Introduction

This topic focuses on finding particular solutions to differential equations using initial conditions and separation of variables. The essential knowledge tells us that a general solution may describe infinitely many solutions to a differential equation, but there is only one particular solution passing through a given point. Additionally, the function $F (x) = y_{0} + \int_{a}^{x} f (t) d t$ is a particular solution to $\frac{d y}{d x} = f (x)$ satisfying $F (a) = y_{0}$ , and solutions may be subject to domain restrictions.

Understanding General vs. Particular Solutions

General Solutions and Families of Functions

A general solution to a differential equation contains an arbitrary constant and represents infinitely many possible solutions.

Example 1:

For $\frac{d y}{d x} = 2 x$ , the general solution is $y = x^{2} + C$ .

This represents a family of parabolas, each shifted vertically by the constant $C$ .

Particular Solutions from Initial Conditions

A particular solution is obtained when we specify an initial condition to determine the value of the arbitrary constant.

Example 2:

For $\frac{d y}{d x} = 2 x$ with $y (1) = 5$ :

General solution: $y = x^{2} + C$

Apply initial condition: $5 = 1^{2} + C$ , so $C = 4$

Particular solution: $y = x^{2} + 4$

Uniqueness of Particular Solutions

For most differential equations with initial conditions, exactly one particular solution passes through the given point.

Example 3:

The initial value problem $\frac{d y}{d x} = 3 y$ , $y (0) = 2$ has exactly one solution.

General solution: $y = C e^{3 x}$

Particular solution: $y = 2 e^{3 x}$

Using Separation of Variables for Particular Solutions

Complete Solution Process

Step 1: Solve the differential equation using separation of variables

Step 2: Apply the initial condition to find the constant

Step 3: Write the particular solution

Example 4:

Solve $\frac{d y}{d x} = x y$ with $y (0) = 3$ .

Step 1: Separate variables: $\frac{d y}{y} = x, d x$

Integrate: $ln ∣ y ∣ = \frac{x ^{2}}{2} + C$

General solution: $y = A e^{\frac{x ^{2}}{2}}$

Step 2: Apply $y (0) = 3$ : $3 = A e^{0} = A$

Step 3: Particular solution: $y = 3 e^{\frac{x ^{2}}{2}}$

Working with Different Initial Conditions

Example 5:

Solve $\frac{d y}{d x} = \frac{y ^{2}}{x}$ with $y (1) = 2$ .

Separate: $\frac{d y}{y ^{2}} = \frac{d x}{x}$

Integrate: $\int y^{- 2} d y = \int x^{- 1} d x$

$- y^{- 1} = ln ∣ x ∣ + C$

$- \frac{1}{y} = ln ∣ x ∣ + C$

General solution: $y = \frac{- 1}{ln ∣ x ∣ + C}$

Apply $y (1) = 2$ : $2 = \frac{- 1}{ln ∣1∣ + C} = \frac{- 1}{0 + C} = \frac{- 1}{C}$

So $C = - \frac{1}{2}$

Particular solution: $y = \frac{- 1}{ln ∣ x ∣ - \frac{1}{2}} = \frac{2}{1 - 2 ln ∣ x ∣}$

The Fundamental Theorem Connection

Direct Integration Form

For differential equations of the form $\frac{d y}{d x} = f (x)$ , the particular solution can be written using definite integration.

Theorem: If $\frac{d y}{d x} = f (x)$ with $y (a) = y_{0}$ , then:

$F (x) = y_{0} + \int_{a}^{x} f (t) d t$

is the particular solution satisfying $F (a) = y_{0}$ .

Example 6:

Solve $\frac{d y}{d x} = sin x$ with $y (0) = 1$ .

Using the theorem: $F (x) = 1 + \int_{0}^{x} sin t d t$

$F (x) = 1 + [- cos t]_{0}^{x}$

$F (x) = 1 + (- cos x - (- cos 0))$

$F (x) = 1 - cos x + 1 = 2 - cos x$

Verification of the Theorem

We can verify that $F (x) = y_{0} + \int_{a}^{x} f (t) d t$ satisfies both the differential equation and initial condition.

Check differential equation: $\frac{d}{d x} F (x) = \frac{d}{d x} [y_{0} + \int_{a}^{x} f (t) d t] = 0 + f (x) = f (x)$ ✓

Check initial condition: $F (a) = y_{0} + \int_{a}^{a} f (t) d t = y_{0} + 0 = y_{0}$ ✓

Applications with Complex Functions

Example 7:

Find the particular solution to $\frac{d y}{d x} = e^{x^{2}}$ with $y (1) = 3$ .

