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    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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    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.

    $2.49

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