# Variation of Parameters Problem

• Jul 15th 2009, 10:32 PM
ecl
Variation of Parameters Problem
Hello,

This is the first time I have posted here and my first time using latex. Hope it comes out OK!

I have a problem to solve for my diff eq's class. It regards using variation of parameters to solve inhomogeneous, second-order DE's. I've been beating my head against the wall with this one and am very frustrated. Hopefully someone can give me some direction.

The problem says to find a particular solution to the DE

$\displaystyle \left( 1-x\right)y'' + xy' -y = g\left( x\right) \quad\quad0<x<1 \quad y_{1}\left( x\right) =e^{x} \quad y_{2}\left( x\right) =x$

where y1 and y2 are solutions to the corresponding homogeneous equation and g(x) is an arbitrary function of x.

Dividing through by $\displaystyle \left( 1-x\right)$ I get

$\displaystyle y'' + \dfrac{x}{ \left( 1-x\right)}y' -\dfrac{y}{ \left( 1-x\right)} = \dfrac{g\left( x\right)}{\left( 1-x\right)}$

My first thought was to use the formula for finding a particular solution

$\displaystyle Y_{p}\left( t\right) =-y_{1}\left( t\right) \int\dfrac{y_{2}\left( t\right)g\left( t\right)}{W\left( y_{1},y_{2}\right) \left( t\right) }\,dt+y_{2}\left( t\right) \int\dfrac{y_{1}\left( t\right)g\left( t\right)}{W\left( y_{1},y_{2}\right) \left( t\right) }\,dt$

Where the denominator in both integrands is the Wronskian of y1 and y2. I calculate the Wronskian for my problem to be

$\displaystyle e^{x}\left( 1-x\right)$

So now I try to set up the integrals to calculate a particular solution as

$\displaystyle -e^{x}\int\dfrac{xg\left( x\right)}{e^{x}\left( 1-x\right)^{2}}\,dx+x\int\dfrac{e^{x}g\left( x\right)}{e^{x}\left( 1-x\right)^{2}}\,dx$

Or

$\displaystyle -e^{x}\int\dfrac{xg\left( x\right)}{e^{x}\left( 1-x\right)^{2}}\,dx+x\int\dfrac{g\left( x\right)}{\left( 1-x\right)^{2}}\,dx$

It is at this point that I am totally confused. I just don't know how to proceed with an arbitrary function of x in the integrand. To make matters worse, I peeked at the answer in the back of the book. It shows

$\displaystyle Y_{p}\left( x\right) =\int\dfrac{xe^{t}te^{x}}{\left( 1-t\right) ^{2}}\,g\left( t\right) \,dt$

How the answer reduces to a single integral with both x and t in the integrand has me completely flustered. It would seem that the fact x is between 0 and 1 would be a clue, but I just can't figure out how to use it.

I am not looking for a complete, worked solution. But if someone could point me in the right direction, I would really appreciate it.

• Aug 1st 2009, 01:34 PM
Media_Man
Almost there...
The answer is a shortcut using one of the fundamental rules of calculus: $\displaystyle F(x)=\int_a^x f(t)dt$. In plain English, you can change the variable without changing the relationship between F and f.

So...

$\displaystyle -e^{x}\int\frac{xg(x)}{e^x(1-x)^2}dx+x\int\frac{g(x)}{(1-x)^2}dx$ $\displaystyle =-e^{x}\int\frac{tg(t)}{e^t(1-t)^2}dt+x\int\frac{g(t)}{(1-t)^2}dt$

This makes the terms involving x constants within the context of the integral...

$\displaystyle =\int\frac{-e^{x}tg(t)}{e^t(1-t)^2}dt+\int\frac{xg(t)}{(1-t)^2}dt$$\displaystyle =\int\frac{(xe^t-te^{x})g(t)}{e^t(1-t)^2}dt$

This accounts for the use of "t" but the answer is still off. I am not sure where the error lies.

P.S. Your LaTex contained no errors, but was quite unnecessarily long. Look at mine and see that it is actually quite simple, much like punching in expressions on a graphing calculator. Might save you some keystrokes and confusion later :)