A solution of the differential equation (x2+xy+4x+2y+4)dydx−y2=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 options are A intersects y=x+2 exactly at one point D does NOT intersect y=(x+3)2 (x2+xy+4x+2y+4)dydx−y2=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