Euler's method
1. y'=1/2 -x + 2y y(0) = 1
Find the exact solution "Sphee"
{0 with line diagonal of the 0}
2. y'=x square + y square
y(0)=1
a. Let h=.1 use Euler and improve to approximate to get
Sphee(.1) Sphee(.2) Sphee(.3)
b. h=.05 so compare with Shee(.1)
Sphee (.2) sphee is a circle
https://brainmass.com/math/integrals/eulers-improved-eulers-method-exact-solution-2600
Solution Preview
Please see attachment.
Problem:
I posted a simular question. I entered the wrong equation.
1. y'=1/2 -x + 2y y(0) = 1
Find the exact solution "Sphee"
{0 with line diagnal of the 0}
2. y'=x square + y square
y(0)=1
a. Let h=.1 use Euler and improve to aproximate to get
Sphee(.1) Sphee(.2) Sphee(.3)
b. h=.05 so compare with Shee(.1)
Sphee (.2) ps. rember sphee is a cirle
Solution:
1. write equation of the form y'=1/2-x+2y and
First we solve y'= 2y
We know dy/y=2dx, integrate both sides to get
ln|y|=2x+C.
Thus
y=+e^(2x+C)=+e^(C)e^(2x) or
y=e^(2x-C)=-e^(C)e^(2x)
Thus
y=ke^(2x) for some constant k
Now we solve
Please note that if y=g(x) is the solution to
y=2y, then the solution for
y'=1/2 -x + 2y is always of the form
y=g(x)f(x) for some function f(x). Thus for
y'=1/2 -x + 2y, we expect solutions to be of the form
y=f(x)(e^(2x)), for some function f(x)
From y'= 1/2 -x + 2y we have
f'(x) e^(2x)+f(x)(2 ...
Solution Summary
This shows how to find an exact solution and how to use Euler's method with Sphee equations.