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Variation of parameters or letâ€™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, letâ€™s say it is desired to determine a specific solution of the equation: - yâ€³ + p(x) yâ€² + q(x) y = g(x).

In order to illustrate the method, letâ€™s say it is desired to determine a specific solution of the equation: - yâ€³ + p(x) yâ€² + 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 equationâ€”i.e., an associated equation where the right-hand side is zero. If y1(x) and y2(x) be 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 uâ€²1y1 + uâ€²2y2 = 0andu1â€²y1â€² + u2â€²y2â€² = g, then u1y1 + u2y2 will satiate the original differential equation. The last two equations can be solved for providing the u1â€² = âˆ’y2g/(y1y2â€² âˆ’ y1â€²y2) and u2â€² = y1g/(y1y2â€² âˆ’ y1â€²y2). These last equations either will identify u1 and u2 or else will cater as an initial point for determining an estimated solution.

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

DÂ²y/dxÂ² + P(x) dy/dx + Q(x)y = f(x)

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

There are mainly 2 methods of solving equations like:

DÂ²y/dxÂ² + P(x) dy/dxÂ + 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.

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

Dx2 + p 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.:

The general solution of the homogeneous equation expressed as d2y/dx2 + p dy/dx + qy = 0.

Particular solutions of the non-homogeneous equation expressed as d2y/dx2+dy/ DxÂ + 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.

Example:

SolveÂ the following equation: d2y/dx2Â âˆ’ 3 dy/dxÂ + 2y = ex

1. Finding the general solution of d2y/dx2 âˆ’ 3 dy/dx + 2y = 0

Solution:

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

Factor: (r âˆ’ 1)(r âˆ’ 2) = 0

r = 1 or 2

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

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

y_{1}(x) = ex

y_{1}'(x) = ex

y_{2}(x) = e2x

y_{2}'(x) = 2e2x

FAQ (Frequently Asked Questions)

Q1. What is the General Solution to Variation of Parameters?

Answer: 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:

d^{2}y/dx^{2} + p dy/dx + qy = 0

and decrease it down to the "characteristic equation":

r^{2} + pr + q = 0

Which is a quadratic equation that consists of three possible solution types based on the discriminant p^{2} âˆ’ 4q. When p^{2} âˆ’ 4q is positive we obtain two real roots, and the solution is

y = Ae^{r}_{1}^{x} + Ber_{2}x

Zero we obtain one real root, and the solution is

y = Ae^{rx} + Bxe^{rx}

Negative we obtain two complex rootsÂ

r_{1} = v + wiÂ

andÂ

r_{2} = v âˆ’ wi,

while the solution is

y = e^{vx} ( Ccos(wx) + iDsin(wx) )

Q2. What is the Fundamental Solution to Variation of Parameters?

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

y = Ae

^{r}_{1}^{x}+ Be^{r}_{2}^{x}is composed of y_{1}= Ae^{r}_{1}^{x}and y_{2}= Be^{r}_{2}^{x}

y = Ae

^{rx}+ Bxe^{rx}is composed of y_{1}= Ae^{rx}and y_{2}= Bxe^{rx}y = e

^{vx}( Ccos(wx) + iDsin(wx) ) is composed of y_{1}= e^{vx}Ccos(wx) and y_{2}= e^{vx}iDsin(wx)

y_{1} and y_{2} are called the fundamental solutions of the equation.

And y_{1} and y_{2} are known to be linearly independent since neither function is a constant multiple of the other.