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

Marks : 8

Year : 2014

Auxiliary equation is

$D^2+5D+6=0 \\ \therefore D=-3,-2\\ \therefore y_c=c_1e^{-3x}+c_2e^{-2x}\\ \text{Here} \space y_1=e^{-3x} \space , y_2=e^{-2x}\\ \space X=e^{2x}\sec^2x(1+2\tan x)$

Now

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

$=\begin{vmatrix} e^{-3x}&e^{-2x}\\ e^{-3x}.-3& e^{-2x}.-2\\ \end{vmatrix} $

$=e^{-5x}(-2)+e^{-5x}(3) \\ =e^{-5x} \\ U=-\int\dfrac {y^2}W\times dx\\ =-\int \dfrac {e^{-2x}}{e^{-5x}}e^{-2x}\sec^2x(1+2\tan x)dx \\=-\int e^{-x}\sec^2x(1+2\tan x)dx$

Integrating by parts we get

$U=-[e^x\int \sec^2(1+2\tan x)dx-\int \dfrac d{dx}e^x\int \sec^2x(1+2\tan x)dx\space dx]\\ Let \space I=\int \sec^2x(1+2\tan x)dx\\ Let\space \tan x=t\\ \therefore \sec^2x dx=dt\\ I=\int (1+2t)dt\\ =t+t^2\\ =\tan x+\tan^2 x -----(1)\\ \therefore U=-[e^x(\tan x+\tan ^2x)-\int e^x(\tan x+\tan^2x)dx]\\U=-e^x(\tan x+\tan ^2x) + \int e^x(\tan x+\tan^2x)dx\\ Let\space f(x)=(\tan x-1)\space then \space f^1(x)=\sec^2$

we have a property of integraton as $\int e^x[f(x)+f^1(x)]dx=e^xf(sc)+c$

$\therefore U=-e^x(\tan x+\tan^2x)+e^x(\tan x-1)+c\\ U=e^x(-\tan x-\tan^2x\tan x-1)\\=-e^x(\tan^2x+1)\\ =-e^x\sec^2x\\ And \space V =\int \dfrac {y_1}W\times dx\\ =\int \dfrac {e^{-3x}}{e^{-5x}}e^{-2x}\sec^2x(1+2\tan x)dx\\ =\tan x+\tan^2x\hspace {2cm} From (1)\\ \therefore P.I \space is \space y_p=uy_1+vy_2\\ =-e^x\sec^2xe^{-3x}+(\tan x+\tan^2x)e^{-2x}$

The complete solution $y=yc+yp\\ y=c_1e^{-3x}+c_2e^{-2x}-e^{-2x}\sec^2xe^{-3x}+(\tan x+\tan^2 x)e^{-2x}$