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Unit 1 - Notes

MTH166

Unit 1: Ordinary Differential Equations

1. Exact Differential Equations

1.1 Definition

A first-order ordinary differential equation of the form:

M(x, y) dx + N(x, y) dy = 0

is called an exact differential equation if the left-hand side represents the exact total differential of some function

u(x, y)

. That is,

du = M dx + N dy = 0

1.2 Necessary and Sufficient Condition

For the differential equation $M dx + N dy = 0$ to be exact, the necessary and sufficient condition is:

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

Where:

$M$ and $N$ are functions of $x$ and $y$ .
$\frac{\partial M}{\partial y}$ is the partial derivative of $M$ with respect to $y$ (treating $x$ as constant).
$\frac{\partial N}{\partial x}$ is the partial derivative of $N$ with respect to $x$ (treating $y$ as constant).

1.3 Method of Solution

If the condition of exactness is satisfied, the general solution is given by:

\int_{y=\text{const}} M \, dx + \int (\text{terms of } N \text{ free from } x) \, dy = C

Step-by-Step Procedure:

Identify $M$ and $N$ from the given equation.
Verify the condition $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ .
Integrate $M$ with respect to $x$ , treating $y$ as a constant.
Integrate only those terms in $N$ that do not contain $x$ with respect to $y$ .
Add the results of steps 3 and 4 and equate to an arbitrary constant $C$ .

2. Equations Reducible to Exact Equations (Integrating Factors)

If $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$ , the equation is not exact. However, it can often be made exact by multiplying it by a function $\mu(x, y)$ called an Integrating Factor (I.F.).

Rule 1: Homogeneous Equations

If $M dx + N dy = 0$ is a homogeneous equation in $x$ and $y$ (i.e., $M$ and $N$ are homogeneous functions of the same degree), and $Mx + Ny \neq 0$ , then:

\text{I.F.} = \frac{1}{Mx + Ny}

Rule 2: Function of the form $y f_1(xy) dx + x f_2(xy) dy = 0$

If the equation can be written in the form $y \cdot f_1(xy) dx + x \cdot f_2(xy) dy = 0$ and $Mx - Ny \neq 0$ , then:

\text{I.F.} = \frac{1}{Mx - Ny}

Rule 3: Function of $x$ only

If $\frac{1}{N} \left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right)$ is a function of $x$ alone, say $f(x)$ , then:

\text{I.F.} = e^{\int f(x) dx}

Rule 4: Function of $y$ only

If $\frac{1}{M} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right)$ is a function of $y$ alone, say $g(y)$ , then:

\text{I.F.} = e^{\int g(y) dy}

Rule 5: Exactness by Inspection

Sometimes, an Integrating Factor can be found by regrouping terms to form exact differentials of known functions. Common exact differentials include:

$d(xy) = x dy + y dx$
$d(\frac{y}{x}) = \frac{x dy - y dx}{x^2}$
$d(\frac{x}{y}) = \frac{y dx - x dy}{y^2}$
$d(\tan^{-1}\frac{y}{x}) = \frac{x dy - y dx}{x^2 + y^2}$
$d(\frac{1}{2} \ln(x^2 + y^2)) = \frac{x dx + y dy}{x^2 + y^2}$

3. Equations of the First Order and Higher Degree

These are differential equations where the highest derivative is of the first order ( $\frac{dy}{dx}$ ), but it may appear with a power higher than 1.
Let $p = \frac{dy}{dx}$ . The general form is:

p^n + P_1 p^{n-1} + P_2 p^{n-2} + \dots + P_n = 0

where

P_1, P_2, \dots, P_n

are functions of

x

and

y

Case I: Equations Solvable for $p$

If the equation can be factored into linear factors of $p$ :

(p - f_1)(p - f_2)\dots(p - f_n) = 0

We equate each factor to zero to get

n

first-order, first-degree equations:

\frac{dy}{dx} = f_1(x,y), \quad \frac{dy}{dx} = f_2(x,y), \dots

Solve each individually to get solutions

F_1(x,y,c) = 0, F_2(x,y,c) = 0, \dots

.
The general solution is the product of these solutions:

F_1(x,y,c) \cdot F_2(x,y,c) \cdots F_n(x,y,c) = 0

Case II: Equations Solvable for $y$

If the equation can be expressed explicitly as $y = f(x, p)$ .
Method:

Differentiate the equation with respect to $x$ .
Replace $\frac{dy}{dx}$ with $p$ .
This results in a differential equation involving $x, p, \text{and } \frac{dp}{dx}$ .
Solve this equation to find a relation between $x$ , $p$ , and an arbitrary constant $c$ .
Eliminate $p$ between the original equation and the result from step 4 to get the general solution. If $p$ cannot be easily eliminated, express $x$ and $y$ in terms of $p$ (parametric solution).

Case III: Equations Solvable for $x$

If the equation can be expressed explicitly as $x = f(y, p)$ .
Method:

Differentiate the equation with respect to $y$ .
Replace $\frac{dx}{dy}$ with $\frac{1}{p}$ .
This results in a differential equation involving $y, p, \text{and } \frac{dp}{dy}$ .
Solve to find a relation between $y, p, c$ .
Eliminate $p$ between the original equation and the result from step 4.

4. Clairaut's Equation

4.1 Standard Form

Clairaut's equation is a specific type of first-order, higher-degree equation of the form:

y = px + f(p)

where

p = \frac{dy}{dx}

4.2 General Solution

To solve Clairaut's equation:

Differentiate the equation with respect to $x$ :
$p = p + x \frac{dp}{dx} + f'(p) \frac{dp}{dx}$
Simplify:
$0 = [x + f'(p)] \frac{dp}{dx}$
Disregarding the factor $[x + f'(p)]$ , we equate $\frac{dp}{dx} = 0$ .
Integrating $\frac{dp}{dx} = 0$ gives $p = c$ (constant).
Substitute $p = c$ into the original equation.

Result: The general solution is obtained directly by replacing $p$ with $c$ :

y = cx + f(c)

This represents a family of straight lines.

4.3 Singular Solution

The factor ignored in the general solution derivation, $x + f'(p) = 0$ , provides the Singular Solution.

Take the general solution: $y = cx + f(c)$ .
Differentiate with respect to $c$ : $0 = x + f'(c)$ .
Eliminate between these two equations.
- Alternatively, eliminate $p$ between the original equation $y = px + f(p)$ and $x + f'(p) = 0$ .
  The singular solution is the envelope of the family of straight lines given by the general solution. It contains no arbitrary constants.

4.4 Equations Reducible to Clairaut's Form

Many non-linear equations can be transformed into Clairaut's form using suitable substitutions.
Common Substitutions:

If the equation involves $x^2$ and $y^2$ or products like $xp, yp$ , try substitutions like $X = x^2, Y = y^2$ or $X = e^x, Y = e^y$ .
Example: For , put .
- Then $P = \frac{dY}{dX} = \frac{2y dy}{2x dx} = \frac{y}{x}p$ .
- Substitute $p = \frac{x}{y}P$ into the original equation to reduce it to $Y = PX + f(P)$ .