Subject: Applied Mathematics 4

Topic: Calculus of Variation

Difficulty: Medium

$ F = y^2 - (y')^2 -2y \, coshx \\ \frac{\partial F}{\partial y} = 2y - 2 \, coshx \hspace{0.25cm}... (1) \\ \frac{\partial F}{\partial y'} = -2y' \hspace{0.25cm}... (2) $

Putting (1) and (2) in $ \frac{\partial F}{\partial y} - \frac{d}{dx}[\frac{\partial F}{\partial y'}] = 0 $, we get,

$ [2y - 2 \, coshx] - \frac{d}{dx}[-2y'] = 0 \\ [2y - 2 \, coshx] + \frac{d}{dx}[2y'] = 0 \\ [2y - 2 \, coshx] + 2 \frac{d}{dx}[\frac{dy}{dx}] = 0 \\ [2y - 2 \, coshx] + 2 \frac{d^2 y}{dx^2} = 0 \\ 2 \frac{d^2 y}{dx^2} +2(y - coshx) = 0 \\ \frac{d^2 y}{dx^2} + (y - coshx) = 0 \\ \frac{d^2 y}{dx^2} + y = coshx \hspace{0.25cm} ...(*) $

We have to solve differential equation $(*)$.

$ (D^2 + 1) = coshx $

The auxiliary equation is D$^2$ + 1 = 0. Therefore, D = +i, -i

The complementary function is given by,

$ y_c = C_1 \, cosx + C_2 \, sinx $

The partial integration is given by,

$ P.I. = \frac{1}{D^2 + 1}[coshx] \\ = \frac{1}{D^2 + 1}[\frac{e^x + e^{-x}}{2}] \\ = \frac{1}{2}[\frac{1}{D^2 + 1}(e^x) + \frac{1}{D^2 + 1}(e^{-x})] \\ = \frac{1}{2}[\frac{e^x}{2} + \frac{e^{-x}}{2}] = \frac{1}{2}[\frac{e^x + e^{-x}}{2}] = \frac{1}{2}coshx $

The complete solution is, $ y_c + y_p = C_1 \, cosx + C_2 \, sinx + \frac{1}{2}coshx $