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Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 12

Answers (1)

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

Hint: To solve this equation we use $\int u v dx$ formula.

Give: $\begin{aligned} &\frac{d y}{d x}+y=\sin x \\ & \end{aligned}$

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

$P=1, Q=\sin x \\$

$\text { If }=e^{\int P d x} \\$

$=e^{\int 1 d x} \\$

$=e^{x}$

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

$y e^{x}=\int \sin x e^{x} d x+C \\$

$=I=\sin \int e^{x}-\int \frac{d}{d x} \sin x \int e^{x} d x d x \\$

$=I=\sin e^{x}-\int \cos x e^{x} d x$

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

$=I=e^{x}[\sin x-\cos x]-I \\$

$=2 I=e^{x}(\sin x-\cos x) \\$

$=I=\frac{e^{x}}{2}(\sin x-\cos x)$

Put in original equation

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

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

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