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.

Thanks in advance.