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