Solving Diff Eq with initial conditions and first 6 coefficients...

I need some help here because I am completely lost!!!

I need to solve the ODE (1-x)y''-y'+xy=0 using the initial conditions of y(0)=1, y'(0)=1.

So far I have (1-x)y''-y'=-xy and (1-x)y''=-xy+y' but this is where I got lost. I know it's something really simple but I still can't seem to find the error and I don't know where to go next!!!

I have to find the general pattern and show my final answer is equal to y=e^x...

Thanks for the help!!!!!!!

Re: Solving Diff Eq with initial conditions and first 6 coefficients...

Your title says "first six coefficients". Coefficients of what? I suspect you were expected to find a power series sollution.

There are **two** ways you can do that:

1) Write the solution as a power series: so the and .

Put those into the differential equation to get a "recursive equation" for the coefficents . From y(0), = 0. From y'(0)= 1, [tex]a_1= 1[tex]. Then determine , , and .

2) Recall that the first 6 terms of a McLaurin series for y is . Then , . For higher derivatives, evaluated at x= 0, solve the equation for y'': y''= (-xy+ y')/(1- x) so y''(0)= (y')/1= 1,

so that , etc.

Re: Solving Diff Eq with initial conditions and first 6 coefficients...

This second order linear ODE is homogeneous with respect to , and that is if is a solution then is also a solution ( ). In this case we may use the following transformation: . From the original differential equation we may conclude to which is the general case of the Riccati-equation. We must find a particular solution to this first order d.e. If we can solve the Riccati differential equation, from we can determine . However - according to Wolframalpha - the solution of the Riccati-equation involves exponential integral, which cannot be expressed with non-elementary functions. Therefore we cannot gain a closed formula for the differential equation, try a series method as HallsofIvy suggested.