#Initial Conditions

#Table of Contents

Introduction to Initial Conditions
Finding Particular Solutions
Worked Example 1: Basic Differential Equation
Finding Particular Solutions Using Separation of Variables
Worked Example 2: Separation of Variables
Practice Questions
Glossary
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

General Solution: Represents an infinite number of possible solutions.
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:

\frac{dy}{dx}=f(x)

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

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

This ensures that: $F(a) = y_0 + \int_a^a f(t) , dt = y_0 + 0 = y_0$

#Worked Example 1: Basic Differential Equation

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

Solution:

Integrate the differential equation: $y = \int \frac{3}{x} , dx = 3 \ln |x| + C$
Apply the initial condition $y(e) = 1$ : $1 = 3 \ln |e| + C$ Since $\ln(e) = 1$ : $1 = 3 \times 1 + C \Rightarrow C = -2$

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

#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: $\frac{dy}{dx} = -x y^3$ with the general solution: $y^2 = \frac{1}{x^2 + C}$

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

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

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

#Worked Example 2: Separation of Variables

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

Solution:

Separate the variables: $\cos y \frac{dy}{dx} = \frac{1}{x + 3}$ $\int \cos y , dy = \int \frac{1}{x + 3} , dx$
Integrate both sides: $\sin y = \ln |x + 3| + C$
Apply the initial condition $y = \frac{3\pi}{2}$ when $x = -2$ : $\sin \left( \frac{3\pi}{2} \right) = \ln |(-2) + 3| + C \Rightarrow -1 = \ln 1 + C \Rightarrow C = -1$

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

#Practice Questions

Practice Question

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

Answer

General Solution:

y = \int x^2 \, dx = \frac{x^3}{3} + C

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

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

Practice Question

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

Answer

General Solution:

y = \int \frac{1}{x} \, dx = \ln |x| + C

Applying initial condition $y = 0$ when $x = 1$ : $0 = \ln 1 + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$

Particular Solution: $y = \ln |x|$

#Glossary

Term	Definition
Initial Condition	Additional information used to determine a particular solution of a differential equation.
General Solution	A solution that contains an arbitrary constant, representing a family of solutions.
Particular Solution	A specific solution that satisfies the initial condition.
Separation of Variables	A 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.

#Initial Conditions

#Table of Contents

Introduction to Initial Conditions
Finding Particular Solutions
Worked Example 1: Basic Differential Equation
Finding Particular Solutions Using Separation of Variables
Worked Example 2: Separation of Variables
Practice Questions
Glossary
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

#Finding Particular Solutions

General Solution: Represents an infinite number of possible solutions.
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:

\frac{dy}{dx}=f(x)

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

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

This ensures that: $F(a) = y_0 + \int_a^a f(t) , dt = y_0 + 0 = y_0$

#Worked Example 1: Basic Differential Equation

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

Solution:

Integrate the differential equation: $y = \int \frac{3}{x} , dx = 3 \ln |x| + C$
Apply the initial condition $y(e) = 1$ : $1 = 3 \ln |e| + C$ Since $\ln(e) = 1$ : $1 = 3 \times 1 + C \Rightarrow C = -2$

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

#Finding Particular Solutions Using Separation of Variables

#Example

Given: $\frac{dy}{dx} = -x y^3$ with the general solution: $y^2 = \frac{1}{x^2 + C}$

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

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

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

#Worked Example 2: Separation of Variables

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

Solution:

Separate the variables: $\cos y \frac{dy}{dx} = \frac{1}{x + 3}$ $\int \cos y , dy = \int \frac{1}{x + 3} , dx$
Integrate both sides: $\sin y = \ln |x + 3| + C$
Apply the initial condition $y = \frac{3\pi}{2}$ when $x = -2$ : $\sin \left( \frac{3\pi}{2} \right) = \ln |(-2) + 3| + C \Rightarrow -1 = \ln 1 + C \Rightarrow C = -1$

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

#Practice Questions

Practice Question

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

Answer

General Solution:

y = \int x^2 \, dx = \frac{x^3}{3} + C

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

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

Practice Question

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

Answer

General Solution:

y = \int \frac{1}{x} \, dx = \ln |x| + C

Applying initial condition $y = 0$ when $x = 1$ : $0 = \ln 1 + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$

Particular Solution: $y = \ln |x|$

#Glossary

Term	Definition
Initial Condition	Additional information used to determine a particular solution of a differential equation.
General Solution	A solution that contains an arbitrary constant, representing a family of solutions.
Particular Solution	A specific solution that satisfies the initial condition.
Separation of Variables	A 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.

First-Order Differential Equations

Emily Davis

#Initial Conditions

#Table of Contents

#Introduction to Initial Conditions

#What are Initial Conditions?

#Finding Particular Solutions

#Example of Initial Condition

#Worked Example 1: Basic Differential Equation

#Finding Particular Solutions Using Separation of Variables

#Example

#Worked Example 2: Separation of Variables

#Practice Questions

#Glossary

#Summary and Key Takeaways

Continue your learning journey

How are we doing?

First-Order Differential Equations

Emily Davis

#Initial Conditions

#Table of Contents

#Introduction to Initial Conditions

#What are Initial Conditions?

#Finding Particular Solutions

#Example of Initial Condition

#Worked Example 1: Basic Differential Equation

#Finding Particular Solutions Using Separation of Variables

#Example

#Worked Example 2: Separation of Variables

#Practice Questions

#Glossary

#Summary and Key Takeaways

Continue your learning journey

How are we doing?