Variation of Parameters for Solving Differential Equations

The Variation in Maths is mostly based on the maths of Class 12 Mathematics which the students need to learn for the chapter named differential equations. This concept is used in many competitive examinations for students pursuing fields related to mathematics. The students will learn about what is meant by the Variation of Parameters. Then, the notes provide the student with a better glimpse by explaining the method with a generic example with step-by-step calculation. Then, you get to understand the method in Variation of parameters and 2 methods in Variation of Parameters. The end is concluded with a solved example so you get to solve it first and then correct your mistakes if there have been any. 

What is Variation in Maths?

Variation of parameters or let’s say variation in mathematics is a general method of finding a specific solution of a differential equation through replacing the constants in the solution of an associated (homogeneous) equation by functions and identifying these functions such that the original differential equation is satisfied.

In order to illustrate the method, let’s say it is desired to determine a specific solution of the equation: - y″ + p(x) y′ + q(x) y = g(x).

Variation of Parameters

In order to illustrate the method, let’s say it is desired to determine a specific solution of the equation: - y″ + p(x) y′ + q(x) y = g(x).

For the purpose of using this method, it is necessary to first know the general solution of the corresponding homogeneous equation—i.e., an associated equation where the right-hand side is zero. If y₁(x) and y₂(x) are two different solutions of the equation, then any combination ay₁(x) + by₂(x) is also a solution, known as the general solution, for any constants a and b.

The variation of parameters involves replacing the constants a and b by functions u₁(x) and u₂(x) and identifying what these functions must be to satiate the actual non-homogeneous equation. After a few manipulations, it can be presented that if the functions u₁(x) and u₂(x) satisfies the mathematical expression u′₁y₁ + u′₂y₂ = 0 and u₁′y₁′ + u₂′y₂′ = g, then u₁y₁ + u₂y₂ will satiate the original differential equation. The last two equations can be solved for providing the u1′ = −y₂g/(y₁y₂′ − y₁′y₂) and u₂′ = y₁g/(y₁y₂′ − y₁′y₂). These last equations either will identify u₁ and u₂ or else will cater as an initial point for determining an estimated solution.

Method of Variation of Parameters

In this lesson, we will discuss the method of variation of parameters with respect to second-order differential equations of this type:

\[\frac {D^2 y}{dx^2} + P(x) \frac {dy}{dx} + Q(x)y = f(x)\]

where P(x), Q(x) and f(x) are said to be functions of x.

Two Methods in Variation of Parameters

There are mainly 2 methods of solving equations like:

\[\frac {D^2 y}{dx^2} + P(x) \frac {dy}{dx} + Q(x)y = f(x)\]

Undetermined Coefficients that only work when f(x) will be a polynomial, exponential, 

sine, cosine, or a linear combination of those.

Variation of Parameters works on a wide range of functions but is a bit messy to use.

Solutions to Variation of Parameters

In order to keep things simple, we will only look at the case: d²y

\[Dx^2 + P \frac {dy}{dx} + qy = f(x)\]

In which, p and q are constants and f(x) is a non-zero function of x.

A full-fledged solution to such an equation can be identified by combining two types of solution i.e.:

1. The general solution of the homogeneous equation expressed as \[\frac{d^2 y}{dx^2} + P \frac {dy}{dx} + qy = 0\]

2. Particular solutions of the non-homogeneous equation expressed as \[\frac {d^2 y}{dx^2} +  \frac {dy}{dx} + qy = f(x)\]

Remember that f(x) can be a single function or a sum of two or more functions.

Once we have determined the general solution and all the particular solutions, then the ultimate complete solution is identified by adding up all the solutions together.

This method depends upon integration.

A minor issue with this method is that, although it may produce a solution, in some cases the solution has to be left as an integral.

Solved Example using Variation of Parameter Formula

Example:  

Solve  the following equation: \[\frac {d^2 y}{dx^2} - 3 \frac {dy}{dx} + 2y = e^x\]

1. Finding the general solution of \[\frac {d^2 y}{dx^2} - 3 \frac {dy}{dx} + 2y = 0\]

Solution:

The characteristic equation will be: r² − 3r + 2 = 0

Factor: (r − 1)(r − 2) = 0

r = 1 or 2

Thus, the general solution of the differential equation is y = Ae^x + Be^2x

Therefore, in this case, the fundamental solutions and their derivatives will be:

y₁(x) = e^x

y₁'(x) = e^x

y₂(x) = e^2x

y₂'(x) = 2e^2x