Ordinary differential equations of first order notes — Unit 1

An equation involving independent variables, dependent variables, and derivatives of dependent variables with respect to independent variables is called a Differential Equation.

⇒Order and Degree

• Order: The highest order derivative present in the differential equation.
• Degree: The highest power (exponent) of the highest order derivative, provided the equation is written as a polynomial in derivatives (i.e., free from radicals and fractions).

If a first-order, first-degree differential equation $\frac{dy}{dx} = f(x, y)$ can be expressed in the form $f(x)dx + g(y)dy = 0$, then it is called variable separable.

1. 1.Group all terms involving $x$ and $dx$ on one side.
2. 2.Group all terms involving $y$ and $dy$ on the other side.
3. 3.Integrate both sides independently.
4. 4.Add an arbitrary constant $C$ to one side of the equation.

Example: Solve $\frac{dy}{dx} = \frac{x}{y}$

$\Rightarrow y\\,dy = x\\,dx$
$\Rightarrow \int y\\,dy = \int x\\,dx$
$\Rightarrow \frac{y^2}{2} = \frac{x^2}{2} + C$

A differential equation of the form $\frac{dy}{dx} = \frac{f(x,y)}{g(x,y)}$ is Homogeneous if $f(x,y)$ and $g(x,y)$ are homogeneous functions of the same degree.

⇒Solution Method:

1. 1.Substitute $y = vx$.
2. 2.Differentiate with respect to $x$: $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
3. 3.Substitute these back into the original equation.
4. 4.The equation will reduce to a Variable Separable form in terms of $v$ and $x$.
5. 5.Integrate and replace $v$ with $\frac{y}{x}$ in the final answer.

Equations of the form: $\frac{dy}{dx} = \frac{a_1x + b_1y + c_1}{a_2x + b_2y + c_2}$

⇒Case 1: If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

Put $x = X + h$ and $y = Y + k$. Choose $h$ and $k$ such that the constants vanish: $a_1h + b_1k + c_1 = 0$ and $a_2h + b_2k + c_2 = 0$. The equation becomes homogeneous in $X, Y$.

⇒Case 2: If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = m$

Put $(a_1x + b_1y) = t$ and differentiate w.r.t $x$. The equation reduces directly to Variable Separable form.

An equation $M(x,y)dx + N(x,y)dy = 0$ is Exact if it is the total differential of some function $U(x,y) = c$ (i.e., $dU = 0$).

⇒Solution of an Exact Equation:

$$\int M\\,dx \text{ (treating y as constant)} \\;+\\; \int (\text{terms in } N \text{ free from } x)\\,dy = C$$

Solve: $(x^2 - y)dx + (y^2 - x)dy = 0$

• Here, $M = x^2 - y$ and $N = y^2 - x$.
• $\frac{\partial M}{\partial y} = -1$ and $\frac{\partial N}{\partial x} = -1$.
• Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is Exact.

Integration:
$\int (x^2 - y)\\,dx \text{ [y is constant]} = \frac{x^3}{3} - xy$

$\int (\text{terms in } N \text{ free from } x)\\,dy = \int y^2\\,dy = \frac{y^3}{3}$

Final Solution:
$$\frac{x^3}{3} - xy + \frac{y^3}{3} = C$$

If $Mdx + Ndy = 0$ is NOT exact ($\partial M / \partial y \neq \partial N / \partial x$), we multiply it by a suitable function $u(x,y)$ called an Integrating Factor (I.F.) to make it exact.

⇒Rule 1: Homogeneous Equations

If $M$ and $N$ are homogeneous functions of the same degree, and $Mx + Ny \neq 0$, then:
$$I.F. = \frac{1}{Mx + Ny}$$

⇒Rule 2: $f_1(xy)y\\,dx + f_2(xy)x\\,dy = 0$

If the equation is of this form and $Mx - Ny \neq 0$, then:
$$I.F. = \frac{1}{Mx - Ny}$$

⇒Rule 3: Function of $x$ alone

If $\frac{1}{N} \left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right) = f(x)$ (a function of $x$ only or a constant), then:
$$I.F. = e^{\int f(x)\\,dx}$$

⇒Rule 4: Function of $y$ alone

If $\frac{1}{M} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) = g(y)$ (a function of $y$ only or a constant), then:
$$I.F. = e^{\int g(y)\\,dy}$$

A first-order differential equation is linear if the dependent variable ($y$) and its derivative ($\frac{dy}{dx}$) appear only in the first degree, and are not multiplied together.

