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  • 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

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