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Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 14 textbook solution.

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

Hint : To solve this equation we use  e^{\int P d x}  formula.

Give : \frac{d y}{d x}+2 y=\sin x

Solution : \frac{d y}{d x}+P y=Q

               P=2,Q=\sin \; x

               \begin{aligned} &I f=e^{\int f d x} \\ &=e^{\int 2 d x} \\ &=e^{2x} \end{aligned}\begin{aligned} &I f=e^{\int f d x} \\ &=e^{\int 2 d x} \\ &=e^{2x} \end{aligned}

                \begin{aligned} &=y I f=\int Q I f d x+C \\ &=y e^{x}=\int e^{2 x} \sin x d x+C \end{aligned}

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

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