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I'm trying to understand this problem:

The differential equation $$\left\{ \begin{align} y'' &= 2x(y − 2)\\ y(0) &= 10\\ y(8) &= 3 \end{align} \right.$$ is used with the step $h = 1$. Then an equation system is generated. How many equations does this system have if $y(0)$ and $y(8)$ have been eliminated with help from the boundary conditions?

The correct answer should be 7, but why? Could you explain it? Usually an ODE of order $2$ generates a system of $2$ first-order equation. What is the corresponding rule for boundary value problems?

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" used with the step $h = 1$." - well, what discretization of the BVP are you using? Some people use forward differences, some use central differences... your book should have mentioned what discretization to use for your problem. – J. M. Aug 14 '12 at 10:11
It's an old exam question that I'm trying to understand. They're asking about stuff we didn't cover, just like engineering school is. I can think that it is one equation per step and there are 9 equations if there is one equation per step and if 2 are "eliminated" then only 7 equations are left but why should there be 9 and not just 2 to begin with? I don't see the formula in general for BVP. The methods we usually use to solve differential equations are Euler's method and the Runga Kutta methods but the choice of method seems to be irrelevant here which is also confusing. Please help clarify. – Programmer 400 Aug 14 '12 at 10:23
First step: are you familiar with the usual limit-based definition of the second derivative? – J. M. Aug 14 '12 at 10:26
Ok the trick is to rewrite y'' with its limit definition but is the answer here really correct since I think it should be 1 instead of 2 at the bottom, since 1²=1 and the formula's here – Programmer 400 Aug 14 '12 at 16:51
up vote 1 down vote accepted

There is one equation at each point. As there are seven points between $0$ and $8$, there are seven equations. The equation at $1$ is $y(2)-2y(1)+y(0)=2\cdot 1(y(1)-2)$. For the other equations, increment the indices of $y$ and make the $1$ the current value of $x$.

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but why 2 in the denominator? shoudn't it be 1? since 1²=1 looking at the formula here I think it's not that obvious. – Programmer 400 Aug 14 '12 at 16:53
@NickRosencrantz: You are right, there should not be a 2 in the denominator. Fixed. – Ross Millikan Aug 14 '12 at 17:33

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