Numerical Differential Equation.

[MathematicalPhysicist]

give a numerical method of order bigger or equals 2 for the numerical solution of:
y'''(x)+2xy'(x)+e^(y(x))=cos(x)
in the interval: [0,1] with boundary conditions:
y(0)=1,y(1)=2,y'(0)=0

Attempt at solving:
well I thought of difference equation:
[/tex]y_{n+3}+2x_ny_{n+1}+e^{y_n}=cos(x_n)[/tex]
with y_0=1 and y_N=2 where we divide the interval [0,1] with N points of difference:
h=1/N.
Now we want the order would be greater than or equals 2, i.e the error of y_n+3-y(x_n+3) would be of O(h^3) at least.

where [/tex]y(x_{n+3})=y((n+3)*h)[/tex]
So now I need to expand y through taylor's expansion and then find the error y_n+3-y(x_n+3), I think that I also need to expand e^(y_n) and cos(x_n) through taylor's expansion, am I missing something or as I stated it, it's that much easy?

thanks in advance.

 

[mighty2000]

I would recommend using the Runge-Kutta method of order 4. This method is commonly used for solving ordinary differential equations and has an error of O(h^4), which satisfies the requirement of having an order greater than or equal to 2.

The steps for using the Runge-Kutta method are as follows:

1. Divide the interval [0,1] into N subintervals with a step size of h=1/N.

2. Set up the initial conditions as y_0=1 and y_N=2.

3. Use the following formula to calculate the values of y at each point in the interval:
y_{n+1} = y_n + (1/6)(k_1 + 2k_2 + 2k_3 + k_4)
where
k_1 = hf(x_n, y_n)
k_2 = hf(x_n + (1/2)h, y_n + (1/2)k_1)
k_3 = hf(x_n + (1/2)h, y_n + (1/2)k_2)
k_4 = hf(x_n + h, y_n + k_3)

4. Repeat this process for each subinterval until you reach the end of the interval, where y_N=2.

5. The resulting values of y_n will give you the numerical solution of the differential equation.

The Runge-Kutta method of order 4 is a reliable and accurate method for solving ordinary differential equations, and it is widely used in scientific and engineering applications. I hope this helps in solving the given problem.