Finding General Solutions Using Separation of Variables - AP Calculus AB Study Guide

The Core Idea: Finding Particular Solutions Using Initial Conditions and Separation of Variables

A differential equation relates a function to its derivatives. The solution to a differential equation is not a single value, but an entire family of functions, often called the "general solution." This family is represented by an equation that includes a constant of integration, $C$. To identify one specific function from this infinite family, we need more information. This information is provided as an "initial condition," which is a specific point $(x,y)$ that the solution curve must pass through.

This topic focuses on a method called "separation of variables" to solve a specific type of differential equation known as a separable differential equation. By separating the variables, we can use antidifferentiation on both sides of the equation to find the general solution. Then, by substituting the values from the initial condition into the general solution, we can determine the precise value of the constant $C$, which in turn gives us the "particular solution" that satisfies the given conditions.

The Method of Separation of Variables

The primary technique for solving separable differential equations is the method of separation of variables. A differential equation is separable if it can be written in the form:

$$\frac{dy}{dx}=g(x)h(y)$$

The process to solve it is as follows:

1. Separate the variables: Algebraically manipulate the equation so that all $y$ terms (including $dy$) are on one side and all $x$ terms (including $dx$) are on the other.

   $$\frac{1}{h(y)}dy=g(x)dx$$

2. Integrate both sides: Find the antiderivative of each side of the equation.

   $$\int \frac{1}{h(y)}dy=\int g(x)dx$$

3. Find the general solution: The result of the integration will be an equation relating $x$ and $y$ that includes a constant of integration, $C$. This is the general solution.

   $$H(y)=G(x)+C$$

   (where $H(y)$ and $G(x)$ are the antiderivatives).

Understanding the Role of the Initial Condition

The general solution $H(y)=G(x)+C$ represents an infinite family of curves, where each value of $C$ corresponds to a different curve. An initial condition, such as $y(x_0)=y_0$, provides a specific point $(x_0,y_0)$ that must lie on our desired solution curve.

The critical step is to use this point to find the specific value of $C$. By substituting $x_0$ for $x$ and $y_0$ for $y$ into the general solution, we create an equation with only one unknown: $C$. Solving for $C$ and substituting its value back into the general solution yields the unique particular solution that satisfies the differential equation and passes through the given point. It is often algebraically simpler to solve for $C$ immediately after integrating, before attempting to isolate $y$ in the final equation.

Core Concepts & Rules

• A particular solution to a differential equation is a specific function that satisfies both the differential equation and a given initial condition.

• Separable differential equations are those that can be algebraically rearranged to isolate all $y$ and $dy$ terms on one side and all $x$ and $dx$ terms on the other.

• The method of separation of variables is the primary technique used to solve these equations.

• The first step in finding a particular solution is to separate the variables and then find the antiderivative of both sides of the equation.

• The process of antidifferentiation introduces a constant of integration, $C$, resulting in a general solution that represents a family of functions.

• The given initial condition (a specific point) is substituted into the general solution to determine the unique value of the constant $C$.

• Once $C$ is found, substituting it back into the general solution provides the specific particular solution.

Step-by-Step Example 1: Basic Application

Problem: Find the particular solution to the differential equation $\frac{dy}{dx}=\frac{x^2}{y}$ with the initial condition $y(3)=-5$.

Step 1: Separate the variables.

Multiply both sides by $y$ and $dx$ to group the $y$ terms with $dy$ and the $x$ terms with $dx$.

$$ydy=x^{2}dx$$

Step 2: Integrate both sides.

Find the antiderivative of each side. Remember to add the constant of integration, $C$, to one side (conventionally the $x$ side).

$$\int ydy=\int x^{2}dx$$

$$\frac{1}{2}{y}^2=\frac{1}{3}{x}^3+C$$

This is the general solution.

Step 3: Use the initial condition to find $C$.

Substitute the values from the initial condition $y(3)=-5$ (i.e., $x=3$ and $y=-5$) into the general solution.

$$\frac{1}{2}(-5{)}^2=\frac{1}{3}(3{)}^3+C$$

$$\frac{1}{2}(25)=\frac{1}{3}(27)+C$$

$$\frac{25}{2}=9+C$$

$$C=\frac{25}{2}-9=\frac{25}{2}-\frac{18}{2}=\frac{7}{2}$$

Step 4: Write the particular solution.

Substitute the value of $C$ back into the general solution. You can leave the solution in this implicit form or solve for $y$.

$$\frac{1}{2}{y}^2=\frac{1}{3}{x}^3+\frac{7}{2}$$

To solve for $y$, multiply by 2 and take the square root.

$$y^2=\frac{2}{3}{x}^3+7$$

$$y=\pm \sqrt{\frac{2}{3}{x}^3+7}$$

Since our initial condition is $y(3)=-5$, the $y$ value is negative. Therefore, we must choose the negative root.

$$y=-\sqrt{\frac{2}{3}{x}^3+7}$$

Step-by-Step Example 2: Exam-Style Application

Problem: Find the particular solution $y=f(x)$ to the differential equation $\frac{dy}{dx}=(y-1)\cos (x)$ with the initial condition $f(\pi )=3$.

