• y1 is a solution with the lsode function.
• y2 is an explicit solution. You may either remember, how to solve such equations, or ask WolframAlpha for the solution.

After all, plot the two solutions and make sure that they are the same. So, you should implement a function that does not have arguments, does not return anything, and just plots a function.

task2 Solve the ODE \(y’‘(x) + y(x) * (cos(x) + 1) = 0, y(0)=-1, y’(0)=1\) on the segment \(0\leq x \leq 10\), plot the solution.

• Create a function task2a that solves the same equation, but allows to change initial values. It should have the three parameters: x coordinates, value for y(0), and a value for y'(0). Create the function task2b to test the function task2a, it should have the following code inside:

       x = linspace(0, 10, 1000);
       plot(x, task2a(x, -1, 1));
       hold on
       plot(x, task2a(x, 0, 1));
       plot(x, task2a(x, 1, 1));
       plot(x, task2a(x, -1, 0));
       plot(x, task2a(x, 0, 0));
       plot(x, task2a(x, 1, 0));
       plot(x, task2a(x, -1, -1));
       plot(x, task2a(x, 0, -1));
       plot(x, task2a(x, 1, -1));
  

The Predator-Prey Model. The link leads to an article about a predator-prey model, that models population changes of, for example, foxes and rabbits when they live together and eat each other. The equation is given in the first paragraph.

Implement a function that gets \(\alpha\), \(\beta\), \(\gamma\) ,\(\delta\) as parameters, evaluates a non-zero equlibrium point \((\overline{x}, \overline{y})\) (find it in the article), and then plots six plots with initial values \(x(0) = \overline{x}, y(0) = \overline{y} + ks\), where \(k\) is an integer from 0 to 5, and \(s\) is a value given as the fifth argument of a function. So, you should have a plot with 6 lines on it. Do not return any result. Make sure that you understand what do these plots mean.