Mumbai University > Electronics and Telecommunication > Sem 4 > Applied Maths 4

Marks: 8M

Year: May 2015

We have $f = (y')^2 - 2y - 2xy$

Now we assume the trial solution to be $\bar{y}(x) = c_{0} + c_{1}x + c_{2}x^2$

But by data y(0) = 2

$\therefore c_{0} = 2$ and $y_{1} = 1$

$\therefore 1 = 2 + 1c_{1} + c_{2} \\ \therefore c_{1} + c_{2} = -1 \\ i,e. c_{2} = -(c_{1} + 1) \\ \therefore \bar{y}(x) = 2 + c_{1}x - c_{1}x^2 - 1x^2 \\ \bar{y^{-1}}(x) = c_{1} - 2c_{1}x - 2x$

Putting these volumes in main equation we get

$$I = \int\limits_{0}^{1} (c_{1} - 2c_{1}x - 2x)^2 - 2[2 + c_{1}x - c_{1}x^2 - x^2] - 2x[2 + c_{1}x - c_{1}x^2 - x^2]dx \\ \therefore I = \int\limits_{0}^{1}[c_{1} - 2x(c_{1} + 1)]^2 - 4 - 2c_{1}x + 2c_{1}x^2 + 2x^2 - 4x - 2c_{1}x^3 + 2x^3 dx \\ \therefore I = \int\limits_{0}^{1} c_{1}^2x - 2c_{1}x^2 + (c_{1} + 1) + 4x^2 (c_{1} + 1)^2 - 4 - 2c_{1}x + 2c_{1}x^2 + 2x^2 - 4x - 2c_{1}x^2 + 2c_{1}x^3 + 2x^3dx \\ I = \bigg[c_{1}^2 x - 2c_{1}x^2 + (c_{1} + 1) + \frac{4x^3}{3} (c_{1} + 1)^2 - 4x - c_{1}x^2 + \frac{2c_{1}x^3}{3} + \frac{2}{3}x^3 - 2x^2 - \frac{2}{3}c_{1}x^3 + \frac{c_{1}x^4}{4} + \frac{1}{2}x^4\bigg]_{0}^{1} \\ I = c_{1}^2 - 2c_{1}^2 - 2c_{1} + \frac{4}{3} c_{1}^2 + \frac{8}{3}c_{1} + 1 - 4 - c_{1} + \frac{2}{3}c_{1} + \frac{2}{3} - 2 - \frac{2}{3}c_{1} + \frac{c_{1}}{4} + \frac{1}{2} \\ I = \frac{1}{3}c_{1}^2 - \frac{1}{12}c_{1} - \frac{23}{6}$$

Its stationary values are given by

$\frac{\partial \bar{I}}{\partial c_{1}} = 0 \\ \therefore \frac{2}{3}c_{1} - \frac{1}{12} = 0$

Thus the approximate solution is

$\bar{y} = 2 + \frac{1}{8}x - \frac{1}{8}x^2 - x^2 \\ = 2 + \frac{1}{8}x - \frac{9}{8}x^2$