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First-Order Differential Equations

Emily Davis

Emily Davis

5 min read

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Study Guide Overview

This study guide covers finding particular solutions of differential equations using initial conditions. It explains general and particular solutions, demonstrates solving with given initial conditions, and using separation of variables. Worked examples and practice questions reinforce applying these techniques. A glossary defines key terms.

Initial Conditions

Table of Contents

  1. Introduction to Initial Conditions
  2. Finding Particular Solutions
  3. Worked Example 1: Basic Differential Equation
  4. Finding Particular Solutions Using Separation of Variables
  5. Worked Example 2: Separation of Variables
  6. Practice Questions
  7. Glossary
  8. Summary and Key Takeaways

Introduction to Initial Conditions

What are Initial Conditions?

Initial conditions are additional pieces of information provided to determine a specific solution from a family of solutions of a differential equation.

Key Concept

The general solution of a differential equation represents a family of solutions. To determine a specific (particular) solution, additional information known as initial conditions (or boundary conditions) is needed.

Finding Particular Solutions

  1. General Solution: Represents an infinite number of possible solutions.
  2. Particular Solution: A unique solution that satisfies the initial condition.

Example of Initial Condition

If you are given a differential equation and a point through which the solution curve passes, you can determine the particular solution:

Consider the differential equation: dydx=f(x)\frac{dy}{dx}=f(x)

General Solution: y=∫f(x),dx+Cy = \int f(x) , dx + C where CC is the constant of integration.

Particular Solution: If the solution curve passes through the point (a,y0)(a, y_0), then: y=y0+∫axf(t),dty = y_0 + \int_a^x f(t) , dt

This ensures that: F(a)=y0+∫aaf(t),dt=y0+0=y0F(a) = y_0 + \int_a^a f(t) , dt = y_0 + 0 = y_0

Worked Example 1: Basic Differential Equation

Given: dydx=3x\frac{dy}{dx} = \frac{3}{x} and the initial condition y=1y = 1 when x=ex = e.

Solution:

  1. Integrate the differential equation: y=∫3x,dx=3ln⁡∣x∣+Cy = \int \frac{3}{x} , dx = 3 \ln |x| + C

  2. Apply the initial condition y(e)=1y(e) = 1: 1=3ln⁡∣e∣+C1 = 3 \ln |e| + C Since ln⁡(e)=1\ln(e) = 1: 1=3×1+C⇒C=−21 = 3 \times 1 + C \Rightarrow C = -2

Thus, the particular solution is: y=3ln⁡∣x∣−2y = 3 \ln |x| - 2

Finding Particular Solutions Using Separation of Variables

When solving differential equations using separation of variables, you can find the particular solution by substituting the initial condition into the general solution to determine the arbitrary constant.

Example

Given: dydx=−xy3\frac{dy}{dx} = -x y^3 with the general solution: y2=1x2+Cy^2 = \frac{1}{x^2 + C}

If the initial condition is y=13y = \frac{1}{3} when x=2x = 2:

  1. Substitute the initial condition: (13)2=14+C⇒19=14+C⇒4+C=9⇒C=5\left(\frac{1}{3}\right)^2 = \frac{1}{4 + C} \Rightarrow \frac{1}{9} = \frac{1}{4 + C} \Rightarrow 4 + C = 9 \Rightarrow C = 5

Thus, the particular solution is: y2=1x2+5y^2 = \frac{1}{x^2 + 5}

Worked Example 2: Separation of Variables

Given the differential equation: (x+3)dydx=sec⁡y(x + 3) \frac{dy}{dx} = \sec y and the initial condition y=3π2y = \frac{3\pi}{2} when x=−2x = -2.

Solution:

  1. Separate the variables: cos⁡ydydx=1x+3\cos y \frac{dy}{dx} = \frac{1}{x + 3} ∫cos⁡y,dy=∫1x+3,dx\int \cos y , dy = \int \frac{1}{x + 3} , dx

  2. Integrate both sides: sin⁡y=ln⁡∣x+3∣+C\sin y = \ln |x + 3| + C

  3. Apply the initial condition y=3π2y = \frac{3\pi}{2} when x=−2x = -2: sin⁡(3π2)=ln⁡∣(−2)+3∣+C⇒−1=ln⁡1+C⇒C=−1\sin \left( \frac{3\pi}{2} \right) = \ln |(-2) + 3| + C \Rightarrow -1 = \ln 1 + C \Rightarrow C = -1

Thus, the particular solution is: sin⁡y=ln⁡∣x+3∣−1\sin y = \ln |x + 3| - 1

Practice Questions

Practice Question

Question 1: Solve the differential equation dydx=x2\frac{dy}{dx} = x^2 given that y=2y = 2 when x=1x = 1.

Answer General Solution: y=∫x2 dx=x33+Cy = \int x^2 \, dx = \frac{x^3}{3} + C

Applying initial condition y=2y = 2 when x=1x = 1: 2=133+C⇒2=13+C⇒C=532 = \frac{1^3}{3} + C \Rightarrow 2 = \frac{1}{3} + C \Rightarrow C = \frac{5}{3}

Particular Solution: y=x33+53y = \frac{x^3}{3} + \frac{5}{3}

Practice Question

Question 2: Given the initial condition y=0y = 0 when x=1x = 1, find the particular solution to dydx=1x\frac{dy}{dx} = \frac{1}{x}.

Answer General Solution: y=∫1x dx=ln⁡∣x∣+Cy = \int \frac{1}{x} \, dx = \ln |x| + C

Applying initial condition y=0y = 0 when x=1x = 1: 0=ln⁡1+C⇒0=0+C⇒C=00 = \ln 1 + C \Rightarrow 0 = 0 + C \Rightarrow C = 0

Particular Solution: y=ln⁡∣x∣y = \ln |x|

Glossary

TermDefinition
Initial ConditionAdditional information used to determine a particular solution of a differential equation.
General SolutionA solution that contains an arbitrary constant, representing a family of solutions.
Particular SolutionA specific solution that satisfies the initial condition.
Separation of VariablesA method to solve differential equations by separating the variables on either side of the equation.

Summary and Key Takeaways

  • The general solution of a differential equation represents a family of solutions.
  • An initial condition is necessary to find a particular solution.
  • Use the initial condition to determine the constant of integration in the general solution.
  • Separation of variables is a powerful method to solve certain types of differential equations.
  • Apply the initial condition after integrating to find the specific solution.
Exam Tip

Always double-check if the solution satisfies the initial condition to ensure accuracy.

Previous Topic - Separation of VariablesNext Topic - Exponential Models

Question 1 of 7

What is the primary purpose of initial conditions in solving a differential equation? 🤔

To find the general solution

To determine a specific solution from a family of solutions

To complicate the integration process

To find arbitrary constants