Standard Form: $$\frac{dy}{dx} + Py = Q$$
where $P$ and $Q$ are functions of $x$ only (or constants).

1. 1.Identify $P$ and $Q$.
2. 2.Calculate the Integrating Factor: $I.F. = e^{\int P\\,dx}$
3. 3.Write the solution directly: $y \times (I.F.) = \int Q \times (I.F.)\\,dx + C$

Solve: $\frac{dy}{dx} + \frac{1}{x}y = x^2$

• Compare with $\frac{dy}{dx} + Py = Q$. Here $P = \frac{1}{x}, Q = x^2$.
• $I.F. = e^{\int \frac{1}{x}\\,dx} = e^{\ln x} = x$.
• Solution: $y \cdot (x) = \int (x^2) \cdot (x)\\,dx + C$
• $xy = \int x^3\\,dx + C$
• $xy = \frac{x^4}{4} + C$

An equation of the form: $$\frac{dy}{dx} + Py = Qy^n$$ where $P$ and $Q$ are functions of $x$ alone, is called Bernoulli's Equation.

It is non-linear but can be reduced to a linear form.

1. 1.Divide the entire equation by $y^n$: $y^{-n}\frac{dy}{dx} + Py^{1-n} = Q$
2. 2.Put $y^{1-n} = v$. Differentiate w.r.t $x$: $(1-n)y^{-n}\frac{dy}{dx} = \frac{dv}{dx}$.
3. 3.Substitute this back. The equation becomes a standard Linear Equation in $v$ and $x$.

During exams, the method is rarely given. Use this checklist in order:

A curve that cuts every member of a given family of curves at a right angle ($90^\circ$) is called an Orthogonal Trajectory.

⇒Cartesian Coordinates ($x, y$):

1. 1.Differentiate the given family equation $f(x,y,c)=0$ w.r.t $x$ to find $\frac{dy}{dx}$.
2. 2.Eliminate the arbitrary constant $c$ to form the differential equation.
3. 3.Replace $\frac{dy}{dx}$ with $-\frac{dx}{dy}$ (since product of perpendicular slopes is $-1$).
4. 4.Solve the new differential equation to get the orthogonal family.

For curves given in polar form $r = f(\theta)$:

1. 1.Differentiate the given equation w.r.t $\theta$ to find $\frac{dr}{d\theta}$.
2. 2.Eliminate the arbitrary constant to form the differential equation.
3. 3.Replace $\frac{dr}{d\theta}$ with $-r^2\frac{d\theta}{dr}$.
4. 4.Solve the resulting differential equation to find the orthogonal trajectories.

The rate of change of temperature of a body is proportional to the difference between its own temperature and the temperature of the surrounding medium.

$$\frac{dT}{dt} = -k(T - T_s)$$

• $T$ = Temperature of the body at time $t$
• $T_s$ = Temperature of the surrounding (constant)
• $k$ = Positive constant of proportionality (the negative sign shows temp is decreasing)

This is a simple Variable Separable ODE. Integrating gives:
$\ln(T - T_s) = -kt + C$

ODEs are heavily used to model transient currents in circuits.

⇒L-R Series Circuit

By Kirchhoff's Voltage Law, the sum of voltage drops across an Inductor ($L$) and a Resistor ($R$) equals the applied voltage $E(t)$:

$$L\frac{di}{dt} + Ri = E(t)$$

This is a standard First-Order Linear Differential Equation.
Divide by $L$: $\frac{di}{dt} + \frac{R}{L}i = \frac{E}{L}$.
Here, $P = \frac{R}{L}$ and $I.F. = e^{Rt/L}$.

For a circuit with a Resistor ($R$) and a Capacitor ($C$):

$$R i + \frac{1}{C}q = E(t)$$

Since current $i = \frac{dq}{dt}$, the equation becomes:

$$R\frac{dq}{dt} + \frac{1}{C}q = E(t)$$

This is a linear ODE in terms of charge $q$.
$I.F. = e^{t/RC}$.
Once $q(t)$ is found by integrating, differentiate it to find the transient current $i(t)$.

⇒Test yourself before the exam:

• Can I determine the order and degree of a complex ODE containing radicals?
• What is the condition for an equation to be Exact?
• If an equation is homogeneous, what substitution must I make?
• Write the Integrating Factor formula for a Linear Differential Equation.
• What substitution converts Bernoulli's equation into a linear equation?
• In polar coordinates, what do you replace $dr/d\theta$ with to find orthogonal trajectories?