Solving First-Order Linear Differential Equations: Integrating Factor Method

Recap: The Solution Process

As a quick refresher from our last lesson, here is the step-by-step process for solving any first-order linear differential equation:

1. Standard Form: Put the differential equation into the correct form: $\frac{dy}{dx}+p(x)y=q(x)$. This step is crucial.
2. Find the Integrating Factor: Identify $p(x)$ from the standard form and calculate the integrating factor: $\mu (x)=e^{\int p(x)dx}$.
3. Multiply and Integrate: Multiply the standard-form equation by $\mu (x)$. The left side will become the derivative of the product $\frac{d}{dx}[\mu (x)y]$. Then, integrate both sides.
4. Solve for y(x): Isolate $y(x)$ to find the general solution. Remember to include the constant of integration, $C$.

Let's now apply this process to our first example.

Example 1: A Straightforward Case

Consider the differential equation:

This equation is already in standard form, making it a great starting point.

1. Standard Form: The equation is already in the form $y^{\prime }+p(x)y=q(x)$. We can directly identify:

   • $p(x)=2$
   • $q(x)=2e^x$

2. Integrating Factor: We calculate $\mu (x)$.

   $$\mu (x)=e^{\int 2dx}=e^{2x}$$

3. Multiply and Integrate: We multiply the entire standard-form equation by $\mu (x)=e^{2x}$.

   $$e^{2x}{y}^{\prime }+2e^{2x}y=e^{2x}(2e^x)$$

   The left side now perfectly matches the product rule for $\frac{d}{dx}[e^{2x}y]$. The right side simplifies to $2e^{3x}$.

   $$\frac{d}{dx}[e^{2x}y]=2e^{3x}$$

   Now, we integrate both sides with respect to $x$:

   $$\int \frac{d}{dx}[e^{2x}y]dx=\int 2e^{3x}dx$$ $$e^{2x}y=\frac{2}{3}{e}^{3x}+C$$

4. Solve for y(x): To get the final solution, we divide by $e^{2x}$ (or multiply by $e^{-2x}$).

   $$y(x)=\frac{2}{3}{e}^x+C{e}^{-2x}$$

This is the general solution. For a video walkthrough of this exact problem, I recommend the following resource.

Example 2: The Need for Standard Form

Let's tackle a problem that requires an initial step of algebraic manipulation.

1. Standard Form: The equation is not in standard form because of the $x$ multiplying $y^{\prime }$. We must divide the entire equation by $x$ to isolate $y^{\prime }$.

   $$y^{\prime }+\frac{4}{x}y=2x^2$$

   Now we can identify:

   • $p(x)=\frac{4}{x}$
   • $q(x)=2x^2$

2. Integrating Factor:

   $$\mu (x)=e^{\int \frac{4}{x}dx}=e^{4\ln |x|}$$

   Using the logarithm property $r\ln (a)=\ln (a^r)$, we can simplify this:

   $$\mu (x)=e^{\ln (x^4)}=x^4$$

   (We can drop the absolute value since $x^4$ is always non-negative).

3. Multiply and Integrate: Multiply the standard form equation by $\mu (x)=x^4$.

   $$x^4\left(y^{\prime }+\frac{4}{x}y\right)=x^4(2x^2)$$ $$x^4{y}^{\prime }+4x^{3}y=2x^6$$

   The left side is now $\frac{d}{dx}[x^{4}y]$.

   $$\frac{d}{dx}[x^{4}y]=2x^6$$

   Integrating both sides:

   $$x^{4}y=\int 2x^{6}dx=\frac{2}{7}{x}^7+C$$

4. Solve for y(x): Divide by $x^4$.

   $$y(x)=\frac{2}{7}{x}^3+\frac{C}{x^4}$$

This example highlights the importance of the first step. The integrating factor method will not work unless the equation is in standard form.

The same video resource also covers this problem in detail.

Example 3: An Example with Trigonometric Functions

Let's try one more example that involves trigonometric functions, which often appear in these problems.

1. Standard Form: Divide by $\cos (x)$.

   $$y^{\prime }+\frac{\sin (x)}{\cos (x)}y=\frac{1}{\cos (x)}$$ $$y^{\prime }+\tan (x)y=\sec (x)$$

   So, $p(x)=\tan (x)$ and $q(x)=\sec (x)$.

2. Integrating Factor: We need to find the integral of $\tan (x)$. Recall that $\int \tan (x)dx=\ln |\sec (x)|$.

   $$\mu (x)=e^{\int \tan (x)dx}=e^{\ln |\sec (x)|}=\sec (x)$$

   (Assuming $\sec (x)>0$ for simplicity).

3. Multiply and Integrate: Multiply the standard form by $\mu (x)=\sec (x)$.

   $$\sec (x)y^{\prime }+\sec (x)\tan (x)y={\sec }^2(x)$$

   The left side is the derivative of the product $\mu (x)y$, which is $\frac{d}{dx}[\sec (x)y]$. The derivative of $\sec (x)$ is $\sec (x)\tan (x)$, so this checks out.

   $$\frac{d}{dx}[\sec (x)y]={\sec }^2(x)$$

   Integrate both sides. The integral of ${\sec }^2(x)$ is $\tan (x)$.

   $$\sec (x)y=\tan (x)+C$$

4. Solve for y(x): Divide by $\sec (x)$.

   $$y(x)=\frac{\tan (x)}{\sec (x)}+\frac{C}{\sec (x)}$$

   Simplifying the trigonometric terms ($\tan (x)=\frac{\sin (x)}{\cos (x)}$ and $\sec (x)=\frac{1}{\cos (x)}$) gives us:

   $$y(x)=\sin (x)+C\cos (x)$$

To see another perspective on this problem, the following video is excellent.

Conclusion

Today, we successfully moved from theory to practice. We applied the four-step integrating factor method to solve several different types of first-order linear differential equations.

Key Takeaways:

• The first and most critical step is always to put the differential equation in standard form: $y^{\prime }+p(x)y=q(x)$.
• The calculation of the integrating factor, $\mu (x)=e^{\int p(x)dx}$, can sometimes involve logarithmic simplifications.
• Once multiplied by $\mu (x)$, the left side of the equation always conveniently becomes $\frac{d}{dx}[\mu (x)y]$, making integration straightforward.
• The process is a reliable, algorithmic procedure for finding the general solution to any first-order linear ODE.

Preview of the next lesson:
The general solutions we found today, like $y(x)=\sin (x)+C\cos (x)$, represent an infinite family of functions (one for each value of $C$). In our next lesson, we will learn how to use an initial condition—a specific point $(x_0,y_0)$ that the solution must pass through—to find the value of $C$ and pin down the one specific, or particular, solution to the problem.