Answer to Question #170943 in Quantitative Methods for Khan

Question

If an ODE is given as

(x²+x+1)dx+(y²-y+3)dy=0 with y(0)=1.

Find the numerical solution with h=0.1.

Answer 1

Solution

For the first-order differential equation with the initial value

y’(x) = F(x,y y(x)),  y(x₀) = y₀  

according to the Euler method

y_n+1 = y_n + h*F(x_n, y_n)

where h – step, x_n = h*n, y_n = y(x_n), n = 0,1,2…N

In this case F(x,y) = -(x²+x+1)/(y²-y+3) , h = 0.1 , x₀ = 0 , y₀ = 1.

Calculations gives such results:

n        x_n           y_n

0        0.0         1.000

1        0.1         0.967

2        0.2         0.929

3        0.3         0.887

4        0.4         0.839

5        0.5         0.785

6        0.6         0.723

7        0.7         0.653

8        0.8         0.574

9        0.9         0.485

10      1.0         0.387