Step 1: Separate the variables.

Divide by $(y-1)$ and multiply by $dx$.

$$\frac{1}{y-1}dy=\cos (x)dx$$

Step 2: Integrate both sides.

The antiderivative of $\frac{1}{y-1}$ is $\ln |y-1|$, and the antiderivative of $\cos (x)$ is $\sin (x)$.

$$\int \frac{1}{y-1}dy=\int \cos (x)dx$$

$$\ln |y-1|=\sin (x)+C$$

This is the general solution.

Step 3: Use the initial condition to find $C$.

Substitute the initial condition $f(\pi )=3$ (i.e., $x=\pi$ and $y=3$) into the general solution.

$$\ln |3-1|=\sin (\pi )+C$$

$$\ln |2|=0+C$$

$$C=\ln (2)$$

Step 4: Write the particular solution.

Substitute $C=\ln (2)$ back into the general solution.

$$\ln |y-1|=\sin (x)+\ln (2)$$

Now, solve for $y$. Exponentiate both sides using base $e$.

$$e^{\ln |y-1|}=e^{\sin (x)+\ln (2)}$$

$$|y-1|=e^{\sin (x)}\cdot e^{\ln (2)}$$

$$|y-1|=2e^{\sin (x)}$$

To remove the absolute value, we check our initial condition $y=3$. Since $3-1=2$ is positive, we can drop the absolute value bars.

$$y-1=2e^{\sin (x)}$$

$$y=2e^{\sin (x)}+1$$

Using Your Calculator

This topic is almost entirely analytical, meaning the steps of separating, integrating, and solving for $C$ must be done by hand. A calculator has a very limited role but can be useful for checking your final answer.

How to Check Your Solution:

Suppose you found the particular solution $y=f(x)$.

1. Graph the solution: Enter your final particular solution (e.g., y = 2e^{\sin(x)} + 1`) into `Y1` on your calculator. 2. **Verify the initial condition:** Use the `CALC` menu (`2nd` + `TRACE`) and select $1: value. Enter the x-value from your initial condition (e.g., $x=\pi$). The calculator should return the y-value from the initial condition (e.g., $y=3$). If it doesn't, you made an error finding $C$ or solving for $y$.

2. Verify the derivative: You can also check if your solution satisfies the original differential equation at the initial point.

   • On the home screen, calculate the value of the derivative of your solution at the initial x-value. Using the nDeriv command (often found in the MATH menu): nDeriv(Y1, X, π).

   • Separately, calculate the value of the original differential equation $(y-1)\cos (x)$ at the initial point $(\pi ,3)$: $(3-1)cos(\pi )$.

   • The two values should be equal. In this case, nDeriv would return $-2$, and (3-1)cos(\pi) = 2(-1) = -2`. This confirms the solution is correct. ## AP Exam Quick Hit ### Common Question Types - **Find the Particular Solution:** A direct question, often in the multiple-choice or free-response section, that provides a separable differential equation and an initial condition and asks for the particular solution $y=f(x).

   • Example: "Find the solution to the differential equation $\frac{dy}{dx}=2x{y}^2$ that passes through the point $(1,1/2)$."

• FRQ with Slope Fields: A common free-response question first asks you to sketch a slope field for a given differential equation, and a subsequent part asks you to find the particular solution that passes through a specific point shown on the field.

  • Example: "Part (a): On the axes provided, sketch a slope field for $\frac{dy}{dx}=\frac{x}{2y}$. Part (b): Find the particular solution $y=f(x)$ to the differential equation with the initial condition $f(2)=3$."

Common Mistakes

• Forgetting + C: The most frequent error is forgetting to include the constant of integration, $C$, immediately after taking the antiderivative. This makes it impossible to find the correct particular solution.

• Adding $+C$ to Both Sides: You only need one constant of integration. Adding $C_1$ to one side and $C_2$ to the other is redundant, as they can be combined into a single constant $C=C_2-C_1$.

• Incorrect Separation: Failing to correctly move all $y$ terms to one side and all $x$ terms to the other. For example, in $\frac{dy}{dx}=x+y$, trying to separate this is incorrect because it is not a separable equation. For a separable equation like $\frac{dy}{dx}=\frac{e^x}{y}$, a mistake would be integrating $ydy=e^{x}dx$ instead of the correct $\int ydy=\int e^{x}dx$.

• Solving for $C$ at the Wrong Time: Plugging in the initial condition values for $x$ and $y$before integrating. The initial condition must be applied to the general solution (the result of integration).

• Algebra Errors with $ln$ and $e$: When solving for $y$ from an equation like $\ln |y|=G(x)+C$, common mistakes include:

  • Forgetting that $e^{a+b}=e^a\cdot e^b$.

  • Incorrectly handling the absolute value. You must use the initial condition to determine whether $y$ is positive or negative to correctly resolve $|y|$.