Section 5.3 Integrating Factor Method
The integrating factor method provides a systematic approach to solving first-order linear differential equations. These equations can be written in the
standard form as:
A general linear differential equation looks like:
\begin{equation*}
a_n(x)y^{(n)} + \cdots + a_2(x)y'' + a_1(x)y' + a_0(x)y = f(x)\text{.}
\end{equation*}
The first-order case simplifies to:
\begin{equation*}
a_1(x)y' + a_0(x)y = f(x)\text{,}
\end{equation*}
By dividing through by \(a_1(x)\text{,}\) this reduces to:
\begin{equation*}
y' + \us{P(x)}{\ub{\frac{a_0(x)}{a_1(x)}}}y = \us{Q(x)}{\ub{\frac{f(x)}{a_1(x)}}}\text{.}
\end{equation*}
This form is called the **standard form** of a first-order linear differential equation:
\begin{equation*}
y' + P(x)y = Q(x)\text{.}
\end{equation*}
\begin{equation}
y' + P(x)y = Q(x)\text{.}\tag{12}
\end{equation}
Now that we’ve introduced the integrating factor method, let’s break down the step-by-step process:
Method 3. Integrating Factor Method.
Here’s the procedure for solving a first-order linear differential equation:
- Get the Standard Form & Identify \(P\)
- \(\ds\hspace{8.5em}\frac{dy}{dx} + P(x) y = Q(x)\)
- Find the Integrating Factor \(\mu\)
- \(\ds\hspace{8.5em}\mu(x) = e^{\int P(x) dx}\)
- Multiply the Equation by \(\mu\) & Reverse the Product Rule
\begin{align*}
\mu(x)\frac{dy}{dx} + \mu(x) P(x) y =\amp\ \mu(x) Q(x) \\
\frac{d}{dx}\left[\mu(x)\cdot y\right] =\amp\ \mu(x) Q(x)
\end{align*}
- Apply Direct Integration
- \(\ds\hspace{8.5em}\int \frac{d}{dx}\left[\mu(x)\cdot y\right]\ dx = \int\mu(x) Q(x)\ dx\)
Now that we’ve laid out the method, let’s apply it to some concrete examples. In each case, we will follow the four steps outlined above: (1) put the equation into standard form, (2) compute the integrating factor, (3) multiply the equation and simplify using the product rule, and (4) integrate to find the solution.
Example 47.
\(\ \ \) Find the general solution to the equation:
\begin{equation*}
y' + 6y = 1\text{.}
\end{equation*}
Solution.
This is a first-order linear equation, so the integrating factor method applies. Since it’s already in standard form, we proceed by identifying \(P(x)\text{.}\)
\begin{gather*}
y' + \os{P}{\boxed{6}}\ y = 1
\end{gather*}
\begin{gather*}
\mu(x) = e^{\int 6\ dx} = e^{6x}
\end{gather*}
Note: We ignore the constant from integration since we only need one integrating factor to reverse the product rule.
\begin{align*}
e^{6x}y' + 6e^{6x}y =\amp\ e^{6x} \\
\frac{d}{dx}\left[e^{6x}\cdot y\right] =\amp\ e^{6x}
\end{align*}
\begin{align*}
\int \frac{d}{dx}\left[e^{6x}\cdot y\right]\ dx =\amp\ \int e^{6x}\ dx \\
e^{6x}\cdot y =\amp\ \frac{1}{6} e^{6x} + c \\
y =\amp\ \frac{1}{6} + ce^{-6x}.
\end{align*}
Example 48.
\(\ \ \) Solve the equation:
\begin{equation*}
x^2y' = 5x^3 + 2x^3y \text{.}
\end{equation*}
Solution.
Rewriting this equation in standard form, we apply the integrating factor method.
\begin{gather*}
y' + \os{P}{\boxed{-2x}} y = 5x
\end{gather*}
\begin{gather*}
\mu(x) = e^{\int -2x\ dx} = e^{-x^2}
\end{gather*}
\begin{align*}
e^{-x^2}y' - 2x e^{-x^2}y =\amp\ 5xe^{-x^2} \\
\frac{d}{dx}\left[e^{-x^2}\cdot y\right] =\amp\ 5xe^{-x^2}
\end{align*}
\begin{align*}
\int \frac{d}{dx}\left[e^{-x^2}\cdot y\right]dx =\amp\ \int 5xe^{-x^2}dx \\
e^{-x^2}\cdot y =\amp\ -\frac{5}{2}e^{-x^2} + c \\
y =\amp\ -\frac{5}{2} + ce^{x^2}
\end{align*}
Solution.
