Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • 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

Answers (2)

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

 

        

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

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}

 

Posted by

infoexpert27

View full answer

Similar Questions

Trending Articles/News

anam-khan

CBSE extends single girl child scholarship registration deadline till November 20
anam-khan

CBSE reminds schools to submit registration data of Class 9, 11 students by October 16
anam-khan

CBSE reopens online portal for submission of LOC for Class 10, 12 board exams 2026; submit forms by Oct 8

Latest Question