A) \[y=c(x+a)(1-ay)\]
B) \[y=c(x+a)(1+ay)\]
C) \[y=c(x-a)(1+ay)\]
D) None of the above
Correct Answer: A
Solution :
Given, \[y-x\frac{dy}{dx}=a\left( {{y}^{2}}+\frac{dy}{dx} \right)\] \[\Rightarrow \] \[\frac{dy}{dx}(a+x)=y-a{{y}^{2}}\] \[\Rightarrow \] \[\int{\frac{dy}{y(1-ay)}}=\int{\frac{dx}{a+x}}\] \[\Rightarrow \] \[\int{\left( \frac{1}{y}+\frac{a}{1-ay} \right)}dy=\int{\frac{dx}{a+x}}\] \[\Rightarrow \] \[\log y-\log (1-ay)=\log (a+x)+\log c\] \[\Rightarrow \] \[\log y=\log (1-ay)(a+x)c\] \[\Rightarrow \] \[y=c(1-ay)(a+x)\]
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