# 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

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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.