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# Variation of Parameters

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Last updated date: 17th Sep 2024
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## Introduction

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 = Aex + Be2x

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

y1(x) = ex

y1'(x) = ex

y2(x) = e2x

y2'(x) = 2e2x

## FAQs on Variation of Parameters

1. What is the general solution to Variation of Parameters?

Upon introduction to the second-order differential equation, we will look into how to identify the general solution.

Usually, we take the equation in the form of:

$\frac {d^2 y}{dx^2} + p \frac {dy}{dx} + qy = 0$

and decrease it down to the "characteristic equation":

r2 + pr + q = 0

Which is a quadratic equation that consists of three possible solution types based on the discriminant p2 − 4q. When p2 − 4q is positive we obtain two real roots, and the solution is

y = Aer1x + Ber2x

Zero we obtain one real root, and the solution is

y = Aerx + Bxerx

Negative we obtain two complex roots

r1 = v + wi

and

r2 = v − wi,

while the solution is

y = evx ( Ccos(wx) + iDsin(wx) )

2. What is the fundamental solution to Variation of Parameters?

In all the three cases taken in the general solution "y" is composed of two parts:

• y = Aer1x + Ber2x is composed of y1 = Aer1x and y2 = Ber2x

• y = Aerx + Bxerx is composed of y1 = Aerx and y2 = Bxerx

• y = evx ( Ccos(wx) + iDsin(wx) ) is composed of y1 = evxCcos(wx) and y2 = evxiDsin(wx)

y1 and y2 are called the fundamental solutions of the equation.

And y1 and y2 are known to be linearly independent since neither function is a constant multiple of the other.

3.  Where can I get the notes for Variation of Parameters?

The students can get the study material for the variation of parameters at Vedantu, your one-stop for all study material. The students can access the unlimited study material for absolutely free by signing in to the Vedantu website or the Vedantu app. These notes are comprehensible and made at your convenience. Made by our master teachers with several years of experience, the students get a chance to learn the smart way. Made for your better understanding and consistent progress, Vedantu.com is the solution.

4. How can I understand Variation of Parameters better?

The students can understand the Variation of Parameters better by having knowledge about differentiation and integration which is collectively known as Calculus. To solve the questions of calculus and Variation of parameters, the students need to also have the knowledge of algebra.

5. Is it important to solve the questions based on Variation of Parameters?

The students must understand the need to solve the questions based on a variety of parameters. They can take help from the solved examples to attempt all the questions after they are done understanding the chapter through the notes available at Vedantu.com. Some questions may seem hard to attempt at first but eventually, with regular practice and guidance available at Vedantu, a student can achieve top rankings in their classes and results. The questions from Variation of Parameters may also appear in competitive exams.