PLEASE USE MATLAB!!!!! PLEASE DO NOT USE SOMEONE ELSES ANSWER!!!

Here is some hints that we were given...

$\%\%\%$ Use finite differences to solve the BVP

$\%\%\%$ Be careful about the shape of the vectors, you may have to transpose to

$\%\%\%$ get the correct shape. It's a good idea to print the solutions out to

$\%\%\%$ make sure the shape is correct.

$\%\%\%$ Part $2$: Shooting Method via Bisection

$\%\%\%$ Use the shooting method to solve the BVP

$\%\%\%$ It's a good idea to test out a few in the command window first to make

$\%\%\%$ sure that your initial conditions gets you to different sides of the right

$\%\%\%$ boundary condition.

$\%\%\%$ Use the plot to help you figure out what your choices of initial

$\%\%\%$ conditions should be

Consider the boundary value problem

$x^{}+x=5cos(4t),$ with $x(0)=1$ and $x(6)=0\mathrm{.}5.$

The true solution to this boundary value problem is

$x(t)=C_{1}sin(t)+C_{2}cos(t)-\frac{1}{3}cos(4t)$

where

$C_1=\frac{\frac{1}{2}+\frac{1}{3}cos(24)-\frac{4}{3}cos(6)}{sin(6)}$ and $C_2=\frac{4}{3}.$

The solution looks like this:

$(1)$ Following the finite difference process we used in lecture, use a second order difference scheme to

approximate $x^{}$ and rewrite this initial value problem as a linear system of equations $Ax=b.$ Use

$t=0\mathrm{.}1.$ Note that there will be $61$ total $t-$values, but we are only trying to approximate $x$ at the

interior times. The matrix A should be $N_{\text{i}\text{n}\text{t}}{N}_{\text{i}\text{n}\text{t}},$ where $N_{\text{i}\text{n}\text{t}}$ is the number of interior times, and

the vector b should be $N_{\text{i}\text{n}\text{t}}1.$ Save a copy of $A$ in a variable named A$9.$ Save a copy of b in a

variable named A$10.$

Solve this linear system using Gaussian elimination $($the backslash operator in MATLAB or the

solve function in python$).$ Make a $611$ column vector containing the $x$ values at all times $($including

the two boundary values$)$ and save a copy of this vector in a variable named A$11.($Hint: if A$11$ is

passing but A$9$ and A$10$ aren't, to be consistent with the autograder don't distribute the $\frac{1}{}t.)$

Find the maximum $($in absolute value$)$ error between this approximation and the true solution. Save

this error in a variable named A$12.$

$(2)$ Now solve the problem using the shooting method. Remember, you need to convert this second order

ODE into a system of two first order ODEs. Apply a bisection scheme to the usual IVP solvers ode$45$

$($MATLAB$)$ or solve$\_$ivp $($Python$)$ using $t=0\mathrm{.}1.$ For the second initial conditions, you can use the

plot to help you test out values that will get you on different sides of the right boundary condition.

Break out of the loop for bisection when the difference between your approximation at the right

boundary point and the exact value at the right boundary point is less than $1$ e$-8$; i$.$e$.,|x_{\text{a}\text{p}\text{p}\text{r}\text{o}\text{x}}(6)-$

Save the numerical solution as a $611$ column vector A$13$ and the

maximum $($in absolute value$)$ error max$|x_{\text{a}\text{p}\text{p}\text{r}\text{o}\text{x}}-x|$ as A$14.$ Further save the maximum difference in

absolute value between the solution from part $1$ and this solution as A$15$ A$13-$ A$11)$