Unit II: First-Order Differential Equations and Advanced Integration Techniques

1. Integrating Factors and Rules

Introduction to Exact Equations

A first-order differential equation of the form M(x, y) dx + N(x, y) dy = 0 is called exact if its left-hand side matches the total differential of some function u(x, y). The mathematical condition for exactness is:

∂M/∂y = ∂N/∂x

If this condition is met, the general solution is obtained by integrating M with respect to x (treating y as a constant) and integrating only those terms in N that do not contain x with respect to y.

What is an Integrating Factor (IF)?

If the differential equation M(x, y) dx + N(x, y) dy = 0 is not exact (i.e., ∂M/∂y ≠ ∂N/∂x), it can often be made exact by multiplying the entire equation by a suitable function μ(x, y). This multiplying factor is called an Integrating Factor (IF).

Rules for Finding Integrating Factors

There are five primary rules used to determine the integrating factor depending on the structure of the differential equation:

Step-by-Step Examples

Example for Rule 1:

Solve: (x²y - 2xy²) dx - (x³ - 3x²y) dy = 0.

1. Identify M and N:
M = x²y - 2xy²
N = -x³ + 3x²y

2. Verify exactness:
∂M/∂y = x² - 4xy
∂N/∂x = -3x² + 6xy
Since ∂M/∂y ≠ ∂N/∂x, the equation is not exact.

3. Notice that both M and N are homogeneous functions of degree 3. Let's calculate Mx + Ny:
Mx + Ny = x(x²y - 2xy²) + y(-x³ + 3x²y)
Mx + Ny = x³y - 2x²y² - x³y + 3x²y² = x²y² ≠ 0.

4. Find the Integrating Factor:
IF = 1 / (x²y²)

5. Multiply the original equation by the IF:
(1 / (x²y²)) × (x²y - 2xy²) dx - (1 / (x²y²)) × (x³ - 3x²y) dy = 0
(1/y - 2/x) dx - (x/y² - 3/y) dy = 0

This new equation is exact. Integrating M' = 1/y - 2/x with respect to x treating y as a constant yields: x/y - 2 ln(x). Integrating the terms in N' that do not contain x with respect to y (which is 3/y) yields: 3 ln(y). Combine them for the general solution:
x/y - 2 ln(x) + 3 ln(y) = C.

2. Linear Differential Equations

A first-order differential equation is linear if the dependent variable and its derivative appear only to the first power and are not multiplied together.

Standard Form (Linear in y)

The standard form of a linear differential equation of the first order is:

dy/dx + P(x) y = Q(x)

where P(x) and Q(x) are functions of x or constants.

Solution Methodology

1. Write the equation in standard form and identify P(x) and Q(x).
2. Calculate the Integrating Factor (IF):

   IF = e^{∫P(x) dx}

3. The general solution is given by:

   y × IF = ∫(Q(x) × IF) dx + C

Alternative Form (Linear in x)

Sometimes an equation is linear with respect to x as the dependent variable:

dx/dy + P₁(y) x = Q₁(y)

For this form, the Integrating Factor is:
IF = e^{∫P₁(y) dy}
And the general solution is:
x × IF = ∫(Q₁(y) × IF) dy + C.

3. Equations Reducible to Linear Form (Bernoulli's Equation)

An equation of the form:

dy/dx + P(x) y = Q(x) yⁿ

where n is any real number (n ≠ 0 and n ≠ 1) is called Bernoulli's Equation. This equation is non-linear but can be reduced to a linear first-order differential equation using a suitable transformation.

Step-by-Step Reduction Method

1. Divide the entire equation by yⁿ to isolate the term with Q(x):

y^-n (dy/dx) + P(x) y^1-n = Q(x)

2. Introduce a new variable v:

Let v = y^1-n

3. Differentiate v with respect to x:

dv/dx = (1-n) y^-n (dy/dx)
=> y^-n (dy/dx) = (1 / (1-n)) (dv/dx)

4. Substitute these expressions back into the modified differential equation:

(1 / (1-n)) (dv/dx) + P(x) v = Q(x)

5. Multiply by (1-n) to obtain the standard linear form:

dv/dx + (1-n) P(x) v = (1-n) Q(x)

6. Solve this linear equation for v using the Integrating Factor method, then substitute back v = y^1-n to obtain the final solution.

