Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 45 sub question (x)

Answer: $e^{x}-\tan y=2$

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

Given: $: e^{x} \tan y d x+\left(2-e^{x}\right) \sec ^{2} y d y=0 ; y(0)=\frac{\pi}{4}$

Solution:

\begin{aligned} &\quad e^{x} \tan y d x+\left(2-e^{x}\right) \sec ^{2} y d y=0 \\\\ &\Rightarrow e^{x} \tan y d x=-\left(2-e^{x}\right) \sec ^{2} y d y \\\\ &\Rightarrow e^{x} \tan y d x=\left(e^{x}-2\right) \sec ^{2} y d y \\\\ &\Rightarrow \frac{e^{x}}{e^{x}-2} d x=\frac{\sec ^{2} y}{\tan y} d y \end{aligned}

Integrating both sides

\begin{aligned} &\int \frac{e^{x}}{e^{x}-2} d x=\int \frac{\sec ^{2} y}{\tan y} d y \\\\ &\Rightarrow \log \left|e^{x}-2\right|=\log |\tan y|+\log |c| \\\\ &{\left[\therefore \int \frac{1}{x} d x=\log |x|+c\right]} \end{aligned}

\begin{aligned} &\Rightarrow \log \left|e^{x}-2\right|-\log |\tan y|=\log |c| \\\\ &\Rightarrow \log \left|\frac{e^{x}-2}{\tan y}\right|=\log |c| \\\\ &{\left[\therefore \log m-\log n=\log \frac{m}{n}\right]} \end{aligned}

Given that    $y(0)=\frac{\pi}{4} \text { i.e when } x=0 ; y=\frac{\pi}{4}$

\begin{aligned} &\Rightarrow \log \left|\frac{e^{0}-2}{\tan \frac{\pi}{4}}\right|=\log |c| \Rightarrow \log \left|\frac{1-2}{1}\right|=\log c \\\\ &{\left[\therefore e^{0}=1, \tan \frac{\pi}{4}=1\right]} \\\\ &\Rightarrow \log 1=\log c \Rightarrow \log c=0 \end{aligned}

Put in (1)

\begin{aligned} &\Rightarrow \log \left|\frac{e^{x}-2}{\tan y}\right|=0 \Rightarrow \frac{e^{x}-2}{\tan y}=e^{0} \\\\ &\Rightarrow e^{x}-2=\tan y(1) \\\\ &\begin{array}{l} {\left[\therefore e^{0}=1\right]\left[\begin{array}{l} \therefore \log _{e} a=x \\ a=e^{x} \end{array}\right]} \\\\ \Rightarrow e^{x}-\tan y=2 \end{array} \end{aligned}

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