A) \[\frac{5}{2}\]
B) 2
C) \[\frac{7}{2}\]
D) 3
Correct Answer: C
Solution :
Given differential equation is \[\frac{\left( 2+\sin x \right)}{\left( 1+y \right)}.\frac{dy}{dx}=\cos x\]which can be rewritten as\[\frac{dy}{1+y}=\frac{\cos x}{2+\sin x}dx\]Integrate both the sides, we get\[\int_{{}}^{{}}{\frac{dy}{1+y}=\int_{{}}^{{}}{\frac{\cosxdx}{2+\sin x}}}\] \[\Rightarrow \]\[\log (1+y)=\log (2+\sin x)+\log C\] \[\Rightarrow \]\[1+y=C(2+\sin x)\] Given\[y(0)=2\] \[\Rightarrow \]\[1+2=C[2+\sin 0]\Rightarrow C=\frac{3}{2}\] Now,\[y\left( \frac{\pi }{2} \right)\]can be found as \[1+y=\frac{3}{2}\left( 2+\sin \frac{\pi }{2} \right)\Rightarrow 1+y=\frac{9}{2}\]\[\Rightarrow y=\frac{7}{2}\] Hence,\[y\left( \frac{\pi }{2} \right)=\frac{7}{2}\]
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