Finding Particular Solutions Using Initial Conditions and Separation of Variables - AP Calculus BC Study Guide

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

A differential equation describes the rate of change of a quantity. Its "general solution" is not a single function, but rather an entire family of functions that all satisfy the equation, typically distinguished by a constant of integration, $C$. The core task of this topic is to move from this infinite family of solutions to a single, specific function known as the "particular solution."

This is accomplished by using a given "initial condition," which is a specific point $(x_0,y_0)$ that the solution curve must pass through. By substituting this point into the general solution, we can determine the precise value of the constant $C$. The primary algebraic method for finding the general solution for a certain class of differential equations is the method of separation of variables. Finally, we must consider the domain of this particular solution, which is defined as the largest continuous interval containing the initial x-value for which the solution function is valid.

The Method of Separation of Variables

For a differential equation to be "separable," it must be possible to write it in the form $\frac{dy}{dx}=g(x)h(y)$. The method to find the general solution involves the following steps:

1. Separate the Variables: Algebraically manipulate the equation so that all terms involving $y$ and $dy$ are on one side, and all terms involving $x$ and $dx$ 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 with respect to its variable.

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

   A single constant of integration, $C$, should be added to one side (typically the $x$ side).

3. Find the General Solution: The resulting equation, which includes the constant $C$, is the general solution. It may define $y$ implicitly or explicitly.

To find the particular solution, continue with these steps:

4. Apply the Initial Condition: Substitute the values from the given initial condition $(x_0,y_0)$ into the general solution.

5. Solve for the Constant $C$: Determine the specific numerical value of $C$.

6. Write the Particular Solution: Substitute the value of $C$ back into the general solution. If possible, solve explicitly for $y$.

Understanding the Domain of a Particular Solution

A critical aspect of finding a particular solution is determining its domain. The domain is not simply all values of $x$ for which the function is mathematically defined.

Essential Knowledge CHA-6.D.3: The domain of a particular solution to a differential equation is the largest open interval containing the initial value on which the solution is defined.

This means you must:

1. Find the explicit particular solution $y=f(x)$.

2. Identify all values of $x$ for which $f(x)$ is not defined (e.g., values that cause division by zero, the argument of a logarithm to be non-positive, or the argument of an even root to be negative).

3. These points of discontinuity will break the number line into several intervals.

4. The correct domain for the particular solution is the single interval that contains the $x$-value from the initial condition.

For example, if a solution is $y=\frac{1}{x-5}$ and the initial condition is $y(2)=-1/3$, the function is undefined at $x=5$. This creates two intervals: $(-\infty ,5)$ and $(5,\infty )$. Since the initial condition has an $x$-value of 2, and 2 is in the interval $(-\infty ,5)$, the domain of this particular solution is $(-\infty ,5)$.

Core Concepts & Rules

• Separation of Variables: This technique is used to find the general solution of a differential equation that can be written in the form $\frac{dy}{dx}=g(x)h(y)$. The process involves separating variables, integrating both sides, and adding a constant of integration, $C$.

• General vs. Particular Solutions: A general solution represents a family of functions satisfying the differential equation and includes a constant $C$. A particular solution is a single function from that family, determined by using an initial condition to find a specific value for $C$.

• Initial Condition: An initial condition is a point, $(x_0,y_0)$, that lies on the graph of the particular solution. It is the key piece of information needed to solve for the constant of integration.

• Domain of the Solution: The domain of a particular solution must be a single, continuous, open interval. This interval must contain the $x$-value of the initial condition and be bounded by any discontinuities (like vertical asymptotes) of the solution function.

Step-by-Step Example 1: Basic Application

Problem: Find the particular solution to the differential equation $\frac{dy}{dx}=-\frac{x}{y}$ with the initial condition $y(4)=-3$.

Step 1: Separate the variables.

Multiply both sides by $y$ and $dx$ to group the variables.

$$ydy=-xdx$$

Step 2: Integrate both sides.

Find the antiderivative of each side.

$$\int ydy=\int -xdx$$

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

Step 3: Apply the initial condition $y(4)=-3$.

Substitute $x=4$ and $y=-3$ into the general solution to find $C$.

$$\frac{1}{2}(-3{)}^2=-\frac{1}{2}(4{)}^2+C$$

$$\frac{9}{2}=-\frac{16}{2}+C$$

$$\frac{9}{2}=-8+C$$

Step 4: Solve for $C$.

$$C=\frac{9}{2}+8=\frac{9}{2}+\frac{16}{2}=\frac{25}{2}$$

Step 5: Write the particular solution.

Substitute $C=\frac{25}{2}$ back into the integrated equation.

$$\frac{1}{2}{y}^2=-\frac{1}{2}{x}^2+\frac{25}{2}$$

Multiply by 2 to simplify:

$$y^2=-x^2+25$$

$$x^2+y^2=25$$

To get an explicit solution for $y$, take the square root:

$$y=\pm \sqrt{25-x^2}$$

Since the initial condition is $y(4)=-3$, the $y$-value is negative. Therefore, we must choose the negative root.

$$y=-\sqrt{25-x^2}$$

Step 6: Determine the domain.

