Hi Guys i am working on the following question
$$y''+2y'=e^{-x}$$ $$y(0)=1,y(1)=4$$
The Question states Obtain a numerical solution to the given boundary value problem when x = 0.25, x =0.5 and x =0.75 by using- one iteration of the shooting method with initial guess $$m_0 = 4 ,m_1 = 5 $$
along with four iterations of Picard's method, now previous to this question i was asked to work out the the same solution using the finite difference method.
This is what I have attempted thus far
$$z = y'$$ $$z' = y''$$ $$z' = e^{-x}+2z$$
Therefore my two main equations are
$$z = y'$$
$$z' = e^{-x}+2z$$
Therefore by picards method for successive approximations i used the folllowing
$$y_{n+1} = y_0 + \int_0^x Z_n dx$$ and in a similar manner $$Z_{n+1} = Z_0 + \int_0^x (e^{-x}+ 2z) dx$$
Therefore evaluating y0
$$y_{1} = y_0 + \int_0^x Z_0 dx$$
I know that $Z_0 = y'_0$ but my confusion here is what to substitute into the equation so i substituted it as m for an initial gradient
$$y_{1} = 1 + \int_0^x m dx$$ $$y_1 = 1+mx$$
Now considering the $Z_{n+1}$ this is how i deal with this part of the problem $$Z_{n+1} = Z_0 + \int_0^x (e^{-x}+ 2z_0) dx$$ $$Z_1 = m + \int_0^x (e^{-x}+ 2m) dx$$ $$Z_1 = 2mx+m+1-e^{-x}$$
Now my thinking is I am required to evaluate 4 iteration so i need to calculate a y2, y3 and y4
$$y_{2} = y_0 + \int_0^x Z_1 dx$$
Now for my $Z_1$ expression that was derived previously therefore i can make a substitution
$$y_2 = y_0 + \int_0^x 2mx + m + 1 -e^{-x}$$
$$y_2 = mx^2+mx+x+e^{-x}+2$$
Now calculating $Z_2$ and im confused as to expression I am integrating to obtain $Z_2$ and this is my expression for obtaining $Z_2$
$$Z_2 = m + \int^x_0 Z_1 dx$$ $$Z_2 = m + \int^x_0 2mx+m+1-e^{-x} dx$$
this will give the following
$$Z_2 = m + mx^2+mx+x+e^{-x}-1$$
Therefore i can obtain an expression for $y_3$ such that
$$y_3 = y_0 + \int^x_0 Z_2 dx$$
Therefore i am currently trying to understand if I am formulating the question properly by trying to obtain the 4 iterations of picards method and then evaluating one iteration of the shooting method. Can anyone clarify or guide to working out this problem. The part of the question is aii)
You made a sign error. And the formula for $z_2$ is mixed up with the formula for $y_2$. As the formula for $y_n$ shows, you do not need any previous iterate of $y$, for $y_4$ you only need $z_3$. Then \begin{align} z_1(x)&=m+\int_0^x (e^{-s}-2z_0(s))\,ds=m-2mx+(1-e^{-x})\\ z_2(x)&=m+\int_0^x (e^{-s}-2z_1(s))\,ds=m-2mx+2mx^2+(3-2x-3e^{-x})\\ z_3(x)&=m+\int_0^x (e^{-s}-2z_2(s))\,ds=m-2mx+2mx^2+...+(7-7e^{-x}+...)\\ y_4(x)&=y(0)+\int_0^xz_3(s)\,ds=1-mx-mx^2+\frac23x^3+...-7(1-x-e^{-x})+... \end{align} However, this is hardly a numerical method, you could as well just integrate, apply the integrating factor and integrate again.
I take it that in a.i you apply divided differences to obtain the 3 linear equations $$ \frac{y_{n+1}-2y_n+y_{n-1}}{h^2}+\frac{y_{n+1}-y_{n-1}}h=e^{-x_n}, ~~n=1,2,3, $$ for the 3 unknowns $y_1,y_2,y_3$ and solved that system.