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  • Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 39

Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 39

Answers (1)

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

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

Hint: Using integrating factor and integration by parts

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

 \begin{aligned} &\\ &=\frac{d y}{d x}+(-1) y=\cos x \end{aligned}

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

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

\begin{aligned} &=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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