Since $e^{x^{2}}$ has no elementary antiderivative, we use:

$F (x) = 3 + \int_{1}^{x} e^{t^{2}} d t$

This is the exact particular solution, even though we cannot simplify the integral further.

Domain Restrictions in Solutions

Identifying Domain Issues

Solutions to differential equations may have domain restrictions due to:

Division by zero
Negative arguments in square roots or logarithms
Undefined function values

Example 8:

For $\frac{d y}{d x} = \frac{1}{y}$ with $y (0) = 1$ :

Separating: $y d y = d x$

Integrating: $\frac{y ^{2}}{2} = x + C$

General solution: $y^{2} = 2 x + 2 C$

Applying $y (0) = 1$ : $1 = 0 + 2 C$ , so $C = \frac{1}{2}$

Particular solution: $y^{2} = 2 x + 1$ , so $y = \pm 2 x + 1$

Since $y (0) = 1 > 0$ , we choose $y = 2 x + 1$

Domain restriction: $2 x + 1 \geq 0$ , so $x \geq - \frac{1}{2}$

Analyzing Solution Domains

Example 9:

For $y = \frac{1}{ln ( x - 2 )}$ with appropriate initial condition:

Domain restrictions:

$x - 2 > 0$ , so $x > 2$ (for logarithm)
$ln (x - 2) \neq = 0$ , so $x - 2 \neq = 1$ , thus $x \neq = 3$

Combined domain: $x \in (2, 3) \cup (3, \infty)$

Interval of Validity

The interval of validity is the largest interval containing the initial point where the solution exists and is differentiable.

Example 10:

For $y = \frac{2}{1 - x ^{2}}$ with $y (0) = 2$ :

The solution is undefined when $1 - x^{2} = 0$ , i.e., when $x = \pm 1$ .

Since the initial condition is at $x = 0$ , the interval of validity is $(- 1, 1)$ .

Advanced Particular Solution Problems

Multiple Step Problems

Example 11:

Solve $\frac{d y}{d x} = y cos x$ with $y (\frac{π}{2}) = 2$ .

Separate: $\frac{d y}{y} = cos x d x$

Integrate: $ln ∣ y ∣ = sin x + C$

General solution: $y = A e^{s i n x}$

Apply initial condition: $2 = A e^{s i n (\frac{π}{2})} = A e^{1}$

So $A = 2 e^{- 1} = \frac{2}{e}$

Particular solution: $y = \frac{2}{e} \cdot e^{s i n x} = 2 e^{s i n x - 1}$

Implicit Solutions with Initial Conditions

Example 12:

Solve $\frac{d y}{d x} = \frac{x}{y}$ with $y (3) = 4$ .

Separate: $y d y = x d x$

Integrate: $\frac{y ^{2}}{2} = \frac{x ^{2}}{2} + C$

General solution: $y^{2} - x^{2} = K$ where $K = 2 C$

Apply $y (3) = 4$ : $16 - 9 = K$ , so $K = 7$

Particular solution: $y^{2} - x^{2} = 7$ or $y^{2} = x^{2} + 7$

Since $y (3) = 4 > 0$ , we have $y = x^{2} + 7$

Verification and Checking

Substitution Verification

Always verify particular solutions by substituting back into the original differential equation and checking the initial condition.

Example 13:

Verify that $y = 3 e^{2 x}$ solves $\frac{d y}{d x} = 2 y$ with $y (0) = 3$ .

Check differential equation: $\frac{d y}{d x} = \frac{d}{d x} (3 e^{2 x}) = 3 \cdot 2 e^{2 x} = 6 e^{2 x}$ $2 y = 2 (3 e^{2 x}) = 6 e^{2 x}$ ✓

Check initial condition: $y (0) = 3 e^{2 (0)} = 3 e^{0} = 3$ ✓

Common Verification Errors

Arithmetic mistakes in differentiation
Incorrect application of chain rule
Sign errors in integration
Wrong substitution of initial values

Problem-Solving Strategies

Systematic Approach

Solve the differential equation for the general solution
Apply the initial condition to find the constant
Write the particular solution explicitly
Check domain restrictions
Verify by substitution

Handling Special Cases

Case 1: When the general solution is implicit, solve for the constant first, then determine explicit form if possible.

Case 2: When integration produces absolute values, use the initial condition to determine the correct sign.

Case 3: When domain restrictions exist, identify the interval of validity containing the initial point.

FiveHive