A) \[{{\sin }^{-1}}\frac{(y-1)}{x}=a\]
B) \[\sin \frac{(y-1)}{x}=a\]
C) \[\sin \frac{(1-y)}{(1+x)}=a\]
D) \[\sin \frac{y}{(x+1)}=a\]
Correct Answer: B
Solution :
Given \[\sin \frac{dy}{dx}=a\]; \[dy={{\sin }^{-1}}a\,dx\] Integrating both sides,\[\int_{{}}^{{}}{dy}=\int_{{}}^{{}}{{{\sin }^{-1}}a\,dx}\] \[y=x{{\sin }^{-1}}a+c\] and \[y(0)=0+c=1\], \[\therefore c=1\] \[\therefore y=x{{\sin }^{-1}}a+1\] Þ \[a=\sin \frac{y-1}{x}\].
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