4. Equations of First Order and Higher Degrees

These are equations where the derivative dy/dx appears with a power higher than 1. Let us denote:

p = dy/dx

A general differential equation of first order and n-th degree is represented as:

pⁿ + P₁p^n-1 + P₂p^n-2 + ... + P_n-1p + P_n = 0

where P₁, P₂, ..., P_n are functions of x and y.

Type A: Equations Solvable for p

If the left-hand side of the equation can be resolved into n linear factors in p, we can write:

(p - F₁(x, y)) (p - F₂(x, y)) ... (p - F_n(x, y)) = 0

This equates to n separate first-order, first-degree differential equations:

p = F₁(x, y), p = F₂(x, y), ..., p = F_n(x, y)

Solve each equation independently to get the solutions:
f₁(x, y, C) = 0, f₂(x, y, C) = 0, ..., f_n(x, y, C) = 0.
Note: The same arbitrary constant C must be used for each solution.

The general solution of the differential equation is the product of these individual solutions:

f₁(x, y, C) × f₂(x, y, C) × ... × f_n(x, y, C) = 0

Type B: Equations Solvable for y

If the equation can be expressed explicitly in terms of y as:

y = f(x, p)

We use the following procedure:

1. Differentiate both sides with respect to x:

   dy/dx = p = ∂f/∂x + (∂f/∂p) (dp/dx)

2. This yields a new first-order differential equation in x and p:

   F(x, p, dp/dx) = 0

3. Solve this equation to find a relation of the form:

   ϕ(x, p, C) = 0

4. Eliminate the parameter p between the original equation y = f(x, p) and the solution relation ϕ(x, p, C) = 0. If elimination is algebraically difficult, express x and y parametrically in terms of p.

Type C: Equations Solvable for x

If the equation can be expressed explicitly in terms of x as:

x = f(y, p)

We use the following procedure:

1. Differentiate both sides with respect to y:

   dx/dy = 1/p = ∂f/∂y + (∂f/∂p) (dp/dy)

2. This yields a new first-order differential equation in y and p:

   G(y, p, dp/dy) = 0

3. Solve this equation to find a relation of the form:

   ψ(y, p, C) = 0

4. Eliminate p between the original equation x = f(y, p) and ψ(y, p, C) = 0 to obtain the general solution.

Type D: Homogeneous Equations in x and y

An equation of first order and higher degree is homogeneous in x and y if it can be written in the form:

f(p, y/x) = 0

To solve such equations, substitute y = vx, which implies p = v + x(dv/dx). This reduces the equation to a separable form, allowing integration.

5. Clairaut's Equation and Reducible Forms

Clairaut's Equation

A differential equation of the form:

y = px + f(p)

where p = dy/dx, is known as Clairaut's Equation.

Method of Solution:

Differentiating both sides of the equation with respect to x:

dy/dx = p = p + x (dp/dx) + f'(p) (dp/dx)
=> [x + f'(p)] (dp/dx) = 0

This gives two possibilities:

1. dp/dx = 0: This implies p = C (where C is a constant). Substituting this back into the original equation yields the General Solution:

   y = Cx + f(C)

2. x + f'(p) = 0: This relation does not contain the constant C. Eliminating p between this relation and the original equation yields a solution called the Singular Solution.

Equations Reducible to Clairaut's Form

Certain equations that are not directly in Clairaut's form can be reduced to it using standard substitutions. Let's look at the most common transformations:

6. Singular Solutions

A Singular Solution of a differential equation is a solution that satisfies the differential equation but cannot be obtained from the general solution by assigning any particular value to the arbitrary constant C.

Geometrically, a singular solution represents the envelope of the family of curves given by the general solution.

The p-Discriminant and c-Discriminant Relations

To find the singular solution, we use the concepts of the p-discriminant and the c-discriminant:

• p-Discriminant Equation: Obtained by eliminating p between the differential equation f(x, y, p) = 0 and its derivative with respect to p, i.e., ∂f/∂p = 0.
• c-Discriminant Equation: Obtained by eliminating the constant c between the general solution Φ(x, y, c) = 0 and its derivative with respect to c, i.e., ∂Φ/∂c = 0.

Important Observations

• The envelope (singular solution) is always a locus contained within both the p-discriminant and the c-discriminant.
• A locus obtained from the discriminants is a singular solution only if it satisfies the original differential equation.
• Common Mistake: Assuming every factor of the p-discriminant is a singular solution. You must verify by substituting back into the original differential equation!

Summary of Loci Types