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When we look at the solution part, there is a statement The PI of the given PDE is obtained as follows

After the statement, I don't really understand all of the calculation. Especially, After the fourth line, I am stuck. please explain what did the writer do?

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  • I think this is a Taylor expansion of $(1+x)^{-1}$, which for the first two terms is $1-x$. In your case, $$x = \frac{3D'}{D} + \frac{2D'^2}{D^2}.$$ – Mark Fantini Dec 07 '13 at 19:49
  • Please can you write a bit more explicit Dear @Fantini – 1190 Dec 07 '13 at 19:51
  • Things in the form $(1+x)^{-1}$, assuming $|x| < 1$ for real numbers, have the expansion as a Taylor series given by $$\frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n.$$ He seemingly applied the same idea to the expression $$\left[ 1 + \left( 3 \frac{D'}{D} + 2 \frac{D'^2}{D^2} \right) \right]^{-1}.$$ I cannot attest it is correct. – Mark Fantini Dec 07 '13 at 20:03

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They consider the solution of the pde

$$ Lu(x,y)=f(x,y). $$

So, $CF$ stands for solution of $Lu=0$ and $PI$ the particular solution for $Lu=f$, then adding the Solutions to get the general solution, since $L$ is a linear operator. See here for a related technique.

Added: I think what they mean by $D$ is $\frac{d}{dx}$, $D'$ is $\frac{d}{dy}$, $\frac{1}{D^2}$ or ($D^{-2}$) is the second integral w.r.to $x$. Now,

$$ \frac{1}{D^2}\left(1-3\frac{D'}{D}-...\right)(x+y)= \frac{1}{D^2}\left( (x+y)-\frac{D'}{D}(x+y)-... \right)$$

$$=\frac{1}{D^2}\left( (x+y)-\frac{1}{D}(1)-0-0-... \right). $$

$$ \implies\frac{1}{D^2}\left(x+y-\frac{3}{D}(1)\right) = \frac{1}{D^2}(x+y-3x)=\frac{1}{D^2}(y-2x)=y\frac{x^2}{2}-\frac{x^3}{3} $$

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