Question

Which of the following is a correct solution of $x\,cosx\,\left(\frac{dy}{dx}\right) + y(x sin x +cos x) = 1 ?$

• A. $yx\, sec\, y\, =\, C \, + tan x$
• B. $y\,x cos\, y = C + tan x$
• C. $y x \,sec \, x \,= \,C \,+ tan x$
• D. $None\, of \, these$

Answer 1

Answer: (C)

$y x \,sec \, x \,= \,C \,+ tan x$

Answer 2

Given, $x\, cos\, x \left(\frac{dy}{dx}\right)+y \left(x\,sin\,x+cos\,x\right)=1$
$\Rightarrow \frac{dy}{dx}+\frac{y \left(x\,sin\,x+cos\,x\right)}{x\, cos\,x}$
$=\frac{1}{x\, cos\,x}$
On comparing with $\frac{dy}{dx}+py=Q$, we get
$P=\frac{x\,sin\,x+cos\,x}{x\,cos\,x}$
$=tan\,x+\frac{1}{x}$
$\therefore IF=e^{\int p\,dx }$
$=e^{\int\left(tan\,x+\frac{1}{x}\right)dx}$
$=e^{log\,sec\,x+log\,x}$
$=e^{log\,x\,sec\,x}=x\,sec\,x$
$\therefore$ Solution is
$y\times x\,sec\,x=\int \frac{x\,sec\,x}{x\,cos\,x} dx$
$=\int sec^{2}x\,dx$
$\Rightarrow xy\, sec\, x =tan\, x+c$