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  • Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 39 textbook solution.

Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 39 textbook solution.

Answers (1)

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

Give : \frac{d y}{d x}-y=\cos x

Hint : Using integrating factor and integration by parts

Explanation : \frac{d y}{d x}-y=\cos x

                     \frac{d y}{d x}+(-1)y=\cos x

This is a linear differential equation of the form

             \begin{aligned} &\frac{d x}{d y}+P y=Q \\ &P=-1 \text { and } Q=\cos 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 d x} \\ &=e^{-x} \quad\left[\int d x=x+C\right] \end{aligned}

Hence, the solution is

            \begin{aligned} &y I f=\int Q I f d x+C \\ &=y\left(e^{-x}\right)=\int \cos x e^{-x} d x+C \ldots(i) \end{aligned}

Using integration by parts

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

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

Substituting in (i)

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

Divide by e^{-x}

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

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