Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 8

Year : 2014

Auxiliary equation is $D^2-4D+4=0$

$\therefore(D-2)^2=0\\ \therefore D=2,2\\ Y_c=(c_1+xc_2)e^{2x} \\ Y_c=e^{2x}c_1+xe^{2x}c_2\\ Let \space y_1=e^{2x}\\ y_2=xe^{2x}\\ x=e^{2x}\sec^2x $

Now $ W=\begin{vmatrix} y_1 & y_2\\ y_1^1 & y_2^1\\ \end{vmatrix} $

$ W=\begin{vmatrix} e^{2x} & xe^{2x}\\ 2e^{2x} & 2xe^{2x}+e^{2x}\\ \end{vmatrix} $

$W=e^{2x}(2xe^{2x}+e^{2x})-2xe^{2x}.e^{2x}\\ =2xe^{4x}+e^{4x}-2xe^{4x}\\ =e^{4x}\\ \therefore u=-\int\dfrac {y_2\times dx}w\\ =-\int \dfrac {xe^{2x}e^{2x}\sec^2x dx}{e^{4x}}\\ =-\int x\sec^2x \space dx$

Integration by parts we get

$u=-\Bigg[x\int \sec^2x\space dx-\int\Bigg(\dfrac {dx}{dx}\int \sec^2x\space dx\Bigg)dx\Bigg] \\ =-\Big[x\tan x-\int \tan x\space dx\Big]\\ =-x\tan x+\log\sec x \\ And \\ v=\int \dfrac {y_1\times dx}2\\ =\int \dfrac {e^{2x}e^{2x}\sec^2x \space dx}{e^{4x}}\\ =\int \sec^2x \space dx\\ =\tan x\\ \therefore y_p=Uy_1+Vy_2\\ =(-x\tan x+\log \sec x)e^{2x}+\tan x.xe^{2x}\\ Y_p=-\tan x.xe^{2x}+\log \sec x(e^{2x})+\tan x(xe^{2x})\\ Y_p=e^{2x}\log\sec x\\ \therefore y=y_c+y_p\\ \therefore y=(c_1+xc_2)e^{2x} +e^{2x}\log \sec x$