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The solution of the differential equation $\frac{d y}{d x}+y=\cos x$ is
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Verified Answer
The correct answer is:
$y=\frac{1}{2}(\cos x+\sin x)+c e^{-x}$
It is linear equation of the form $\frac{d y}{d x}+P y=Q$
So, I.F. $=e^{\int 1 d x}=e^x$
Hence solution is $y \cdot e^x=\int \cos x \cdot e^x d x+c$
$\Rightarrow y=\frac{1}{2}(\cos x+\sin x)+c e^{-x}$
So, I.F. $=e^{\int 1 d x}=e^x$
Hence solution is $y \cdot e^x=\int \cos x \cdot e^x d x+c$
$\Rightarrow y=\frac{1}{2}(\cos x+\sin x)+c e^{-x}$
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