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

Marks: 6M

Year: Dec 2015

We have $f(x,y,y') = 2xy + y^2 - y^{12}$ does satisfies rule as equation i.e. $\frac{\partial f}{\partial y} - \frac{\partial}{\partial x} \bigg(\frac{\partial f}{\partial y'}\bigg) = 0$

Hence we have to extremise

$\bar{I} = \int\limits_{0}^{1}fdx$

We assume the trial solution

$\bar{y}(x) = c_{0} + c_{1}x c_{1}x^2$

By data $\bar{y}(0) = 0 \therefore c_{0} = 0 \\ \bar{y}(1) = 0 \therefore 0 = c_{1} + c_{2} \therefore c_{2} = -c_{1} \\ \bar{y}(x) = c_{1}x - c_{1}x^2 = c_{1}x(1 - x) \\ y^9 (x) = c_{1} - 2c_{1}x = c_{1}(1 - 2x)$$

Putting these values in I

$$I = \int\limits_{0}^{1} (2xy + y^2 - y'^2)dx$$ we get $$I = \int\limits_{0}^{1} 2x[c_{1}x (1 - x)] + c_{1}^2 x^2 (1 - x)^2 - c_{1}^2 (1 - 2x)^2 dx \ = c_{1} \int\limits_{0}^{1} (2(x^2 - x^3) + c_{1} [ x^2 - 2x^3 + x^4 - ( 1 - 4x + 4x^2)])dx \ = c_{1} \int\limits_{0}^{1} [2(x^2 - x^3) + c_{1} ( - 1 + 4x - 3x^2 - 2x^3 + x^4)]dx \ = c_{1} \bigg[2 \bigg(\frac{x^3}{3} - \frac{x^4}{4}\bigg) + c_{1}\bigg[- x + 2x^2 - x^3 - \frac{x^4}{2} + \frac{x^5}{5} \bigg]\bigg]_{0}^{1} \ = c_{1}\bigg[2\bigg(\frac{1}{3} - \frac{1}{4}bigg) + c_{1} \bigg[- 1 + 2 - 1 - \frac{1}{2} + \frac{1}{5} \bigg] \bigg] \ = c_{1} \bigg(\frac{1}{6} - \frac{3}{10}c_{1}\bigg) \ = \frac{c_{1}}{6} - \frac{3}{10}c_{1}^2$$

Its stationary values are given by

$\frac{d\bar{I}}{dc_{1}} = 0 \hspace{0.5cm} \frac{1}{6} - \frac{3}{5}c_{1} = 0$

Hence Approximate solution is

$\bar{y}(x) = \frac{5}{18}x(1 + x)$