# Particular solutions with initial conditions

> AP Calculus BC · Differential Equations
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u7-particular-solutions-with-initial-conditions/

This module covers finding unique particular solutions to first-order initial value problems (IVPs) for indefinite integrals and separable differential equations, plus verification techniques aligned with AP Calculus BC exam requirements.

**Prerequisites:** Basic integration and antidifferentiation rules; Separation of variables for first-order differential equations; Derivative evaluation techniques

## Learning objectives

- Solve for unique particular solutions using initial conditions for indefinite integrals
- Find particular solutions for separable first-order differential equations
- Verify candidate solutions satisfy both the DE and initial condition
- Avoid common exam pitfalls when solving initial value problems

## Particular Solutions for Indefinite Integrals

When given a derivative $\frac{dy}{dx} = f(x)$ (a function of $x$ only), the indefinite integral produces a general antiderivative $F(x) + C$, where $C$ is an arbitrary constant. The initial condition $y(x_0) = y_0$ lets you substitute known values to solve for $C$, resulting in exactly one unique particular solution for a well-posed IVP.

**Worked example:** Given $\frac{dy}{dx} = 3x^2 + 2\sin x$ and $y(0) = 5$, find the particular solution for $y$.

1. Find the general antiderivative of $\frac{dy}{dx}$ with respect to $x$:

   $$y = \int (3x^2 + 2\sin x) dx = x^3 - 2\cos x + C$$
2. Substitute the initial condition $x=0, y=5$ into the general solution:

   $$5 = (0)^3 - 2\cos(0) + C$$
3. Simplify to solve for $C$, using $\cos(0)=1$:

   $$5 = -2 + C \implies C=7$$
4. Substitute $C=7$ back to get the final particular solution, then verify it matches both the derivative and initial condition:

   $$y = x^3 - 2\cos x + 7$$

> **tip**
>
> Always verify your value of $C$ by plugging $x_0$ back into your final particular solution before moving on—this catches 90% of algebraic errors on this type of problem.

## Particular Solutions for Separable Differential Equations

Most first-order differential equations on the AP exam are separable, meaning they can be rewritten as $g(y) dy = f(x) dx$. After integrating both sides, combine the constants from each side into a single arbitrary constant $C$ to get a general solution. Substitute the initial condition to solve for $C$ and get your unique particular solution.

**Worked example:** Find the particular solution to $\frac{dy}{dx} = \frac{xy}{2}$ that satisfies $y(1) = 4$.

1. Separate variables (valid for $y \neq 0$, which holds here since $y(1)=4$):

   $$\frac{2}{y} dy = x dx$$
2. Integrate both sides:

   $$\int \frac{2}{y} dy = \int x dx \implies 2\ln|y| = \frac{1}{2}x^2 + C_1$$
3. Simplify to get the general explicit solution, combining constants:

   $$\ln|y| = \frac{x^2}{4} + \frac{C_1}{2} \implies y = Ce^{\frac{x^2}{4}}$$
4. Substitute the initial condition $y(1)=4$ to solve for $C$:

   $$4 = Ce^{\frac{1^2}{4}} = Ce^{1/4} \implies C = 4e^{-1/4}$$
5. Write the final simplified particular solution:

   $$y = 4e^{\frac{x^2 - 1}{4}}$$

> **tip**
>
> When exponentiating after integrating $\frac{1}{y}$, always check the sign of the initial condition to fix the sign of $C$. AP graders penalize leaving $\pm$ in final answers when the sign is fixed by the initial condition.

## Verifying a Candidate Particular Solution

AP exams often ask you to confirm if a given function is the correct particular solution to an IVP. Two independent checks are required: (1) the function must satisfy the initial condition, and (2) it must satisfy the original differential equation. Checking the initial condition first eliminates wrong options quickly in multiple choice.

**Worked example:** Is $y = 3e^{3x} + \sin x$ the correct particular solution to $\frac{dy}{dx} = 3y - 3\sin x + \cos x$ with initial condition $y(0)=3$?

1. Check the initial condition first (fastest elimination step):

   $$y(0) = 3e^{0} + \sin 0 = 3(1) + 0 = 3$$
2. This matches the initial condition, so compute the derivative of the candidate:

   $$\frac{dy}{dx} = 9e^{3x} + \cos x$$
3. Substitute the candidate into the right-hand side of the DE and simplify:

   $$3y - 3\sin x + \cos x = 3(3e^{3x} + \sin x) - 3\sin x + \cos x = 9e^{3x} + \cos x$$
4. Compare: $\frac{dy}{dx}$ equals the right-hand side, so the candidate satisfies the DE. Conclusion: Yes, this is the correct particular solution.

> **tip**
>
> In MCQ, always check the initial condition first to eliminate wrong options in seconds without doing any differentiation.

## AP-Style Practice Problems

**Worked example:** Multiple Choice: Which of the following is the particular solution to $\frac{dy}{dx} = 6xy$ with initial condition $y(0) = 3$?

1. Separate variables to get $\frac{1}{y} dy = 6x dx$, then integrate both sides:

   $$\ln|y| = 3x^2 + C_1 \implies y = Ce^{3x^2}$$
2. Substitute initial condition $y(0)=3$:

   $$3 = Ce^{0} \implies C = 3$$
3. Final particular solution is $y = 3e^{3x^2}$, which matches option A.

**Worked example:** Free Response: Consider $\frac{dy}{dx} = 1 + y^2$ with $y(0) = 1$. (a) Find general solution, (b) Find particular solution, (c) Find $y\left(\frac{\pi}{4}\right)$.

