#### Please solve RD Sharma class 12 chapter Differential Equations exercise 21.7 question 49 maths textbook solution

Answer: $x \log (x+1)-x+\log (x+1)+3$

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

Given: $e^{\frac{d y}{d x}}=x+1 \text { given that } y=3 \text { when } x=0$

Solution: $e^{\frac{d y}{d x}}=x+1$

Integrating both sides

\begin{aligned} &\log e^{\frac{dy}{dx}}=\log (x+1) \\\\ &\Rightarrow \frac{d y}{d x}=\log (x+1) d x\left[\because \log _{e} e=1\right] \\\\ &\Rightarrow d y=\log (x+1) d x \end{aligned}

Integrating both sides

\begin{aligned} &\int d y=\int \log (x+1) \cdot 1 d x \\ &\Rightarrow y=\log (x+1) x-\int \frac{1}{x+1} x d x \\\\ &\Rightarrow y=\log (x+1) x-\int \frac{x+1-1}{x+1} d x \\\\ &\Rightarrow y=x \log (x+1)-\int\left(1-\frac{1}{x+1}\right) d x \end{aligned}    [Integration by parts]

\begin{aligned} &\Rightarrow y=x \log (x+1)-[x-\log |x+1|]+c \\\\ &\Rightarrow y=x \log (x+1)-x+\log |x+1|+c \end{aligned}            ..............(1)

Now $y=3$ when $x = 0$

\begin{aligned} &3=0 \cdot \log (0+1)-0+\log (0+1)+c \\\\ &\Rightarrow 3=0-0+0+c \Rightarrow c=3[\because \log 1=0] \end{aligned}

Put in (1)

$\Rightarrow y=x \log (x+1)-x+\log |x+1|+3$

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Answer:  $\tan ^{-1}2-\frac{\pi}{4}$

Hint: To solve this we assume cosec x and cot x in t

Given:  $\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\cos e c x \cot x}{1+\operatorname{cosec}^{2} x} d x$

Solution:

\begin{aligned} &\cos e c x=t \\ & \end{aligned}

$\operatorname{cosec} x \cdot \cot x d x=d t \\$

$\int_{2}^{1} \frac{-d t}{1+t^{2}}=-\left[\tan ^{-1} t\right]_{0}^{1} \\$

$=-\left(\frac{\pi}{4}-\tan ^{-1} 2\right) \\$

$=\tan ^{-1} 2-\frac{\pi}{4}$