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

Marks: 4M

Year: May 2014

Given $F = 16y^2 - (y")^2 + x^2$

$$\therefore \frac{\partial F}{\partial y} = 32y, \frac{\partial F}{\partial y'} = 0, \frac{\partial F}{\partial y'} = -2y"$$

The necessary condition for this functional to be extremum is

$$\frac{\partial F}{\partial y} - \frac{d}{dx} \bigg(\frac{\partial f}{\partial y'}\bigg) + \frac{d^2}{dx^2} \bigg(\frac{\partial F}{\partial y"}\bigg) = 0$$

Hence the equation becomes,

$$32y - 0 + \frac{d^2}{dx^2}\bigg(-2 \bigg(\frac{d^2y}{dx^2}\bigg)\bigg) = 0 \\ \therefore \frac{d^4 y}{dx^4} - 16y = 0$$

Above is linear differential equation of the fourth order.

Its Auxillary Equation is $D^4 - 16 = 0 \\ \therefore (D^2 - 4)(D^2 + 4) = 0 \\ \therefore D = \pm 2, \pm 2i$

Hence it's complimentary function (C.F.) is

$$y = c_{1}e^2x + c_{2}e^{-2x} + c_{3}cos2x + c_{4}sin2x$$

Thus,

Extremal of $\int\limits_{x_{1}}^{x_{2}} (16y^2 - (y")^2 + x^2)dx$ is $$y = c_{1}e^2x + c_{2}e^{-2x} + c_{3}cos2x + c_{4}sin2x$$