Solve With Matlab

Bacteria population: boundary value problem #$2$

Consider the following differential equation used to help determine population for a bacteria over time:

$($d$^(2)$P$)/($dt$^(2))=(1-($dP$)/($dt$))$sqrtP

where t is time in days and P is population in hundred thousands. The population on day $5$ is $4\mathrm{.}2$ hundred thousand. On day $7\mathrm{.}8$ the population is $19\mathrm{.}6$ hundred thousand. Solve the boundary value

problem

A$.$ Solve the problem with the shooting method. Use fprintf statement to display the calculated guess for the initial condition $($e$.$g$.$ Calculated P$^(\text{'})(5)=$dots and the calculated value for P at the end point

$($e$.$g$.$ Calculated P$(7\mathrm{.}8)=$dots $)$

B$.$ Solve the problem using the finite difference approach with Delta x$=0\mathrm{.}4$ and a magnitude percent relative approximate error tolerance of $0\mathrm{.}04\%.$ For your initial guess values make all your y points

linearly distributed between the boundary points. Display your x and y values formatted like the table below:

Sample output

t $($in days$)$ P $($in $100,000$s$)$

x$.$x xx$.$xxxx

C$.$ Plot both solutions on the same axes and use a legend to distinguish data. Use a solid blue line for the shooting method and red stars for the finite difference solution. Turn the grid on$.$ Don't forget

units.