Question
$\frac{{dy}}{{dx}} + \frac{y}{3} = 1$ का हल है
अत: $y\,.\,{e^{x/3}} = \int {1\,.\,{e^{x/3}}\,dx + c} $
$y\,.\,{e^{x/3}} = 3\,{e^{x/3}} + c$; $y = 3 + c{e^{ - x/3}}$.
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$( S 1): f ^{\prime}\left(-\frac{3}{2}\right)+ f ^{\prime}\left(-\frac{1}{2}\right)+ f ^{\prime}\left(\frac{1}{2}\right)+ f ^{\prime}\left(\frac{3}{2}\right)=2$
$(S2): \int_{-2}^2 f ( x ) dx =12$ है। तब