ORDINARY DIFFERENTIAL EQUATION Use Eulers method to obtain a

ORDINARY DIFFERENTIAL EQUATION

Use Euler\'s method to obtain a four-decimal approximation of the indicated value. Carry out the recursion of (3) in Section 2.6 yn + 1 = yn + hf(xn, yn) (3) by hand, first using h = 0.1 and then using h = 0.05. y\' = 2x - 3y + 1, y(1) = 9; y(1.2)

Solution

y\' = 2x - 3y + 1 ; y(1) = 9

y\' + 3y = 2x + 1

Integrating factor = e^(integral of 3*dx) = e^(3x)

Multiply the given DE by e^(3x) on both sides :

e^(3x) * [y\' + 3y = 2x + 1]

d/dx(y*e^(3x)) = (2x + 1)*e^(3x)

Integrating both sides :

y*e^(3x) = 2xe^(3x)/3 + e^(3x)/9 + C

y = 2x/3 + 1/9 + Ce^(-3x)

y(1) = 9 :

9 = 2/3 + 1/9 + Ce^(-3)

74/9 = C/e^3

C = 74e^3/9

So, y = 2x/3 + 1/9 + Ce^(-3x)
becomes
y = 2x/3 + 1/9 + (74e^3/9)e^(-3x)

y = (74/9)e^(3 - 3x) + 2x/3 + 1/9 --- SOLUTION

y\' = 2x - 3y + 1

y(1) = 9 --> x = 1 , y = 9

f0 = 2(1) - 3(9) + 1 --> -24

y1 = y0 + hf0
y1 = 9 + (-24)(0.1) = 9 - 2.4 = 6.6

Now, when x = 1.1 , y = 6.6

f1 = 2(1.1) - 3(6.6) + 1 ---> -16.6

y2 = y1 + hf1

y2 = 6.6 + (0.1)(-16.6)

y2 = 4.94 ----> FIRST ANSWER

------------------------------------------------------------

y\' = 2x - 3y + 1

y(1) = 9 --> x = 1 , y = 9

f0 = 2(1) - 3(9) + 1 --> -24

y1 = y0 + hf0

y1 = 9 + 0.05 * -24 = 7.8

f1 = 2(1.05) - 3(7.8) + 1 --> -20.3

y2 = y1 + hf1
y2 = 7.8 + 0.05(-20.3)
y2 = 6.785

f2 = 2(1.1) - 3(6.785) + 1 --> -17.155

y3 = y2 + hf2
y3 = 6.785 + (0.05)(-17.155)
y3 = 5.92725

f3 = 2(1.15) - 3(5.92725) + 1 --> -14.48175

y4 = y3 + hf3
y4 = 5.92725 + (0.05)(-14.48175)
y4 = 5.2031625 ---> SECOND ANSWER

So, the answers are :

y(1.2) = 4.94 --> FIRST BLANK
y(1.2) = 5.2031625 --> SECOND BLANK