(d^(2)T)/(dr^(2))+5*(dT)/(dr)+100*T-r+1, T(0)=1, T(1.0)=5
Determine T(0.75) by employing one of the following methods for a stepsize of \Delta r=0.25
(a) Int order Runge-Kutta method:
RK1: ,y_(i+1)=y_(i)+k_(1)\Delta x,k_(1)=f(x_(i),y_(i))
Note: T(1.0)=23.85 in case (dT)/(dx)|_(x)=0=0,T(1.0)=2.414 in case (dT)/(dx)|_(x)=0=10
(b) The method of finite-differences
[A]{T}={b},(d^(2)y)/(dx^(2))~=(y_(i+1)-2y_(i)+y_(i-1))/(\Delta x^(2)),(dy)/(dx)~=(y_(i+1)-y_(i-1))/(2\Delta x)
I did not understand how to use the note section given in the question. I would be glad if you could provide information on that subject.When solving according to option A, we will first give 0 to the variable we call (dT)/(dr) band find the t(1) value there. Since five does not come out, we will give 1 to the variable we call(dT)/(dr) and again find the T(1) value. This result will not come out as 5 either. We will find what we will give to the variable we call (dT)/(dr) by making an interpolation from these and from there we will find T(0.75). Because at that value, T(1)=5 will come out.