1. Solve the following problem using the forward Euler’s method with stepsizes of h = 0.2 and 0.1 using Matlab programs. Compute the error and relative error using the true solution $Y\left(x\right)$. For selected values of $x$, observe the ratio by which the error decreases when $h$ is halved. Plot the numerical solution and true solution on the same figure for each step size.

   $Y'(x)=xe^{-x}-Y(x),\;0\leq x\leq10,\;Y(0)=1;$

   The true solution is

   $Y(x)=(1+\frac{x^2}2)e^{-x}$

   Solution:

   clear
   clc
   h = 0.2;
   x = 0:h:10;
   f = (1 + (x.^2) / 2) .* exp(-1 .* x);
   df = x .* exp(-1 .* x) - (1 + (x.^2) / 2) .* exp(-1 .* x);
   df_fd(1) = NaN;
   
   for i = 2:length(x)
       df_fd(i) = (f(i) - f(i - 1)) / h;
   end
   
   plot(x, df, 'r', x, df_fd, 'bo');
   hold on
   
   h = 0.1;
   
   x = 0:h:10;
   f = (1 + (x.^2) / 2) .* exp(-1 .* x);
   df = x .* exp(-1 .* x) - (1 + (x.^2) / 2) .* exp(-1 .* x);
   df_fd(1) = NaN;
   
   for i = 2:length(x)
       df_fd(i) = (f(i) - f(i - 1)) / h;
   end
   
   plot(x, df_fd, 'k*');

2. Use the backward Euler’s method to solve the following problem: Consider the linear equation

   $Y'(x)=\lambda Y(x)+(1-\lambda)\cos(x)-(1+x)\sin(x),\;Y(0)=1;$

   The true solution is $Y(x)=\sin(x)+cos(x)$. Solve this problem with several values of $\lambda$ and $h$, for $0\;\leq\;x\;\leq\;10$. Plot the numerical solution and true solution on the same figure for each stepsize and comment on the results.
   (a) $\lambda\;=\;-1;\;h\;=\;0.25\;and\;0.125.$

   Solution:

   clear
   clc
   
   lamada = -1;
   
   h = 0.25;
   x = 0:h:10;
   
   df = lamada .* (sin(x) + cos(x)) + (1 - lamada) .* cos(x) - (1 + lamada) .* sin(x);
   f = sin(x) + cos(x);
   df_fd(length(x)) = NaN;
   
   
   for i = 1:length(x) - 1
       df_fd(i) = (f(i + 1) - f(i)) / h;
   end
   
   plot(x, df, 'r', x, df_fd, 'bo');
   hold on
   
   h = 0.125;
   x = 0:h:10;
   df = lamada .* (sin(x) + cos(x)) + (1 - lamada) .* cos(x) - (1 + lamada) .* sin(x);
   f = sin(x) + cos(x);
   df_fd(length(x)) = NaN;
   
   for i = 1:length(x) - 1
       df_fd(i) = (f(i + 1) - f(i)) / h;
   end
   
   plot(x, df_fd, 'k*');

   (b) $\lambda\;=\;1;\;h\;=\;0.25\;and\;0.125.$

   Solution:

   clear
   clc
   
   lamada = 1;
   
   h = 0.25;
   x = 0:h:10;
   
   df = lamada .* (sin(x) + cos(x)) + (1 - lamada) .* cos(x) - (1 + lamada) .* sin(x);
   f = sin(x) + cos(x);
   df_fd(length(x)) = NaN;
   
   
   for i = 1:length(x) - 1
       df_fd(i) = (f(i + 1) - f(i)) / h;
   end
   
   plot(x, df, 'r', x, df_fd, 'bo');
   hold on
   
   h = 0.125;
   x = 0:h:10;
   df = lamada .* (sin(x) + cos(x)) + (1 - lamada) .* cos(x) - (1 + lamada) .* sin(x);
   f = sin(x) + cos(x);
   df_fd(length(x)) = NaN;
   
   for i = 1:length(x) - 1
       df_fd(i) = (f(i + 1) - f(i)) / h;
   end
   
   plot(x, df_fd, 'k*');