Subject: Applied Mathematics 4

Topic: Calculus of Variation

Difficulty: Medium

We have, $ f = 16y^2 - (y'')^2 + x^2 \\ \frac{\partial f}{\partial y} = 32 y \\ \frac{\partial f}{\partial y'} = 0 \\ \frac{\partial f}{\partial y''} = -2y'' $

Hence the equation,

$ \frac{\partial f}{\partial y} - \frac{d}{dx}[\frac{\partial f}{\partial y'}] + \frac{d^2}{dx^2}[\frac{\partial f}{\partial y''}] = 0 \\ \therefore 32y + \frac{d^2}{dx^2}[-2y''] = 0 \\ 32y - 2y^{''''} = 0 \\ 32y - 2 \frac{d^4 y}{dx^4} = 0 \\ \frac{d^4 y}{dx^4} - 16y = 0 \\ (D^4 - 16)y = 0 \\ (D^2 - 4)(D^2 + 4) = 0 $

D = 2,-2 and D = 2i, -2i

The complementary function is, $ y_c = C_1 e^{2x} + C_2 e^{-2x} + C_3 \, cos(2x) + C_4 \, sin(2x) $

The general equation is, $ y = C_1 e^{2x} + C_2 e^{-2x} + C_3 \, cos(2x) + C_4 \, sin(2x) \hspace{0.5cm} [\because \frac{d^4 y}{dx^4} = 0] $