First we note that this equation is first order and linear, so the integrating factor method applies.
\begin{gather*}
y' + \os{P}{\boxed{-2x}} y = 5x
\end{gather*}
\begin{gather*}
\mu = e^{\int -2x dx} = e^{-x^2}
\end{gather*}
\begin{align*}
e^{-x^2}y' - 2x e^{-x^2}y =\amp\ 5xe^{-x^2} \\
\frac{d}{dx}\left[e^{-x^2}\cdot y\right] =\amp\ 5xe^{-x^2}
\end{align*}
\begin{align*}
\int \frac{d}{dx}\left[e^{-x^2}\cdot y\right]dx =\amp\ \int 5xe^{-x^2}dx \\
e^{-x^2}\cdot y =\amp\ -\frac{5}{2}e^{-x^2} + c \\
y =\amp\ -\frac{5}{2} + ce^{x^2}
\end{align*}
Solution. Verify the Solution
We have found the general solution, but it is worth remembering that we can always verify that we have the correct solution by substituting back into the original equation.
\begin{align*}
\mbox{LHS} \amp = x^2y' \\
\amp = x^2 \cdot \frac{d}{dx}\left( -\frac{5}{2} + ce^{x^2} \right) \\
\amp = x^2 \cdot \left( 0 + 2cxe^{x^2} \right) \\
\amp = 2c x^3 e^{x^2}
\end{align*}
\begin{align*}
\mbox{RHS} \amp = 5x^3 + 2x^3y \\
\amp = 5x^3 + 2x^3\left( -\frac{5}{2} + ce^{x^2} \right) \\
\amp = 5x^3 - 5x^3+ 2c x^3 e^{x^2} \\
\amp = 2c x^3 e^{x^2} = \mbox{LHS}
\end{align*}
Since we get the same result when we substitute our solution in on the right and left hand sides of the DE, we confirm that we did find the solution.
Example 49.
\(\ \ \) Find the general solution to the differential equation
\begin{equation*}
-\frac{4}{t^2} \cdot M = t^5 - \frac{1}{t} \cdot\frac{dM}{dt} + 1.
\end{equation*}
Solution.
Since the equation is first order and linear, we can apply the integrating factor method. Here are the steps to solve for \(M(t)\text{.}\)
\begin{align*}
\frac{1}{t}\cdot \frac{dM}{dt} - \frac{4}{t^2} \cdot M \amp = t^5 + 1 \\
\frac{dM}{dt} + \us{P}{\boxed{-\frac{4}{t}}} \cdot M \amp = t^6 + t
\end{align*}
\begin{gather*}
\mu = e^{\large \int \sfrac{-4}{t}\ dt} = e^{-4\ln t} = e^{\ln t^{-4}} = t^{-4} = \frac{1}{t^4}
\end{gather*}
Note: Since we only need one integrating factor, we ignore absolute value you would normally get from integrating \(\sfrac{1}{t}\text{.}\)
\begin{align*}
\frac{1}{t^4} \cdot \frac{dM}{dt} - \frac{1}{t^4}\cdot \frac{4}{t} \cdot M \amp = \frac{1}{t^4}(t^6 + t)\\
\frac{1}{t^4}\cdot \frac{dM}{dt} - \frac{4}{t^5} \cdot M \amp = t^2 + t^{-3}\\
\frac{d}{dt}\left[ \frac{1}{t^4}\cdot M \right] \amp = t^2 + t^{-3}
\end{align*}
\begin{align*}
\int \frac{d}{dt}\left[ \frac{1}{t^4}\cdot M \right]\ dt \amp = \int t^2 + t^{-3} dt\\
\frac{1}{t^4}\cdot M \amp = \frac{1}{3}t^{3} - \frac{1}{2}t^{-2} + c\\
M \amp = \frac{1}{3}t^{7} - \frac{1}{2}t^2 + ct^4
\end{align*}
In summary, the integrating factor method can be used to solve any first-order linear equation by transforming it into a form that can be easily solved through direct integration.
