Given the differential equation:
dy/dx = y(1+x), y(0)=1,
Use Euler's method with step size .1 to approximate y(.3).
y' = y(1+x), y'(0) = 1(1+0)=1
->the solution has slope 1 at the point
(0,1); x0=0, y0=1, h=0.1, F(x,y)=y(1+x)
y1=y0+h*F(x0,y0)
y1=1 + 0.1(1(1+0))
y1=1.01
y2=1.01 + 0.1(1.01(1+0.1)
y2=1.02111
y3=1.02111 + 0.1(1.02111(1+0.2))
y3=1.1436432
Is this the answer or am I going about this the wrong way? If so, please steer me in the right direction. Thanks.
You are on the right track! Euler's method is a numerical method commonly used to approximate solutions to ordinary differential equations. Your calculations for the approximations are correct.
To recap, you have correctly set up the initial conditions: y(0) = 1 and y'(0) = 1.
The formula for Euler's method is:
y_(n+1) = y_n + h * F(x_n, y_n)
where h is the step size, x_n is the current x-value, y_n is the current y-value, and F(x,y) is the derivative of y with respect to x.
In your case, the differential equation is dy/dx = y(1+x), so F(x, y) = y(1+x).
Using a step size of h = 0.1, you correctly found:
- y_1 = 1 + 0.1 * (1 * (1+0)) = 1.01
- y_2 = 1.01 + 0.1 * (1.01 * (1+0.1)) = 1.02111
- y_3 = 1.02111 + 0.1 * (1.02111 * (1+0.2)) = 1.1436432
So, according to your calculations, the approximation of y(0.3) using Euler's method with a step size of 0.1 is approximately 1.1436432.
You are following the correct procedure. Keep in mind that as the step size gets smaller, the approximation becomes more accurate. If you desire a more accurate approximation, you can try reducing the step size further.