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I really appreciate it if anyone helps me with this:

How can I solve this ODE and plot the answer for $x$ on $[0.6,5]$:

$$ \begin{align*} -2xy'[x] = y''[x]+ 47.21 (-.0025 x^6 & + 0.0614 x^5- 0.6087 x^4+ 3.048 x^3-8.0588 x^2 \\ & + 10.586 x - 3.9582)^2\operatorname{Erfc}[x] \end{align*} $$

With the following boundary conditions: $y[0.6]=0$ and $y[\infty]=0$

I used NDsolve, but its answer was:

Cannot find starting value for the variable y'[x]


Original source of the equation:

-2xy'[x] = y''[x] + 47.21 (-.0025 x^6 + 0.0614 x^5 - 0.6087 x^4 + 3.048 x^3
                          - 8.0588 x^2 + 10.586 x - 3.9582)^2 Erfc[x]
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1 Answer

If you are willing to settle for $y(5) = 0$ instead of $y(\infty) = 0$, the commands to solve it are

sol = First@NDSolve[
  {-2 x y'[x] == 47.21` (-3.9582` + 10.586` x - 8.0588` x^2 + 3.048` x^3 - 
         0.6087` x^4 + 0.0614` x^5 - 0.0025` x^6)^2 Erfc[x] + y''[x], 
   y[0.6] == 0, y[5] == 0}, 
   y, {x, 0.6, 5}]


Plot[y[x] /. sol // Evaluate, {x, 0.6, 5}]

Mathematica graphics

Increasing the upper bound from 5 to a larger number won't change much, so I believe using $y(5) = 0$ might be a good enough approximation.

There are completely analogous examples in the NDSolve documentation. Please check them.

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@ Szabolcs: thank you so much –  sirous Apr 9 '12 at 12:14
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