The function $y=-\sqrt{25-x^2}$ is defined when $25-x^2\ge 0$, which means $x^2\le 25$, or $-5\le x\le 5$. The CED specifies an open interval. The initial condition is at $x=4$. The largest open interval containing $x=4$ on which the solution is defined is $-5<x<5$.

Final Answer: The particular solution is $y=-\sqrt{25-x^2}$ on the domain $(-5,5)$.

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

Problem: Find the solution $y=f(x)$ to the differential equation $\frac{dy}{dx}=(y-1{)}^2\cos (\pi x)$ with the initial condition $f(1)=0$. State the domain of the solution.

Step 1: Separate the variables.

Divide by $(y-1{)}^2$ and multiply by $dx$.

$$\frac{1}{(y-1{)}^2}dy=\cos (\pi x)dx$$

Step 2: Integrate both sides.

$$\int (y-1{)}^{-2}dy=\int \cos (\pi x)dx$$

Using u-substitution for both sides (or by inspection):

$$-(y-1{)}^{-1}=\frac{1}{\pi }\sin (\pi x)+C$$

$$-\frac{1}{y-1}=\frac{1}{\pi }\sin (\pi x)+C$$

Step 3: Apply the initial condition $f(1)=0$.

Substitute $x=1$ and $y=0$.

$$-\frac{1}{0-1}=\frac{1}{\pi }\sin (\pi \cdot 1)+C$$

$$-\frac{1}{-1}=\frac{1}{\pi }(0)+C$$

$$1=0+C$$

Step 4: Solve for $C$.

$$C=1$$

Step 5: Write the particular solution and solve for $y$.

Substitute $C=1$ back into the equation.

$$-\frac{1}{y-1}=\frac{1}{\pi }\sin (\pi x)+1$$

Now, isolate $y$.

$$\frac{1}{y-1}=-\left(\frac{1}{\pi }\sin (\pi x)+1\right)$$

$$y-1=-\frac{1}{\frac{1}{\pi }\sin (\pi x)+1}$$

$$y=1-\frac{1}{\frac{1}{\pi }\sin (\pi x)+1}$$

Step 6: Determine the domain.

The solution is undefined when the denominator is zero.

$$\frac{1}{\pi }\sin (\pi x)+1=0$$

$$\sin (\pi x)=-\pi$$

Since the range of $\sin (\theta )$ is $[-1,1]$ and $-\pi \approx -3.14159$, the expression $\sin (\pi x)$ can never equal $-\pi$. Therefore, the denominator is never zero, and the function $y(x)$ has no discontinuities. The domain is all real numbers.

Final Answer: The particular solution is $y=1-\frac{1}{\frac{1}{\pi }\sin (\pi x)+1}$ on the domain $(-\infty ,\infty )$.

Using Your Calculator

The process of finding a particular solution via separation of variables is purely analytical and symbolic. A calculator cannot perform the symbolic integration or algebraic manipulation required.

However, a graphing calculator is an excellent tool for verifying your answer.

1. Graph the Slope Field: Use your calculator's graphing utility to draw the slope field for the original differential equation (e.g., $\frac{dy}{dx}=-\frac{x}{y}$ from Example 1).

2. Graph the Initial Condition: Plot the initial point (e.g., $(4,-3)$).

3. Overlay Your Solution: Graph your final particular solution (e.g., $y=-\sqrt{25-x^2}$) on top of the slope field.

4. Check for Agreement: The graph of your particular solution should "follow" the flow lines of the slope field and must pass directly through the initial condition point. This provides strong visual confirmation that your solution is correct.

AP Exam Quick Hit

Common Question Types

• Standard FRQ Part (b) or (c): Given a differential equation $\frac{dy}{dx}=f(x,y)$ and an initial condition, find the particular solution $y=f(x)$. This is one of the most predictable questions on the free-response section.

  • Example: For $\frac{dy}{dx}=2x(y+1)$, find the particular solution $y=f(x)$ with initial condition $f(1)=0$.

• Solution with Domain Restriction: A question that, after finding the particular solution, explicitly asks for the domain. This tests understanding of EK CHA-6.D.3.

  • Example: Find the solution $y=f(x)$ to $\frac{dy}{dx}=\frac{y^2}{x-1}$ passing through $(2,-1)$. State the domain of the solution. (The solution will have a vertical asymptote at $x=1$, so the domain must be $(1,\infty )$.)

Common Mistakes

• Forgetting +C: The most common error is forgetting to add the constant of integration immediately after finding the antiderivatives.

• Solving for $C$ Too Early: Substituting the initial condition into the equation before integrating. The constant $C$ only appears after integration.

• **Incorrectly Handling `ln|y|$:\ast \ast Whenintegrating$\frac{1}{y} dy$, the result is $\ln |y|$. When solving for $y$, this becomes $|y|=e^{Kx+C}=e^C{e}^{Kx}$. Students often incorrectly write $y=C{e}^{Kx}$ without considering that $C=\pm e^C$, which must be determined by the sign of the $y$-value in the initial condition.

• Domain Errors: Stating the domain as all points where the function is defined (e.g., $x\ne 1,x\ne 5$) instead of the single, continuous open interval containing the initial condition's $x$-value (e.g., $(1,5)$).

• Algebraic Errors in Isolating $y$: After finding $C$, simple mistakes in algebra when solving the equation for $y$ are very common. Be careful with reciprocals, signs, and exponents.