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

Marks: 5M

Year: Dec 2015

We have $f = y^2 - y^12 - 2ycoshx \\ \frac{\partial f}{\partial y} = 2y - 2coshx \\ \frac{\partial f}{\partial y'} = - 2y'$

Putting these values in rules equation

$\frac{\partial f}{\partial y} - \frac{\partial}{\partial x} \bigg(\frac{\partial f}{\partial y'}\bigg) = 0$ we get

$2y - 2coshx - \frac{\partial}{\partial x} (-2y') = 0 \\ 2y - 2coshx + 2 \frac{\partial^2 y}{\partial x^2} = 0 \\ \frac{\partial^2 y}{\partial x^2} + y = coshx..................(1)$

We have to solve this equation

It can be written as

$(D^2 + 1)y = coshx$

A.E. equation is

$D^2 + 1 = 0 -\gt D = \pm i$

C.F. os $C_{1}cosx + C_{2}sinx$

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

Thus the complete solution is

$y = c_{1}cosx + c_{2}sinx + \frac{1}{2}cosx$