end{align*}

\begin{align*}
\int \frac{d}{dt} \left[t^2 y\right] dt =\amp\ \int t^3 \ln t\ dt \\
t^2 y =\amp\ \int t^3 \ln t\ dt
\end{align*}

\begin{align*}
u = \ln t, \amp \quad v = \frac14t^4 \\
du = \frac{1}{t}\ dt, \amp \quad dv = t^3\ dt
\end{align*}

gives us

\begin{align*}
t^2 y =\amp\ \frac14 t^4 \ln t - \int \frac14 t^4 \cdot \frac{1}{t}\ dt \\
t^2 y =\amp\ \frac14 t^4 \ln t - \frac14 \int t^3\ dt \\
t^2 y =\amp\ \frac{t^4}{4} \ln t - \frac{t^4}{16} + c \\
y =\amp\ \frac{t^2}{4} \ln t - \frac{t^2}{16} + c\,t^2
\end{align*}

##
Example 24.

Solve the initial value problem.

\begin{equation*}
A = \sec\theta A' - \frac{\theta e^{\theta^2 + \sin\theta}}{\cos \theta}, \hspace{1cm} A(0) = -\frac{1}{2}\text{.}
\end{equation*}

## Solution. Solution

Before applying the integration factor method, we note that the equation is first order and linear.

\begin{align*}
\sec\theta A' - A
=\amp\ \frac{\theta e^{\theta^2 + \sin\theta}}{\cos \theta} \qquad \text{Note: } \sec\theta = \frac{1}{\cos\theta}\\
A' + \us{P}{\boxed{- \cos\theta}}\ A
=\amp\ \theta e^{\theta^2 + \sin\theta}
\end{align*}

\begin{gather*}
\mu = e^{\int -\cos\theta d\theta} = e^{-\sin\theta}
\end{gather*}

\begin{gather*}
e^{-\sin\theta}A' - \cos\theta e^{-\sin\theta}A = \theta e^{\theta^2 + \sin\theta} \\
\frac{d}{d\theta}\left[ e^{-\sin\theta}A \right] = \theta e^{\theta^2}
\end{gather*}

\begin{gather*}
\int \frac{d}{d\theta}\left[ e^{-\sin\theta}A \right] d\theta = \int \theta e^{\theta^2} d\theta\\
e^{-\sin\theta}A = \frac{1}{2}e^{\theta^2} + c\\
A = \frac{1}{2}e^{\theta^2 + \sin\theta} + ce^{\sin\theta}
\end{gather*}

Finally, we apply the initial condition to determine the constant \(c\text{.}\)

\begin{gather*}
-\frac{1}{2} = A(0) = \frac{1}{2}e^{0^2 + \sin(0)} + ce^{\sin(0)} = \frac{1}{2}e^0 + ce^0= \frac{1}{2} + c \\
-1 = c
\end{gather*}

We can then write the particular solution to the initial value problem.

\begin{equation*}
A(\theta) = \frac{1}{2}e^{\theta^2 + \sin\theta} - e^{\sin\theta}
\end{equation*}

The integrating factor method is a versatile tool for solving first-order linear differential equations. By reducing the equation to a form that allows for direct integration, we can tackle a wide variety of problems with confidence. The examples in this chapter illustrate the broad applicability of this technique, and practice with a range of equations will solidify your understanding of how to use it effectively.

###
Reading Questions Check-Point Questions

###
1. *Which of the following steps comes first when solving a differential equation using the integration factor method?*

Which of the following steps comes first when solving a differential equation using the integration factor method?

- Identify the integrating factor.
Correct! The first step is always to identify the integrating factor based on the equation’s form.

- Multiply both sides by the integrating factor.
Incorrect. You need to identify the integrating factor before multiplying it by both sides of the equation.

- Integrate both sides.
Incorrect. Integration occurs later in the process after manipulating the equation using the integrating factor.

- Find the constant of integration using initial conditions.
Incorrect. This step comes after solving the equation using the integrating factor.

###
2. *After determining the integration factor, what is the next step in the process?*

After determining the integration factor, what is the next step in the process?

- The original equation is multiplied by the integrating factor.
Incorrect. You should apply the integrating factor to the equation after transforming it into the standard form.

- The standard form of the equation is multiplied by the integrating factor.
Correct! Once the equation is in the standard form, you multiply both sides by the integrating factor.

- Apply direct integration to solve the equation.
Incorrect. Direct integration is the final step, after applying the integrating factor.

- The reverse product rule is used to rewrite the equation.
Incorrect. The reverse product rule is not part of this process. Use the integrating factor first.

###
3. *In solving the equation \(\ds y' + P(x)y = Q(x)\text{,}\) what form does the general solution take after direct integration?*

In solving the equation \(\ds y' + P(x)y = Q(x)\text{,}\) what form does the general solution take after direct integration?

- \(\ds y(x) = ∫Q(x)dx\text{.}\)
Incorrect. This is not the correct form. You need to account for the integrating factor as well.

- \(\ds y(x) = e^{∫P(x) dx} ∫e^{-∫P(x)dx} Q(x) dx\text{.}\)
Correct! This is the general solution after applying the integrating factor and performing direct integration.

- \(\ds y(x) = ∫e^{P(x)} Q(x) dx\text{.}\)
Incorrect. The correct general solution includes the integrating factor in both the exponent and the integral.

- \(\ds y(x) = Ce^{Q(x)}\text{.}\)
Incorrect. This form does not represent the general solution to this type of equation.

###
4. *What is the purpose of the integrating factor in the first-order linear differential equation \(\ds x^2 y' - y = 1\text{?}\)*

What is the purpose of the integrating factor in the first-order linear differential equation \(\ds x^2 y' - y = 1\text{?}\)

- To simplify the differential equation into a separable form.
Incorrect. The integrating factor is not used to separate variables in this case.

- To eliminate the non-linear terms in the equation.
Incorrect. This equation is already linear, and the integrating factor does not eliminate non-linear terms.

- To transform the equation into a direct integration problem.
Correct! The integrating factor transforms the equation into a form that allows for direct integration.

- To find the value of the constant of integration.
Incorrect. The constant of integration is found after the equation is solved, not by using the integrating factor.

###
5. *What technique is used to evaluate the integral \(\ds \int t^3 \ln t \, dt\text{?}\)*

What technique is used to evaluate the integral \(\ds \int t^3 \ln t \, dt\text{?}\)

- Substitution.
Incorrect. Substitution is not the technique used here.

- Integration by parts.
Correct! Integration by parts is the appropriate technique for this integral.

- Partial fraction decomposition.
Incorrect. Partial fraction decomposition is not useful for this integral.

- The Product Rule.
Incorrect. The product rule is a differentiation technique, not an integration technique.

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