00:01
In this question, we have been given an ordinary differential equation of the form d2y divided by dx square is equals to 5y with initial condition, y at 0 equal to 1 and d of y 1 divided by d x is given to be equal to 0.
00:21
We need to solve this using the finite using this approach, which is the finite difference approach.
00:31
With its domain discreetized into five segments okay so let us see how can we do this so first of all what we have to do we have to draw the digitized discreetized domain with the corresponding notes and segment length delta x and using that difference we need to write an approximate for the ordinary differential equation at the interior notes xi okay so let us start first draw the diagram so what i will do here these are my domain which is being discretized in five segments so x0 till x5 so this will be our x1 x2 x3 and x4 correct so these are the five segment now we need discretizing okay so what we are going to get we can write this d2 by divided by dx2 equal to this formula y i plus 1 minus 2 y i plus y minus 2 y plus y minus 1 divided by okay h square this is the formula discretizing formula so what i can say from here y i plus 1 minus 2 y i plus y i minus 1 divided by delta x square it is given to be equals to 5 y d2 divided by dx2 so here it becomes 5 y i with the condition y 0 equal to 1 correct if i assume my delta x to be equal to 0 .2 so what i am going to get y i plus 1 minus y i divided by delta x equal to so now we know that here what we know is x not equals to 0 and x1 is x not plus delta of x so it becomes 0 .2.
02:40
Similarly x2 is given by x0 plus 2 times of delta x which is 0 .4 x similarly if you calculate it will become 0 .6.
02:51
X4 will be 0 .8 and x5 it is going to be equal to 1.
02:58
So this is what we get so i can write down hence y of xi plus 1 minus 2y of xi plus y of x i minus 1 will be equal to delta x square times 5y of x i correct so from this what i'm going to get y of x i plus 1 equals to 2 y x i minus y of x i minus 1 plus delta x squared multiplied with 5 y x i correct i will call this as equation number 1 and this is what we need here so let me just encircle this equation now we move to the next part and in the next part what we have to do so this is part b and this was part a okay so now let us see what we have to do in this case so write down the boundary condition at the boundary nodes x not an x5 and user -centered difference and an imaginary node x6 then we have to express x6 in terms of the function y.
04:24
Okay and let us see how can we do this.
04:27
So since we know that y x not is equals to one which was given as a boundary condition and this is also given dy one divided by d x equals to zero.
04:40
So this will and we know x5 value is given to be one.
04:43
So from this i can write y of x5 minus y of x4 divided by delta of x which is equal to zero so from this what i am going to get is y of x5 equals to y of x4 correct then if i use this equation which is equation one so from equation number one i can write down the value of y of x6 correct so y of x6 will be what so this one we got as the boundary condition then we have to write x6 in terms of the function y...