1. (a) Separate and integrate:

   $$\int \frac{1}{1+y^2} dy = \int dx \implies \arctan(y) = x + C \implies y = \tan(x + C)$$
2. (b) Substitute initial condition $y(0)=1$:

   $$\arctan(1) = 0 + C \implies C = \frac{\pi}{4} \implies y = \tan\left(x + \frac{\pi}{4}\right)$$
3. (c) Substitute $x = \frac{\pi}{4}$:

   $$y\left(\frac{\pi}{4}\right) = \tan\left(\frac{\pi}{2}\right) \text{, which is undefined}$$$

**Worked example:** Applied Problem: Population $P$ (hundreds of bacteria) grows as $\frac{dP}{dt} = 0.2P$, $t$ is hours. At $t=0$, population is 250 bacteria. Find $P(t)$ and population after 5 hours.

1. Initial condition: $P(0) = 2.5$ (250 = 2.5 hundred). Separate and integrate:

   $$\int \frac{1}{P} dP = \int 0.2 dt \implies \ln|P| = 0.2t + C \implies P(t) = Ce^{0.2t}$$
2. Solve for $C$ using initial condition: $C = 2.5$, so particular solution is:

   $$P(t) = 2.5e^{0.2t}$$
3. Evaluate at $t=5$:

   $$P(5) = 2.5e^{1} \approx 6.7957 \text{ hundred} = 680 \text{ bacteria}$$

## Common pitfalls

- **Wrong:** Adding the constant of integration to only one side after integrating both sides of a separable DE.
  - Why it fails: Students are used to adding $+C$ only to the right for antiderivatives, and forget both integrals produce constants.
  - Correct: After integrating both sides, combine all constants into a single arbitrary constant before substituting the initial condition.
- **Wrong:** Leaving $\pm$ in the final particular solution after the initial condition fixes the sign of $y$.
  - Why it fails: Students keep the general solution's ambiguity even after substituting a known positive or negative initial value.
  - Correct: After substituting the initial condition, explicitly drop the wrong sign and do not leave ambiguous $\pm$ in the final answer.
- **Wrong:** Substituting the initial condition immediately after separating variables, before integrating.
  - Why it fails: Students rush to use the initial condition and skip finding the general solution entirely.
  - Correct: Always complete separation of variables, integrate both sides, and get the general solution with the arbitrary constant before substituting the initial condition.
- **Wrong:** Dropping absolute value when integrating $\frac{1}{y}$ before finding $C$, leading to an incorrect constant for negative $y$.
  - Why it fails: Students drop the absolute value out of habit before using the initial condition to set the sign of $C$.
  - Correct: Keep the absolute value until you exponentiate, then use the initial condition's sign to set the sign of your combined constant $C$.
- **Wrong:** Rounding $C$ too early when working with decimal values in applied problems.
  - Why it fails: Rounding $C$ early accumulates error, leading to a final answer outside the acceptable error range.
  - Correct: Keep $C$ as an exact value through all intermediate steps, and only round the final particular solution to the required decimal places.

## Cheatsheet

| Category | Steps/Formula | Notes |
| --- | --- | --- |
| Indefinite Integral IVP | 1. Integrate $y'(x) = f(x)$ → $y = F(x) + C$<br>2. Substitute $y(x_0)=y_0$ to solve for $C$<br>3. Substitute $C$ back | One constant for all first-order problems |
| Separable DE General Solution | 1. Rewrite $\frac{dy}{dx} = f(x)g(y)$ → $\frac{1}{g(y)} dy = f(x) dx$<br>2. Integrate both sides, combine constants into one $C$ | Only valid for $g(y) \neq 0$; check $g(y)=0$ separately if needed |
| Solve for Particular Solution (Separable) | 1. Get general solution with $C$<br>2. Substitute $y(x_0)=y_0$<br>3. Solve for $C$<br>4. Substitute $C$ back | Simplify to explicit form if possible |
| Verify Particular Solution | Two checks: (1) $y(x_0) = y_0$; (2) Substitute $y, y'$ into DE, confirm LHS=RHS | Check initial condition first to eliminate wrong candidates fast |
| Log DE Constant Simplification | $\ln\|y\| = kx + C_1 \implies \|y\| = e^{C_1}e^{kx} = Ce^{kx}$ | $C$ matches the sign of initial $y$ |
| Exponential IVP Standard Result | $\frac{dP}{dt} = kP$, $P(0)=P_0$ → $P(t) = P_0 e^{kt}$ | Can use to skip integration steps for exponential models |
| Initial Condition Notation | $y(x_0) = y_0$ | Means when $x=x_0$, $y=y_0$; do not mix up input and output |

## What's next

Mastering particular solutions with initial conditions is a prerequisite for nearly all advanced differential equation topics on AP Calculus BC, including slope fields, Euler's method, and all real-world differential equation modeling. Next, you will learn to approximate solutions for IVPs with Euler's method, which relies on the same initial $(x_0, y_0)$ used for exact particular solutions, but produces a numeric approximation instead of an exact function. You will also apply particular solutions to real-world problems for population growth, Newton's law of cooling, and logistic growth, all of which require you to use initial conditions to calibrate a model that matches observed data. Without correctly solving for the constant of integration, all subsequent work will be incorrect.

- [Exponential models with differential equations](https://www.owlsprep.com/study/ap-calculus-bc-u7-exponential-models-with-differential-equations/)
- [Logistic models with differential equations (BC only)](https://www.owlsprep.com/study/ap-calculus-bc-u7-logistic-models-with-differential-equations/)
- [Applications of Integration Overview](https://www.owlsprep.com/study/ap-calculus-bc-u8-overview/)

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