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I was looking speedy and simple algorithm to solve second order differential equation in numarical way. I thought an idea to solve it via geometrical equalities.

See my approach to the problem $$A''(x)=f(x) A(x)$$ I will try to solve $A'(1)$ via numerical method. Initial condition $A'(0)=y(0)=k$ and $A(0)=0$

$$A(x)=\int _0^x y(x) dx$$

$$A''(x)=y'(x) $$

$$y'(x)=f(x) A(x)$$

$A(x)$ is total area of $y(x)$ from $0$ to $x$

$y'(x)$ is $tan(\alpha)$ on $x$

If we divide n point between $0$ and $1$

for x=1/n

$y'(1/n)$~$\frac{y(\frac{1}{n})-y(0))}{(1/n)}$

$A(1/n)$~$y(0)\frac{1}{n}$

if $y'(x)=f(x)A(x)$ then $\frac{y(\frac{1}{n})-y(0))}{(1/n)}=y(0)f(\frac{1}{n})\frac{1}{n}$

$y(1/n)=y(0)[\frac{1}{n^2}f(\frac{1}{n})+1]$

for x=2/n

$y'(2/n)$~$\frac{y(\frac{2}{n})-y(\frac{1}{n}))}{(1/n)}$

$A(2/n)$~$y(0)\frac{1}{n}+y(\frac{1}{n})\frac{1}{n}$

if $y'(x)=f(x)A(x)$ then $\frac{y(\frac{2}{n})-y(\frac{1}{n}))}{(1/n)}=f(\frac{2}{n})y(0)\frac{1}{n}+f(\frac{2}{n})y(\frac{1}{n})\frac{1}{n}$

$y(2/n)=y(\frac{1}{n})[\frac{1}{n^2}f(\frac{2}{n})+1]+f(\frac{2}{n})y(0)\frac{1}{n^2}$

for x=3/n

$y(3/n)=y(\frac{2}{n})[\frac{1}{n^2}f(\frac{3}{n})+1]+f(\frac{3}{n})y(\frac{1}{n})\frac{1}{n^2}+f(\frac{3}{n})y(0)\frac{1}{n^2}$ .

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for x=n/n=1

$A'(1)=y(1)=y(\frac{n-1}{n})[\frac{1}{n^2}f(1)+1]+f(1)y(\frac{n-2}{n})\frac{1}{n^2}+f(1)y(\frac{n-3}{n})\frac{1}{n^2}+....+f(1)y(0)\frac{1}{n^2}$

if we increase n, I think we will be near to value.

Is my approach correct or not? I don't have matlab or other tool to simulate how successfull the idea to solve the equation. Could you please also share your experiences and theories about numerical methods to solve such second order differantial equations?

Thanks in advice and times for me and your shares.

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Please exercise restraint while using $$...$$. (I have now edited it!) –  user21436 Mar 3 '12 at 22:10

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