In calculus, we can read different types of differential equations and their degree, and order, along with their solutions. Like general linear equations, differential equations can also be written as a system of linear differential equations. Similarly, we can define the systems of linear first order ordinary differential equations. To learn how to write the systems of linear 1st order ODEs, we require a basic understanding of what are ordinary differential equations, first order ODEs and systems of linear differential equations. These terms can be understood with the help of a brief introduction given here.

Ordinary differential equations:

We can define the ordinary differential equations (ODEs) as the relation containing only one independent variable, say x, the real dependent variable, say y, with some of its derivatives, i.e., y’, y”, ….y^n, with respect to x. Thus, F(x, y’, y”, ….yn) = 0

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First order ODEs:

First order ODEs (First order ordinary differential equations) have the highest order 1.

For example, y’ – y = 4x is the first order ODE.

Learn about first order differential equations here.

Linear First order ordinary differential equations:

The linear first order ODEs are of the form (x – y)dx + 3xdy = 0. That means the first order linear ODE contains the highest order 1 and the degree 1.

System of Linear Differential equations:

As we know, a linear differential is of the form y’ = Ax + b. The system of linear differential equations is written as:

y’₁(x) = b₁(x) + a_1,1(x) y₁ + a_1,2(x) y₂ + …. + a_1,n(x) y_n

y’₂(x) = b₂(x) + a_2,1(x) y₁ + a_2,2(x) y₂ + …. + a_2,n (x) y_n

…………………………………..

…………………………………..

y’_n(x) = b_n(x) + a_n,1(x) y₁ + a_n,2(x) y₂ + …. + a_n,n(x) y_n

Here, b_n and a_ij are the functions of the variable x.

Systems of First Order Linear Ordinary Differential Equations

The first order linear system of ordinary differential equations is of the form,

x’₁ = a₁₁(t)x₁ + a₁₂(t)x₂ + …. + a_1n(t)x_n + b₁(t)

x’₂ = a₂₁(t)x₁ + a₂₂(t)x₂ + …. + a_2n(t)x_n + b₂(t)

x’₃ = a₃₁(t)x₁ + a₃₂(t)x₂ + …. + a_3n(t)x_n + b₃(t)

: : : :

x’_n = a_n1(t)x₁ + a_n2(t)x₂ + …. + a_nn(t)x_n + b_n(t)

This system of equations can be expressed in the form of matrices as:

x’ = A(t) x + b(t)

Here,

A(t) = [a_ij (T)] is the coefficient matrix.

This can be written as:

And

This can be written in more simplified form as:

This is a non-homogeneous system. If it is free of b(t), i.e.,

Solution of a Systems of Linear First Order Ordinary Differential Equations

The solution of a system of linear first-order ordinary differential equations is the column vector x(t) subjected to the IVP.

The initial value problem (IVM) for the system of a linear first order ODEs, i.e.,

Then, we can find the solution using the concepts of eigenvalues and eigenvectors.

If we are given the system of linear ordinary differential equations of the first order in two variables, then we follow a different approach. That means, we can find the general solution for both of these variable functions. This can be understood in a better way using the solved problem given below.

Solved Example

Question:

Consider the system of two first order linear ODEs:

x′ = 3x – 2y

y′ = 2x – y

Solution:

Given system of equations is:

x′ = 3x – 2y ……..(1)

y′ = 2x – y ……..(2)

First, we need to reduce these equations.

This reduction can be made by differentiating the equations and successive replacement of the unknown functions until we get a differential equation for only one unknown function.

Consider the equation (2).

y′ = 2x – y

From this, we can write x as:

y′ + y = 2x

x = (½)y′ + (½)y……(3)

Differentiating equation equation (2), we get;

y′′ = 2x′ – y′

This can be rearranged as:

x′ = (½)y′′ + (½)y′…….(4)

Substituting equations (3) and (4) in the equation (1), we get;

x′ = 3x – 2y

(½) (y′′ + y′) = (½) [3(y′ + y) – 4y]

y′′ + y′ = 3y′ + 3y – 4y

y′′ + y′ – 3y′ + y = 0

y′′ – 2y′ + y = 0

This is the second-order linear ODE and is homogeneous with constant coefficients.

This can be solved using the standard method.

Thus, get the auxiliary equation as:

m² – 2m + I = 0 such that m_1,2 = I

Therefore, we get the general solution to this equation as:

y = c₁e^t + c₂te^t…….(5)

Now, substituting equation (5) in equation (3), we get;

x = (½)y′ + (½)y = (½) (c₁e^t + c₂te^t)′ + (½) (c₁e^t + c₂te^t)

= (½) [(c₁e^t + c₂te^t)′ + (c₁e^t + c₂te^t)]

= (½) [c₁e^t + c₂(te^t + e^t) + c₁e^t + c₂te^t]

= (½) [2c₁e^t + 2c₂te^t + c₂e^t]

= (½) 2e^t [c₁ + c₂{(½) + t}]

= e^t [c₁ + c₂{(½) + t}]

Hence, the general solution of the given system of equations is:

y(t) = c₁e^t + c₂te^t

x(t) = e^t [c₁ + c₂{(½) + t}]

Practice Problems

1. Find the general solution of the system of ordinary differential equations given below:

   x′₁ = x₁ + 2x₂ + 12e^3t

   x′₂ = 4x₁ + 3x₂ + 18e^2t

2. Solve the following system of equations.

   x′₁ = t²x₁ – e^tx₂ + 1 – t
   x′₂ = x₁ + cos t x₂