Consider the ode, dy dx = y 2 1 + x (2) subject to condition y = 1 when x = 0, use your euler code from class (modified if necessary) to find the solution to the above equation. the equation has the solution: y = 1 1 − ln(1 − x) (3) compare the analytical and euler result. explain any differences

Computation.

Step-by-step explanation:

I'm not really sure if that's the analytical solution of the inital value problem,

because y(0)=11-ln(1-0)(3)=11. Howevwer, let us procede with the given values...

Let us assume that we are going to use euler with n=2 (two steps) and h=0.2(the size of each step)

The update rules of the Euler Methode are

Since the initial value problem tells us that Y=1 when X=0, we know that

and that . Then, we have

and

which gives us the points (0,1), (0.2, 5.2) and (0.4, 27.12).

Now, since we want to compare the analyticaland the Euler result, we first compute the value of y=11-ln(1-x)(3) for the values x=0, 0.2 and 0.4. We get that

and we compute for each i.

It holds

which tells us that we have a really bad approximation, as I already stated there must be a mistake in the analytical solution since the intial values don't coincide. Also note that the curve that we get using the euler methose is growing faster than the analitical solution.

468

step-by-step explanation: