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

**landers** · March 18th 2013, 09:27 AM

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!!!!!!!

**HallsofIvy** · March 18th 2013, 10:27 AM

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: $y= \sum_{n=0}^\infty a_nx^n$ so the $y'= \sum_{n=0}^\infty na_n x^{n-1}$ and $y''= \sum_{n=0}^{n-1}$ .
Put those into the differential equation to get a "recursive equation" for the coefficents $a_n$ . From y(0), $a_0$ = 0. From y'(0)= 1, [tex]a_1= 1[tex]. Then determine $a_2$ , $a_3$ , and $a_4$ .

2) Recall that the first 6 terms of a McLaurin series for y is $y(0)+ y'(0)x+ (y''(0)/2) x^2+ (y'''(0)/3!) x^3+ (y''''(0)4!)x^4+ (y'''''(0)/5!)x^5 ...$ . Then $y(0)= 1$ , $y'(0)= 1$ . For higher derivatives, evaluated at x= 0, solve the equation for y'': y''= (-xy+ y')/(1- x) so y''(0)= (y')/1= 1,

$y'''= (y'')'= ((-xy+ y')/(1- x))'= [(1-x)(-y- xy'+ y'')- (xy+ y')(-1))]/(1-x)^2$

so that $y'''(0)= [1(-0- 0+ 1)- (0+ 1)(-1)]/1= 0$ , etc.

**Zoli** · March 20th 2013, 11:58 PM

This second order linear ODE is homogeneous with respect to $y$ , $y'$ and $y''$ that is if $F(x,y,y',y'')$ is a solution then $F(x,ky,ky',ky'')$ is also a solution ( $k \neq 0$ ). In this case we may use the following transformation: $y(x) = e^{\int u(x) \mathrm{d}x}$ . From the original differential equation we may conclude to $(1-x)u'+(1-x)u^2-u+x=0$ 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 $y(x) = e^{\int u(x) \mathrm{d}x}$ we can determine $y(x)$ . 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.

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