Using Euler's method, find an approximate value of y when x=0.3, given that dy/dx=2xy and y=0.5 when x=0. Take h=0.1
Answers
Summary of Euler's Method
In order to use Euler's Method to generate a numerical solution to an initial value problem of the form:
y′ = f(x, y)
y(xo) = yo
we decide upon what interval, starting at the initial condition, we desire to find the solution. We chop this interval into small subdivisions of length h. Then, using the initial condition as our starting point, we generate the rest of the solution by using the iterative formulas:
xn+1 = xn + h
yn+1 = yn + h f(xn, yn)
to find the coordinates of the points in our numerical solution. We terminate this process when we have reached the right end of the desired interval.
A Preliminary Example
Just to get a feel for the method in action, let's work a preliminary example completely by hand. Say you were asked to solve the initial value problem:
y′ = x + 2y
y(0) = 0
numerically, finding a value for the solution at x = 1, and using steps of size h = 0.25.
Applying the Method
Clearly, the description of the problem implies that the interval we'll be finding a solution on is [0,1]. The differential equation given tells us the formula for f(x, y) required by the Euler Method, namely:
f(x, y) = x + 2y
and the initial condition tells us the values of the coordinates of our starting point:
xo = 0
yo = 0
We now use the Euler method formulas to generate values for x1 and y1.
The x-iteration formula, with n = 0 gives us:
x1 = xo + h
or:
x1 = 0 + 0.25
So:
x1 = 0.25
And the y-iteration formula, with n = 0 gives us:
y1 = yo + h f(xo, yo)
or:
y1 = yo + h (xo + 2yo)
or:
y1 = 0 + 0.25 (0 + 2*0)
So:
y1 = 0
Summarizing, the second point in our numerical solution is:
x1 = 0.25
y1 = 0
We now move on to get the next point in the solution, (x2, y2).
The x-iteration formula, with n=1 gives us:
x2 = x1 + h
or:
x2 = 0.25 + 0.25
So:
x2 = 0.5
And the y-iteration formula, with n = 1 gives us:
y2 = y1 + h f(x1, y1)
or:
y2 = y1 + h (x1 + 2y1)
or:
y2 = 0 + 0.25 (0.25 + 2*0)
So:
y2 = 0.0625
Summarizing, the third point in our numerical solution is:
x2 = 0.5
y2 = 0.0625
We now move on to get the fourth point in the solution, (x3, y3).
The x-iteration formula, with n = 2 gives us:
x3 = x2 + h
or:
x3 = 0.5 + 0.25
So:
x3 = 0.75
And the y-iteration formula, with n = 2 gives us:
y3 = y2 + h f(x2, y2)
or:
y3 = y2 + h (x2 + 2y2)
or:
y3 = 0.0625 + 0.25 (0.5 + 2*0.0625)
So:
y3 = 0.21875
Summarizing, the fourth point in our numerical solution is:
x3 = 0.75
y3 = 0.21875
We now move on to get the fifth point in the solution, (x4, y4).
The x-iteration formula, with n = 3 gives us:
x4 = x3 + h
or:
x4 = 0.75 + 0.25
So:
x4 = 1
And the y-iteration formula, with n = 3 gives us:
y4 = y3 + h f(x3, y3)
or:
y4 = y3 + h (x3 + 2y3)
or:
y4 = 0.21875 + 0.25 (0.75 + 2*0.21875)
So:
y4 = 0.515625
Summarizing, the fourth point in our numerical solution is:
x4 = 1
y4 = 0.515625
Answer:
Given the equation 3 1 2 = x +
dx
dy with 2 y(1) =
estimate ) y(2 by Euler's method using.
(a) h = 0.5 and
(b) 25