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 50.
\(\ \ \) Solve the differential equation:
\begin{equation*}
x^2 y' - y = 1, \quad x > 0\text{.}
\end{equation*}
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 51.
\(\ \ \) Solve the equation:
\begin{equation*}
t^2 y' + 2ty = t^3 \ln t \text{.}
\end{equation*}
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 52.
\(\ \ \) 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.
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 your Understanding
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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