Applied Mathematics for Electrical Engineering (3130908)

Question-4c ( i )-OR

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

Q4) (c ( i ))

The Ricatti’s equation is,

y' = x^{2} + y^{2}

Therefore,

f (x, y) = x^{2} + y^{2}

Now,

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

By Taylor’s method,

y (x_{n}) = y_{n} = y_{n - 1} + \frac{h}{1!} y_{n - 1}^{'} + \frac{h^{2}}{2!} y_{n - 1}^{''} + \dots

Where,

x_{n} = x_{0} + nh; n = 1, 2, 3, . . .

Now,

y_{1} = y_{0} + \frac{h}{1!} y_{0}^{'} + \frac{h^{2}}{2!} y_{0}^{''} + \frac{h^{3}}{3!} y_{0}^{'''} + \dots

y^{'} = f (x, y) = x^{2} + y^{2} \Rightarrow y_{0}^{'} = 0 + 0 = 0

y^{''} = 2 x + 2 y y' \Rightarrow y_{0}^{''} = 2 (0) + 2 (0) (0) = 0

y^{'''} = 2 + 2 y y'' + 2 {(y')}^{2} \Rightarrow y_{0}^{'''} = 2 + 2 (0) (0) + 2 {(0)}^{2} = 2

Then by Taylor’s series,

y (x_{1}) = y_{1} = y_{0} + \frac{h}{1!} y_{0}^{'} + \frac{h^{2}}{2!} y_{0}^{''} + \frac{h^{3}}{3!} y_{0}^{'''} + \dots

y (x_{1}) = y_{1} = 1 + \frac{0.2}{1!} (0) + \frac{{(0.2)}^{2}}{2!} (0) + \frac{{(0.2)}^{3}}{3!} (2) + \dots

y (x_{1}) = 0.0027