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

  1. #1
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    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!!!!!!!
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  2. #2
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    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: $\displaystyle y= \sum_{n=0}^\infty a_nx^n$ so the $\displaystyle y'= \sum_{n=0}^\infty na_n x^{n-1}$ and $\displaystyle y''= \sum_{n=0}^{n-1}$.
    Put those into the differential equation to get a "recursive equation" for the coefficents $\displaystyle a_n$. From y(0), $\displaystyle a_0$= 0. From y'(0)= 1, [tex]a_1= 1[tex]. Then determine $\displaystyle a_2$, $\displaystyle a_3$, and $\displaystyle a_4$.

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

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

    so that $\displaystyle y'''(0)= [1(-0- 0+ 1)- (0+ 1)(-1)]/1= 0$, etc.
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  3. #3
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    Re: Solving Diff Eq with initial conditions and first 6 coefficients...

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