A) \[y=1+cxy\]
B) \[y=\log \{cxy\}\]
C) \[y+1=cxy\]
D) \[y=c+xy\]
Correct Answer: A
Solution :
\[x\frac{dy}{dx}+y={{y}^{2}}\] Þ \[x\frac{dy}{dx}={{y}^{2}}-y\] Þ \[\frac{dy}{{{y}^{2}}-y}=\frac{dx}{x}\] Þ \[\left[ \frac{1}{y-1}-\frac{1}{y} \right]dy=\frac{dx}{x}\] On integrating, we get \[\log (y-1)-\log y=\log x+\log c\] Þ \[\frac{y-1}{y}=xc\] Þ \[y=1+cxy\].You need to login to perform this action.
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