Reading Questions Check your Understanding
1. The differential equation \(y'+2y=3x\) can be solved using integrating factor.
The differential equation \(y'+2y=3x\) can be solved using integrating factor.
True.
This differential equation is first order and linear, so it can be solved using integrating factor.
False.
This differential equation is first order and linear, so it can be solved using integrating factor.
Hint.
Check to see that the differential equation is first order and linear.
2. Why is it important to put the differential equation in standard form before using the integrating factor method?
Why is it important to put the differential equation in standard form before using the integrating factor method?
It allows us to easily identify the coefficient of \(y\) and compute the integrating factor.
- Correct! The standard form highlights \(P(x)\text{,}\) which is needed to calculate the integrating factor.
It makes it easier to find the general solution directly without using integration.
- Incorrect. The standard form doesn’t eliminate the need for integration; it helps set up the equation for the integrating factor method.
It ensures the differential equation has a unique solution.
- Incorrect. Putting the equation in standard form does not guarantee uniqueness; that depends on initial conditions.
It reduces the degree of the differential equation, making it easier to solve.
- Incorrect. The standard form doesn’t reduce the degree of the equation but helps organize terms for applying the integrating factor.
3. The differential equation \(y''+3xy=e^{5x}\) can be solved using integrating factor.
The differential equation \(y''+2y=3x\) can be solved using integrating factor.
True.
This differential equation is linear but not first order, so it cannot be solved using integrating factor.
False.
This differential equation is linear but not first order, so it cannot be solved using integrating factor.
Hint.
Check to see that the differential equation is first order and linear.
4. The differential equation \(xy'+2xy=x^2\) is in standard form for the integrating factor method.
The differential equation \(xy'+2xy=x^2\) is in standard form for the integrating factor method.
True.
This differential equation is not in the form \(y'+P(x)y=Q(x)\text{.}\)
False.
This differential equation is not in the form \(y'+P(x)y=Q(x)\text{.}\)
Hint.
Check to see that the differential equation is in the form \(y'+P(x)y=Q(x)\text{.}\)
5. Determine the integrating factor for the differential equation \(\ds y'=\frac{y}{x}+2x+1\).
Determine the integrating factor for the differential equation \(\ds y'=\frac{y}{x}+2x+1\text{.}\)
- \(x\)
The integrating factor is \(e^{\int-\frac{1}{x}dx}=e^{-\ln(x)}=x^{-1}\text{.}\)
- \(-x\)
TThe integrating factor is \(e^{\int-\frac{1}{x}dx}=e^{-\ln(x)}=x^{-1}\text{.}\)
- \(x^{-1}\)
TThe integrating factor is \(e^{\int-\frac{1}{x}dx}=e^{-\ln(x)}=x^{-1}\text{.}\)
Hint.
Write the differential equation in the form \(y'+P(x)y=Q(x)\) and compute \(e^{\int P(x)}dx\text{.}\)
6. Why do we need to compute the integrating factor?
Why is it necessary to compute the integrating factor when using the integrating factor method?
It allows us to rewrite the left-hand side of the differential equation as a single derivative.
- Correct! The integrating factor simplifies the equation into a form where direct integration can be applied.
It eliminates the need for initial conditions in the equation.
- Incorrect. The integrating factor does not eliminate the need for initial conditions; it helps simplify the equation.
It ensures the equation has a unique solution.
- Incorrect. The integrating factor doesn’t guarantee uniqueness; initial conditions determine uniqueness.
It changes the equation into a second-order differential equation.
- Incorrect. The integrating factor keeps the equation as a first-order differential equation.
7. What is the final step in solving a differential equation using the integrating factor method?
What is the final step after multiplying the differential equation by the integrating factor?
Apply direct integration to both sides of the equation.
- Correct! The final step is to integrate both sides to find the solution.
Factor out the integrating factor from the equation.
- Incorrect. The integrating factor helps rewrite the equation, but it is not factored out at the final step.
Solve for the integrating factor using initial conditions.
- Incorrect. The integrating factor is computed before applying initial conditions.
Substitute \(y\) into the original equation to check for errors.
- Incorrect. Verifying the solution is optional, but the final required step is integration.
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