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  • Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 36 subquestion (ii) textbook solution.

Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 36 subquestion (ii) textbook solution.

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 :  \frac{d y}{d x}-y=\cos 2 x

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

Hence, the solution of differential equation is

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

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

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

            \begin{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 \end{aligned}

            \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\} \\ &\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\} \end{aligned}

           \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\} \\ &\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 \end{aligned}

           \begin{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) \end{aligned}

By substituting the value of I in the original integral we get 

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

           \begin{aligned} &\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} \end{aligned}

 

          

 

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