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Question

A solution of the differential equation (x2+xy+4x+2y+4)dydxy2=0, x>0, passes through the point (1,3). Then the solution curve:

A
intersects y=x+2 exactly at one point
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B
intersects y=x+2 exactly at two points
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C
intersects y=(x+2)2
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D
does NOT intersect y=(x+3)2
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Solution

The correct option is D does NOT intersect y=(x+3)2
(x2+xy+4x+2y+4)dydxy2=0, x>0
(x+2)(x+y+2)dydx=y2

Let x+2=t
t(t+y)dydt=y2
dydt=(y/t)21+y/t

Let yt=k
k+tdkdt=k21+k
(k+1k)dk+dtt=0
k+ln|kt|=c
yx+2+ln|y|=c

Now, Given that the curve passes through the point (1,3).
c=ln3e
yx+2+ln|y|=ln3e

Option 1 & 2:
For the curve y=x+2
1+ln(x+2)=ln3e
LHS is an increasing function & RHS is a constant
Min(1+ln(x+2)=ln2e<ln3e
So, there will be only one intersection point of the curves

Option 3:
For the curve y=(x+2)2
x+2+2ln(x+2)=ln3e
LHS is an increasing function & RHS is a constant
Min(x+2+2ln(x+2))=2+2ln2>ln3e
So, the curve won't intersect eachother

Option 4:
For the curve y=(x+2)3
(x+2)2+3ln(x+2)=ln3e
LHS is an increasing function & RHS is a constant
Min((x+2)2+3ln(x+2)=4+3ln2>ln3e
So, the curve won't intersect eachother

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