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 21 Differential Equation Exercise Revision Exercise (RE) Question 51 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 51 maths textbook solution.

Answers (1)

Answer :    y=\frac{1}{13}(2 \sin 3 x-3 \cos 3 x)+C e^{-2 x}

Hint               : You must know the rules of solving differential equation and integration.

Given            :      \frac{d y}{d x}+2 y=\sin 3 x

Solution        :    \frac{d y}{d x}+2 y=\sin 3 x

Compare with,

                  \frac{d y}{d x}+P y=Q

where, P=2 and Q= \sin 3x

Therefore,

\begin{aligned} \text { Integrating factor }=\text { I.F } &=e^{\int p d x} \\ &=e^{\int 2 d x} \\ &=e^{2 x} \end{aligned}

The solution is,

\begin{aligned} &y \times I . F=\int(Q \times I . F) d x+C \\ &y e^{2 x}=\int e^{2 x} \times \sin 3 x d x+C \\ &y e^{2 x}=I+C \end{aligned}           ......(i)

Where,    I=\int e^{2 x} \sin 3 x d x                    ......(ii)

Apply integrating by parts,

\begin{aligned} I &=e^{2 x} \int \sin 3 x d x-\int\left[\frac{d e^{2 x}}{d x} \int \sin 3 x d x\right] d x \\ I &=\frac{-e^{2 x} \cos 3 x}{3}+\frac{2}{3} \int e^{2 x} \cos 3 x d x \\ I &=\frac{-e^{2 x} \cos 3 x}{3}+\frac{2}{3}\left[e^{2 x} \int \cos 3 x d x-\int\left\{\frac{d e^{2 x}}{d x} \int \cos 3 x d x\right\} d x\right] \end{aligned}

\begin{aligned} I &=\frac{-e^{2 x} \cos 3 x}{3}+\frac{2}{3}\left[\frac{e^{2 x} \sin 3 x}{3}-\frac{2}{3} \int e^{2 x} \sin 3 x d x\right] \\ I &=\left[\frac{2}{9} e^{2 x} \sin 3 x-\frac{e^{2 x} \cos 3 x}{3}\right]-\frac{4}{9} I \end{aligned}

\begin{aligned} I+\frac{4}{9} I &=e^{2 x}\left(\frac{2}{9} \sin 3 x-\frac{\cos 3 x}{3}\right) \\ \frac{13}{9} I &=e^{2 x}\left(\frac{2}{9} \sin 3 x-\frac{\cos 3 x}{3}\right) \\ I &=\frac{9}{13} e^{2 x}\left(\frac{2}{9} \sin 3 x-\frac{\cos 3 x}{3}\right) \\ I &=\frac{1}{13} e^{2 x}(2 \sin 3 x-3 \cos 3 x)+C \end{aligned}                                         .....(iii)

From (i) and (iii), we get,

\begin{aligned} y e^{2 x} &=\frac{e^{2 x}}{13}(2 \sin 3 x-3 \cos 3 x)+C \\ y \quad &=\frac{1}{13}(2 \sin 3 x-3 \cos 3 x)+C e^{-2 x} \end{aligned}     Is required solution.

Posted by

infoexpert23

View full answer

Crack CUET with india's "Best Teachers"

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

Similar Questions

Trending Articles/News

anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question