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

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

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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}

 

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