A) 3
B) 2
C) 1
D) 0
Correct Answer: A
Solution :
\[x\,\,dy=y(dx+y\,\,dy),\,\,y>0\] \[\Rightarrow \] \[x\,\,dy-y\,\,dx={{y}^{2}}dy\] \[\Rightarrow \] \[\frac{x\,\,dy-y\,\,dx}{{{y}^{2}}}=dy\] \[\Rightarrow \] \[-d\left( \frac{x}{y} \right)=dy\] On integrating both sides, we get \[\frac{x}{y}=-y+C\] ? (i) As \[y(1)=1\] \[\Rightarrow \] \[x=1,\,\,y=1\] \[\therefore \] \[C=2\] \[\therefore \] Eq. (i) becomes \[\frac{x}{y}+y=2\] Again, for \[x=-3\] \[\Rightarrow \] \[-3+{{y}^{2}}=2y\] \[\Rightarrow \] \[{{y}^{2}}-2y-3=0\] \[\Rightarrow \] \[(y+1)(y-3)=0\] As\[y>0\], we take\[y=3\], (neglecting\[y=-1\])
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