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= 2. We wish to 7. Consider the differential equation y' + y = 2.. with...

= 2. We wish to 7. Consider the differential equation y + y = 2.. with the initial conditions y(0) approximate y(1). (a) Set

= 2. We wish to 7. Consider the differential equation y' + y = 2.. with the initial conditions y(0) approximate y(1). (a) Set up Euler's method by hand and use it to approximate y(1) with a step size of h = 25. (You may use only a calculator for this question) (b) Use Mathematica to do Euler's method with the step sizes of h= .1,.01, .001, and .0001.
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ANSSWER :

Consider the differential equation

y'+y=\frac{1}{2-x} with the inital conditions y(0) =2 we wish to approximate y(1).

Didfferntial equation

\frac{dy}{dx}+y=\frac{1}{2-x}

\frac{dy}{dx}=-y+\frac{1}{2-x}

y(0) = 2

y(1) = 2, L = 0.25,20 = 0

Apply Euler's Method

f(x,y)=-y+\frac{1}{(2-x)}

y_{1}=y_{0}+h(x_{0},y_{0})

\Rightarrow y_{1}=2+(0.25)\left ( -2+\frac{1}{2-0} \right )

\Rightarrow y_{1}=2+(0.25)\left ( \frac{-4+1}{2} \right )

\Rightarrow y_{1}=1.625

y_{2}=y_{1}+h(x_{1},y_{1})

\Rightarrow y_{2}=1.625+(0.25)\left ( -1.625+\frac{1}{2-0.25} \right )

\Rightarrow y_{2}=1.625+(0.25)\left ( -1.625+\frac{1}{1.75} \right )

\Rightarrow y_{2}=1.361607

y_{3}=y(0.75)+h(x_{2},y_{2})

\Rightarrow y_{3}(0.75)=1.361607+(0.25)\left ( -1.361602+\frac{1}{2-0.5} \right )

=1.1878719

\Rightarrow y_{4} (1)=y_{3}+h(x_{3},y_{3})

=1.1878719+(0.25)\left ( -1.1878719+\frac{1}{2-0.75} \right )

y_{1}=1.090903925

b)

h=0.1,0.01,0.001,0.0001

\Rightarrow y_1=y_0+h(x_0,y_0)

\Rightarrow y_1=2+(0.1)\left(-2+\frac{1}{2-0}\right)

\Rightarrow y_1=2+\frac{(0.1)(-4+1)}{2}=1.85

h=0.01

\Rightarrow y_1=y_0+h(x_0,y_0)

\Rightarrow y_1=2+(0.01)\left(-2+\frac{1}{2-0}\right)

\Rightarrow y_1=2+(0.01)(-1.5)

\Rightarrow y_1=2-0.015

\Rightarrow y_1=1.985

h=0.001

\Rightarrow y_1=y_0+h(x_0,y_0)

\Rightarrow y_1=2+(0.001)\left(-2+\frac{1}{2-0}\right)

\Rightarrow y_1=1.9985

h=0.0001

\Rightarrow y_1=y_0+h(x_0,y_0)

\Rightarrow y_1=2+(0.0001) \left(-2+\frac{1}{2-0}\right)

\Rightarrow y_1=1.99985

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