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  • Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 36 subquetsion (ii)

Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 36 subquetsion (ii)

Answers (1)

Answer:  y=\frac{1}{5}(2 \sin 2 x-\cos 2 x)+C e^{x}

Give:  \frac{d y}{d x}-y=\cos 2 x

Hint: Using integration by parts

Explanation: \begin{aligned} & \frac{d y}{d x}-y=\cos 2 x \\ & \end{aligned}

 \frac{d y}{d x}+(-1) y=\cos 2 x

This is a first order linear differential equation of the form

 \begin{aligned} &\frac{d y}{d x}+P y=Q \\ &P=-1 \text { and } Q=\cos 2 x \end{aligned}

The integrating factor If  of this differential equation is

 \begin{aligned} &I f=e^{\int P d x} \\ & \end{aligned}

=e^{\int-1 d x} \\

=e^{-\int 1 d x} \quad\left[\int d c=x+C\right] \\

=e^{-x}

Hence, the solution of differential equation is

 \begin{aligned} &y(I f)=\int Q I f d x+C \\ & \end{aligned}

\Rightarrow y\left(e^{-x}\right)=\int(\cos 2 x) e^{-x} d x+C \\

\Rightarrow y e^{-x}=\int e^{-x} \cos 2 x d x+C \\

\Rightarrow y e^{-x}=\int\left(e^{-x}\right)(\cos 2 x) d x+C

\begin{aligned} \text { Let } I=& \int\left(e^{-x}\right)(\cos 2 x) d x+C \\ \end{aligned}

\Rightarrow I=e^{-x} \int \cos 2 x d x-\int \frac{d}{d x}\left(e^{-x}\right)\left(\int \cos 2 x d x\right) d x+C \\

\Rightarrow I=e^{-x}\left(\frac{\sin 2 x}{2}\right)-\int-e^{-x}\left(\frac{\sin 2 x}{2}\right) d x+C \\

\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2} \int e^{-x} \sin 2 x d x+C

\begin{aligned} &\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2}\left\{\left[e^{-x}\left(\int \sin 2 x d x\right)\right]-\int \frac{d}{d x} e^{-x}\left(\int \sin 2 x d x\right) d x+C\right\} \\ & \end{aligned}

\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2}\left\{\left[e^{-x}\left(\frac{-\cos 2 x}{2}\right)\right]-\int-e^{-x}\left(-\frac{\cos 2 x}{2}\right) d x+c\right\} \\

\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2}\left\{\left[-\frac{1}{2} e^{-x} \cos 2 x\right]-\frac{1}{2} \int-e^{-x}(-\cos 2 x) d x+C\right\} \\

\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2}\left\{\left[-\frac{1}{2} e^{-x} \cos 2 x\right]-\frac{1}{2} \int e^{-x} \cos 2 x d x+c\right\}

\begin{aligned} &\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2}\left\{\left[-\frac{1}{2} e^{-x} \cos 2 x\right]-\frac{1}{2} I\right\} \\ & \end{aligned}

\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2}\left(-\frac{1}{2} e^{-x} \cos 2 x\right)+\frac{1}{2}\left(-\frac{1}{2} I\right) \\

\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x-\frac{1}{4} e^{-x} \cos 2 x-\frac{1}{4} I \\

\Rightarrow I+\frac{1}{4} I=\frac{1}{2} e^{-x} \sin 2 x-\frac{1}{4} e^{-x} \cos 2 x

\begin{aligned} &\Rightarrow I+\frac{1}{4} I=\frac{1}{2} e^{-x} \sin 2 x-\frac{1}{4} e^{-x} \cos 2 x \\ & \end{aligned}

\Rightarrow \frac{5}{4} I=\frac{1}{4} 2 e^{-x} \sin 2 x-\frac{1}{4} e^{-x} \cos 2 x \\

\Rightarrow \frac{5}{4} I=\frac{1}{4} e^{-x}(2 \sin 2 x-\cos 2 x) \\

\Rightarrow 5 I=e^{-x}(2 \sin 2 x-\cos 2 x) \\

\therefore I=\frac{e^{-x}}{5}(2 \sin 2 x-\cos 2 x)

By substituting the value of   in the original integral we get

 \begin{aligned} &\Rightarrow y e^{-x}=\frac{e^{-x}}{5}(2 \sin 2 x-\cos 2 x)+C \\ & \end{aligned}

\Rightarrow y e^{-x} e^{x}=e^{x}\left[\frac{e^{-x}}{5}(2 \sin 2 x-\cos 2 x)+C\right] \\

\Rightarrow y e^{-x} e^{x}=\frac{e^{x-x}}{5}(2 \sin 2 x-\cos 2 x)+C e^{x} \\

\therefore y=\frac{1}{5}(2 \sin 2 x-\cos 2 x)+C e^{x}

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