Applied Mathematics for Electrical Engineering (3130908)

Question-3c ( ii )

BE | Semester-3 Winter-2019 | 26-11-2019

Q3) (c ( ii ))

y^{'} = x + y = f (x, y)

Now,

y (0) = 1 \Rightarrow y_{0} = 1, x_{0} = 0

By Euler’s method :

y_{n + 1} = y_{n} + \frac{1}{6} [k_{1} + 2 k_{2} + 2 k_{3} + k_{4}]

Where,

x_{n} = x_{0} + n h, n = 0, 1, 2, 3, \dots

k_{1} = h f (x_{n}, y_{n})

k_{2} = h f (x_{n} + \frac{h}{2}, y_{n} + \frac{k_{1}}{2})

k_{3} = h f (x_{n} + \frac{h}{2}, y_{n} + \frac{k_{2}}{2})

k_{4} = h f (x_{n} + h, y_{n} + k_{3})

For,

n = 0

k_{1} = h f (x_{0}, y_{0})

k_{1} = (0.1) f (0, 1)

k_{1} = (0.1) (0 + 1)

k_{1} = 0.1

k_{2} = h f (x_{0} + \frac{h}{2}, y_{0} + \frac{k_{1}}{2})

k_{2} = (0.1) f (0 + \frac{0.1}{2}, 1 + \frac{0.1}{2})

k_{1} = (0.1) (0.05, 1.05)

k_{1} = (0.1) (0.05 + 1.05)

k_{1} = 0.11

k_{3} = h f (x_{0} + \frac{h}{2}, y_{0} + \frac{k_{2}}{2})

k_{2} = (0.1) f (0 + \frac{0.1}{2}, 1 + \frac{0.11}{2})

k_{1} = (0.1) (0.05, 1.055)

k_{1} = (0.1) (0.05 + 1.055)

k_{1} = 0.1105

k_{4} = h f (x_{0} + h, y_{0} + k_{3})

k_{1} = (0.1) f (0 + 0.1, 1 + 0.1105)

k_{1} = (0.1) f (0.1, 1.1105)

k_{1} = (0.1) (0.1 + 1.1105)

k_{1} = 0.1211

Now,

y_{1} = y_{0} + \frac{1}{6} [k_{1} + 2 k_{2} + 2 k_{3} + k_{4}]

Now,

k_{1} = 1 + \frac{1}{6} (0.1 + 2 (0.11) + 2 (0.1105) + 0.1211)

Now,

k_{1} = 1 + \frac{1}{6} (0.6631)

Now,

k_{1} = 1.1105

Now,

x_{1} = x_{0} + h = 0 + 0.1 = 0.1

For,

n = 1

k_{1} = h f (x_{1}, y_{1})

k_{1} = (0.1) f (0.1, 1.1105)

k_{1} = (0.1) (0.1 + 1.1105)

k_{1} = 0.1211

k_{2} = h f (x_{1} + \frac{h}{2}, y_{1} + \frac{k_{1}}{2})

k_{2} = (0.1) f (0.1 + \frac{0.1}{2}, 1.1105 + \frac{0.1211}{2})

k_{1} = (0.1) (0.15, 1.1711)

k_{1} = (0.1) (0.15 + 1.1711)

k_{1} = 0.1321

k_{3} = h f (x_{1} + \frac{h}{2}, y_{1} + \frac{k_{2}}{2})

k_{2} = (0.1) f (0.1 + \frac{0.1}{2}, 1.1105 + \frac{0.1321}{2})

k_{1} = (0.1) (0.15, 1.1766)

k_{1} = (0.1) (0.15 + 1.1766)

k_{1} = 0.1327

k_{4} = h f (x_{1} + h, y_{1} + k_{3})

k_{1} = (0.1) f (0.1 + 0.1, 1.1105 + 0.1327)

k_{1} = (0.1) f (0.2, 1.2432)

k_{1} = (0.1) (0.2 + 1.2432)

k_{1} = 0.1443

Now,

y_{2} = y_{1} + \frac{1}{6} [k_{1} + 2 k_{2} + 2 k_{3} + k_{4}]

Now,

k_{1} = 1.1105 + \frac{1}{6} (0.1211 + 2 (0.1321) + 2 (0.1327) + 0.1443)

Now,

k_{1} = 1.1105 + \frac{1}{6} (0.7950)

Now,

k_{1} = 1.2430

Therefore,

y (0.2) = 1.2430