Linear and Non-Linear First Order Differential Equations (ODE)

Differential Equations

In mathematics, dy/dx + My = N takes the form of a linear differential equation given M and N are numeric constants or functions in x. It includes a y and its derivative. Here, the differential is of the first order, therefore it is called a first-order linear differential equation. The motive of solving an ordinary differential equation is to find out the function or function(s) that satisfy the given equation. There is three fundamental classifications of differential equation namely: Ordinary (ODE), Partial (PDE), or differential-algebraic (DAE). They can further be secluded based on their order, degree, and linearity. Hence, here are some ordinary differential equations examples: (dy/dx) = sin x (d²y/dx²) + c²y = 0 (rdr/dα) + cos α = 12 Generally, we come across three types of ordinary differential equations: autonomous ODEs, linear ODEs, and non-linear ODEs. In this field, several scientists like Newton, Leibniz, Riccati, and Bernoulli have put forward their theories. The elementary depiction of an ordinary differential equation is Newton’s Second Law of Motion which says – the link between the time t of an object and the displacement x under the force F is given by:

1. d²x(t)/dt² = F (x(t))

Here, F is referred to as a function of the particle’s position x(t) at time t. In modern agriculture, the implementation of an ordinary differential equation is necessary to assess the length of the trajectory of agricultural vehicles operating on sloping land. This constitutes a critical planning stage in optimizing agricultural mechanisms.

Ordinary Differential Equation

ODEs are explained as differential equations having one or more functions of only one independent variable and the derivatives of those functions. For the other two types of differential equations, it is possible to have derivatives of functions of more than one variable. These include partial, homogenous, and nonhomogenous differential equations. Provided, F is a function of p, q, and derivatives of q, then an equation can be noted as: F (p, q, q’, . . . , q^(n-1)) = q⁽ⁿ⁾. The equation is of the order n.

Ordinary Differential Equation Order

The highest derivative’s order is considered as the order of differential equations. An n^th order ordinary differential equation is written as:  F(x, y, y’,…,yⁿ) = 0. Here y’ is either dy/dx or dy/dt and yⁿ can be dⁿy/dxⁿ or dⁿy/dtⁿ) For linear ODE, the equation is: p₀(x)yⁿ + p₁(x)yⁿ-1 + … +p_n(x)y = q(x). The function p₁(x) is a co-efficient of the linear equation.

First Order Differential Equations

First-order DEs are of the form dy/dx = f(x,y). The function f(x,y) is defined on a region of the XY plane. Two techniques are devised to solve the linear first-order differential equations namely: integrating method and method of variation constant. First-order DEs find use in solving problems based on Electrical circuits, falling bodies, Newton’s law of cooling, and dilution. Classification of differential equation of first order is given as follows:

• Linear DEs
• Homogenous DEs
• Exact Equations
• Separable Equations
• Integrating Factor or IF

Two characteristics of first-order DEs are:

• Logarithmic and trigonometric functions cannot be included.
• The product of y and any of its derivatives cannot be there.

<br. Problem: Solve the equation y’ + 2xy = x
Here, the equation is already presented in standard form, y’ + P(x)y + Q(x)
So, P(x) = 2x and Q(x) = x
Let us multiply both sides by;
α(x) = e ^∫Pdx = e ^∫2xdx = e^x2
e^x2 + 2xe^x2y = xe^x2
d/dx(e^x2y) = xe^x2
Integrate both sides,
e^x2y = ∫ xe^x2dx
e^x2y = ½. e^x2+ k
y = ½ + ke^-x2

Ordinary Differential Equations Examples

Problem  dx/dt+e^tx(t)=t²cos(t) 
(1) Initial condition be x(0)=5x(0)=5.
Solution:
To begin with, let’s calculate the integrating factor
μ(t)=e^∫etdt=e^et=exp(e^t)  [exp(x) is another way of writing e^x].
Multiplying equation (1) by μ(t), now the left-hand side is the derivative of μ(t)x(t).
This can written as:
d/dt(exp(e^t)x(t)) = t²cos(t)exp(e^t).
To solve the ODE in terms of the initial conditions x(0), we integrate from 0 to t,
∫₀^t d/ds(exp(e⁸)x(s))ds = ∫₀^t s²cos(s)exp(e^s)ds
exp(e^t)x(t) – exp(e0)x(0) =∫₀^t s²cos(s)exp(e^s)ds
As we cannot find the integral analytically, we still consider the ordinary differential equation solved. Referring to the initial conditions x(0) = 5, we can write the solution of the ODE (1) as:
x(t)=5exp(1−e^t) +∫₀^t  s2cos(s)exp(e⁸−e^t)ds
You can find that x(t) satisfies the ODE (1) and the initial conditions x(0) = 5.
Problem
Cos² 𝑥 sin(𝑥) 𝑑𝑦/𝑑𝑥 + cos³ (𝑥) 𝑦 = 1
Solution:
First, let us change the equation to standard form from general.
𝑑𝑦/𝑑𝑥 + cos(𝑥)/sin(𝑥) 𝑦 = 1 /cos²(𝑥) sin(𝑥)
The integrating factor is:
𝑒 ^{∫cos(𝑥)/sin(𝑥) 𝑑𝑥} = 𝑒 ^{ln |sin(𝑥)|} = sin(𝑥)
Multiply the standard form with sin(𝑥),
sin(𝑥) 𝑑𝑦/𝑑𝑥 + cos(𝑥)𝑦 = 1/cos² 𝑥
∴ 𝑑/𝑑𝑥 [𝑦. sin(𝑥)] = 1/cos²(𝑥)
∴ 𝑦. sin(𝑥) = ∫sec² (𝑥) 𝑑𝑥 = tan 𝑥 + 𝑐
∴ (𝑥) = 1/cos(𝑥) + 𝑐/sin(𝑥)
Or, ( 𝑥) = sec (𝑥) + 𝑐. csc(𝑥)

Conclusion

There are two types of ordinary differential equations: linear and non-linear. Linear DEs are linear in respect of both x(t) and its first-order derivative dx/dt (t). While non-linear DEs are not linear in the unknown function and their derivatives. Therefore, it cannot be written as x=Bx for some matrix B.