#### Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 46

Answer: $x=2 \cos y$

Hint: Separate the terms of x and y and then integrate them.

Given: $x \frac{d y}{d x}+\cot y=0, y=\frac{\pi}{4} \text { when } x=\sqrt{2}$

Solution:

\begin{aligned} &x \frac{d y}{d x}+\cot y=0 \\\\ &\Rightarrow x \frac{d y}{d x}=-\cot y \\\\ &\Rightarrow \frac{d y}{\cot y}=-\frac{d x}{x} \end{aligned}

Integrating both sides

\begin{aligned} &\int \frac{1}{\cot y} d y=-\int \frac{1}{x} d x \\\\ &-\int \frac{\sin y}{\cos y} d y=\int \frac{1}{x} d x \\\\ &\Rightarrow \log |\cos y|=\log |x|+c \end{aligned}

\begin{aligned} &{\left[\therefore \int \frac{1}{x} d x=\log |x|+c\right]} \\\\ &\Rightarrow \log |\cos y|-\log |x|=c \end{aligned}                ...............(1)

Now, when $x=\sqrt{2}, y=\frac{\pi}{4}$

\begin{aligned} &\therefore \log \left|\cos \frac{\pi}{4}\right|-\log |\sqrt{2}|=c \\\\ &\Rightarrow \log \frac{1}{\sqrt{2}}-\log \sqrt{2}=c \Rightarrow \log (2)^{-\frac{1}{2}}-\log (2)^{\frac{1}{2}}=c \end{aligned}

\begin{aligned} &\Rightarrow-\frac{1}{2} \log 2-\frac{1}{2} \log 2=c \Rightarrow-2\left(\frac{1}{2} \log 2\right)=c \\\\ &\Rightarrow-\log 2=c \end{aligned}

Put in (1)

\begin{aligned} &\Rightarrow \log |\cos y|-\log |x|=-\log 2 \\\\\ &\Rightarrow \log \left|\frac{\cos y}{x}\right|+\log 2=0 \end{aligned}

$\left[\begin{array}{l} \therefore \log (m)+\log (n)=\log m n \\\\ \log (m)-\log (n)=\log \frac{m}{n} \end{array}\right]$

\begin{aligned} &\Rightarrow \log \left|\frac{2 \cos y}{x}\right|=0 \\\\ &{\left[\begin{array}{l} \therefore \log _{\varepsilon} a=x \\\\ x=a^{x} \end{array}\right]} \\\\ &\Rightarrow \frac{2 \cos y}{x}=e^{0} \end{aligned}

\begin{aligned} &{\left[e^{0}=1\right]} \\\\ &\Rightarrow 2 \cos y=x(1) \\\\ &\Rightarrow 2 \cos y=x \end{aligned}

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