Differential Equations Question 15
Question 15 - 2024 (31 Jan Shift 1)
Let $y=y(x)$ be the solution of the differential equation
$\frac{d y}{d x}=\frac{(\tan x)+y}{\sin x(\sec x-\sin x \tan x)}$,
$x \in\left(0, \frac{\pi}{2}\right)$ satisfying the condition $y\left(\frac{\pi}{4}\right)=2$.
Then, $y\left(\frac{\pi}{3}\right)$ is
(1) $\sqrt{3}\left(2+\log _e \sqrt{3}\right)$
(2) $\frac{\sqrt{3}}{2}\left(2+\log _e 3\right)$
(3) $\sqrt{3}\left(1+2 \log _e 3\right)$
(4) $\sqrt{3}\left(2+\log _e 3\right)$
Show Answer
Answer (1)
Solution
$\frac{d y}{d x}=\frac{\sin x+y \cos x}{\sin x \cdot \cos x\left(\frac{1}{\cos x}-\sin x \cdot \frac{\sin x}{\cos x}\right)}$
$=\frac{\sin x+y \cos x}{\sin x\left(1-\sin ^{2} x\right)}$
$\frac{d y}{d x}=\sec ^{2} x+y \cdot 2(\operatorname{cosec} 2 x)$
$\frac{d y}{d x}-2 \operatorname{cosec}(2 x) \cdot y=\sec ^{2} x$
$\frac{d y}{d x}+p \cdot y=Q$
IF. $=e^{\int xdx}=e^{\int-2 \cos x(2 x) dx}$
Let $2 x=t$
$2 \frac{dx}{dt}=1$
$dx=\frac{dt}{2}$
$=e^{-\int \cos \cos (t) a}$
$=e^{-\left|\tan \frac{t}{2}\right|}$
$=e^{-t^{\mid \sin x}}=\frac{1}{|\tan x|}$
$y(I F)=\int Q(I F) d x+c$
$\Rightarrow y \frac{1}{|\tan x|}=\int \sec ^{2} x \cdot \frac{1}{|\tan x|}+c$
$y \cdot \frac{1}{|\tan x|}=\int \frac{d t}{|t|}+c \quad$ for $\tan x=t$
$y \cdot \frac{1}{|\tan x|}=\ln |t|+c$
$y=|\tan x|(\ln |\tan x|+c)$
Put $x=\frac{\pi}{4}, y=2$
$2=\ln 1+c \Rightarrow c=2$
$y=|\tan x|(\ln |\tan x|+2)$
$y\left(\frac{\pi}{3}\right)=\sqrt{3}(\ln \sqrt{3}+2)$
