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.

y > 1/3x

y < = -x + 4

hope this : )