- #1
- 301
- 0
Homework Statement
Use the Runge-Kutta method to find approximate values of the solution of the initial-value problem y'+(x^2)y=sin xy, y(1)=pi; h=0.2 at the points xi=x0+ih, where x0 is the point where the initial condition is imposed and I=1, 2.
Homework Equations
yn+1=yn+hf(xn+1/2h, yn+(1/2)hf(xn, yn))
f(x, y)=sin xy-(x^2)y
h=0.2, x0=1, y0=pi.
The Attempt at a Solution
y1=pi+0.2f(1+0.1, pi+0.1f(1, pi))
=pi+0.2f(1.1, 2.83291)
=2.4669
y2=2.4669+0.2f(1.3, 2.4669+0.1f(1.2, 2.4669))
=2.4669+0.2f(1.3, 2.11683)
=1.76101
But the answer is:
y1=2.475605264, y2=1.825992433.
I got y1=2.4669 and y2=1.76101.
Which is the correct answer? Mine or the book's answer? If I'm wrong, please correct me.