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

Answers (1)

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

Hint: To solve this equation we use \int u v\; d x formula.

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

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

\begin{aligned} &P=1, Q=\sin x \\ \end{aligned}

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

\begin{aligned} &=e^{\int 1 d x} \\ \end{aligned}

\begin{aligned} &=e^{x} \end{aligned}

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

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

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

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