Finding General Solutions Using Separation of Variables

7.6 Finding General Solutions Using Separation of Variables

Welcome back to AP Calculus with Fiveable! This topic focuses on separation of variables in differential equations. We’ve worked through modeling differential equations with slope fields and verifying solutions, so lets take the final step and learn how to find the solution to a differential equation. 🙌

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🎯 Solving Differential Equations

The solution to a differential equation is any continuous function that satisfies the differential equation. This solution set can include many equations! We learned how to go from solutions to equations in a previous lesson. To brush up on verifying solutions, check out this Fiveable guide: Verifying Solutions to Differential Equations.

But how can we go from an equation to a solution? Many differential equations, such as 2xy' + y = 3x^2 are in fact difficult to solve complex, and are out of the scope of this course. You’ll likely encounter them again in college! But we can solve other types of differential equations: separable differential equations.

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👐 Separable Differential Equations

Separable differential equations are first-order differential equations characterized by having two variables, the independent and dependent variables, that can be integrated separately to give a solution to the differential equation.

They are usually written in the form $\frac{dy}{dx}​=g(x)h(y)$, where $g(x)$ is a function of $x$ and $h(y)$ is a function of $y$.

🪜 The general approach to solving a separable differential equation involves integrating both sides of the equation to eliminate the derivative. Here are the steps:

1. 👐 Separate Variables: Express the equation in the form $\frac{dy}{dx}​=g(x)h(y)$
2. 📏 Integrate: Move terms involving $y$ to one side and terms involving $x$ to the other side. Integrate both sides with respect to their respective variables. Don’t forget the constant of integration!
3. ✏️ Solve for y: If needed, solve the equation for $y$ explicitly.
4. 🧠 Include Initial Conditions (if given): If initial conditions are provided, use them to determine any constants of integration.

We won’t be tackling this last step in this key topic, 7.6, so check out the next guide to get some practice for this level: Finding Particular Solutions Using Initial Conditions.

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📝 Practice Problems

Let’s do some practice with solving differential equations and finding general solutions, or sets of families of functions that satisfy the conditions specified by the equation.

Practice Problem #1

$$\frac{dy}{dx} = \frac{2x}{y}$$

To solve the differential equation $\frac{dy}{dx} = \frac{2x}{y}$ , we'll use separation of variables. Buckle up! 🚘

We start by separating the variables:

$$y , dy = 2x , dx$$

Now, we can integrate both sides:

$$\int y , dy = \int 2x , dx$$

After integrating, we get the following equation, where $C$ is the constant of integration.

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

Our next step is to explicitly solve for $y$:

$$y^2 = 2x^2 + 2C$$

$$y = \pm \sqrt{2x^2 + 2C}$$

Therefore, the general solution to the differential equation $\frac{dy}{dx} = \frac{2x}{y}$ is

$$y = \pm \sqrt{2x^2 + 2C}$$

The constant $C$ can be determined later if an initial condition is provided, which will be covered in the next guide: Finding Particular Solutions Using Initial Conditions.

Practice Problem #2

$$\frac{dy}{dx} + xy = y^2$$

We can start by moving the term $xy$ to the right side of the equation. That gives us:

$$\frac{dy}{dx} = y^2 - xy$$

However, we cannot separate this differential equation! There is no way to solve this one using the methods we just learned. Nice try, let's move on to the next question! 🙌

Practice Problem #3

$$\frac{dy}{dx} = x^2y$$

To solve the differential equation $\frac{dy}{dx} = x^2y$ we can use the method of separation of variables. The idea is to rearrange the equation so that all terms involving $dy$ and $y$ are on one side, and all terms involving $dx$ and $x$ are on the other side. Let’s get to work!

To begin to solve this differential equation, we use the algebra to separate our variables:

$$\frac{dy}{y} = x^2 ,dx$$

Then we can integrate both sides:

$$\int \frac{1}{y} ,dy = \int x^2 ,dx$$

This gives us:

$$\ln|y| = \frac{1}{3}x^3 + C$$

Next, we need to solve for $y$.

$$|y| = e^{\frac{1}{3}x^3 + C} = e^{C}\cdot e^{\frac{1}{3}x^3}$$

Since $,e^C$ is just another constant, we can rewrite the term as just $C$ to get:

$$|y| = Ce^{\frac{1}{3}x^3}$$

Considering the absolute value, we now have two solutions where $C$ is an arbitrary constant.

$$y = Ce^{\frac{1}{3}x^3} \quad \text{or} \quad y = -Ce^{\frac{1}{3}x^3}$$

These two are the general solutions to the given differential equation. The constant $C$ can be determined afterward with an initial condition to give a specific answer.

Therefore, the solution to the differential equation $\frac{dy}{dx} = x^2y$ is:

$$y = \pm ,Ce^{\frac{1}{3}x^3}$$

Amazing work! This is actually an exponential function, which is recognizable by the fact that the derivative is based on the dependent variable, just like $e^x$.

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🌟 Closing

Congratulations! You've made a great start to tackling separable differential equations. These equations often model various real-world scenarios, making them a powerful tool in calculus. You can expect to see these types problems on the AP Calculus exam in both the Free Response and Multiple Choice sections.

Now, go tackle those practice problems to solidify your knowledge. Happy studying! 🎉

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