# Variation of Parameters

• February 10th 2010, 08:07 PM
Aryth
Variation of Parameters
Use variation of parameters to obtain a particular solution of the non-homogeneous problem

$\frac{d^2 y}{dx^2} + p(x)\frac{dy}{dx} + q(x)y(x) = \frac{1}{x}$

With y(1) = 1, y'(1) = -1, x > 0

With the fundamental (meaning it solves the homogeneous equation) set:

$\{y_1(x), y_2(x)\} = \{x, 1 - x\}$

I cannot get this problem started as I do not remember how to use variation of parameters, if someone could give me a head start or an example it would be much appreciated.
• February 10th 2010, 10:37 PM
Calculus26
See attachment
• February 11th 2010, 03:42 AM
HallsofIvy
Quote:

Originally Posted by Aryth
Use variation of parameters to obtain a particular solution of the non-homogeneous problem

$\frac{d^2 y}{dx^2} + p(x)\frac{dy}{dx} + q(x)y(x) = \frac{1}{x}$

With y(1) = 1, y'(1) = -1, x > 0

With the fundamental (meaning it solves the homogeneous equation) set:

$\{y_1(x), y_2(x)\} = \{x, 1 - x\}$

I cannot get this problem started as I do not remember how to use variation of parameters, if someone could give me a head start or an example it would be much appreciated.

For this particular problem you begin by looking for a solution of the form y(x)= u(x)x+ v(x)(1- x).

The reason this is called "variation of parameters" is that you are allowing the constants in the general solution to the homogeneous equation to vary.

Now by the product rule, of course, y'= u(x)+ u'(x)x- v(x)+ v'(x)(1- x).

Important point: there are an infinite number of function u and v that would work. For example, given any actual solution, y, we could let u= 0 and v= y/(1-x) or let v= 0 and u= y/x. Because there are an infinite number of possible solutions, we can "narrow" the search a bit, and simplify our work, by requiring that u'(x)x+ v'(x)(1- x)= 0.

That way, the equation for y' becomes y'= u(x)- v(x) and involves no derivatives of u or v.

Differentiating again, y"= u'(x)- v'(x). Now put y, y', and y" into the equation: $u'(x)- v'(x)+ p(x)(u'(x)x+ v'(x)(1- x))+ q(x)(u(x)x+ v(x)(1- x)= \frac{1}{x}$.

Now you have two equations, the one just above and u'(x)x+ v'(x)(1- x)= 0, that we can solve, algebraically, for u'(x) and v'(x). Integrate those to find u(x) and v(x) and put into y(x)= u(x)x+ v(x)(1- x).

Of course, in order that the differential equation have y(x)= x and y(x)= 1- x as solutions to the homogeneous equation, you must have specific values of p(x) and q(x) that you have not given here.
• February 11th 2010, 01:25 PM
Aryth
Thank you very much. It was a great help indeed.
• February 11th 2010, 05:16 PM
Aryth
Sorry for the double post but I fear that no one would see it if I simply edited it...

I noticed that next to p(x) you put what you had already set equal to zero, should the DE not be:

$u'(x) - v'(x) + p(x)(u(x) - v(x)) + q(x)(u(x)x + v(x)(1 - x)) = \frac{1}{x}$