Let y=y(x) be the solution of the differential equation [(d y)/(d x)=(( x)+y)/( x( x- x x)…

Question

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

Accepted Answer

$$\begin{aligned}
& \frac{d y}{d x}=\frac{\sin x+y \cos x}{\sin x \cdot \cos x\left(\frac{1}{\cos x}-\sin x \cdot \frac{\sin x}{\cos x}\right)} \
& =\frac{\sin x+y \cos x}{\sin x\left(1-\sin ^2 x\right)} \
& \frac{d y}{d x}=\sec ^2 x+y \cdot 2(\operatorname{cosec} 2 x) \
& \frac{d y}{d x}-2 \operatorname{cosec}(2 x) \cdot y=\sec ^2 x \
& \frac{d y}{d x}+p \cdot y=Q
\end{aligned}$$

$$\text { I.F. }=\mathrm{e}^{\int p \mathrm{dx}}=\mathrm{e}^{\int-2 \operatorname{cosec}(2 \mathrm{x}) \mathrm{dx}}$$

Let $$2 \mathrm{x}=\mathrm{t}$$

$$2 \frac{\mathrm{dx}}{\mathrm{dt}}=1$$

$$\mathrm{dx}=\frac{\mathrm{dt}}{2}$$

$$=\mathrm{e}^{-\int \operatorname{cosec}(\mathrm{t}) \mathrm{dt}}$$

$$=\mathrm{e}^{-\ln \left|\tan \frac{t}{2}\right|}$$

$$=\mathrm{e}^{-\ln |\tan x|}=\frac{1}{|\tan x|}$$

$$\begin{aligned}
& y(\text { IF })=\int Q(I F) d x+c \
& \Rightarrow y \frac{1}{|\tan x|}=\int \sec ^2 x \cdot \frac{1}{|\tan x|}+c \
& y \cdot \frac{1}{|\tan x|}=\int \frac{d t}{|t|}+c \quad \text { for } \tan x=t \
& y \cdot \frac{1}{|\tan x|}=\ln |t|+c \
& y=|\tan x|(\ln |\tan x|+c) \
& \text { Put } x=\frac{\pi}{4}, y=2 \
& 2=\ln 1+c \Rightarrow c=2 \
& y=|\tan x|(\ln |\tan x|+2) \
& y\left(\frac{\pi}{3}\right)=\sqrt{3}(\ln \sqrt{3}+2)
\end{aligned}$$

Solutions · 1