A solution curve of the differential equation $\left(x^{2}+x y+4 x+2 y+4\right) \frac{d y}{d x}-y^{2}=0, x>0$, passes through the point (1, 3). Then the solution curve

(A) intersects $y=x+2$ exactly at one point; (D) does NOT intersect $y=(x+3)^{2}$

(B) intersects $y=x+2$ exactly at two points

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Differential Equations

A solution curve of the differential equation

(x^{2} + x y + 4 x + 2 y + 4) \frac{d y}{d x} - y^{2} = 0, x > 0

, passes through the point (1, 3). Then the solution curve

Answer: (A), (D)

Rewrite the equation as

[(x + 2)^{2} + y (x + 2)] \frac{d y}{d x} = y^{2}

.

This can be rearranged to

\frac{d y}{y} = \frac{y d ( x + 2 ) - ( x + 2 ) d y}{( x + 2 ) ^{2}}

.

Integrating:

ln y = - \frac{y}{x + 2} + c

.

Using the initial condition

(1, 3)

ln 3 = - \frac{3}{3} + c = - 1 + c

, so

c = 1 + ln 3

.

The solution is

ln (\frac{y}{3}) + \frac{y}{x + 2} = 1

.

(A) For

y = x + 2

ln (\frac{x + 2}{3}) + 1 = 1

, so

ln (\frac{x + 2}{3}) = 0

, giving

x + 2 = 3

, thus

x = 1

. One intersection. True.

(C) For

y = (x + 2)^{2}

ln (\frac{( x + 2 ) ^{2}}{3}) + (x + 2) = 1

. For

x > 0

x + 2 > 2

, so

(x + 2)^{2} > 4

and

ln (\frac{( x + 2 ) ^{2}}{3}) > ln \frac{4}{3} > 0

. Thus LHS

> 2 > 1

. No solution. False.

(D) For

y = (x + 3)^{2}

: similar analysis shows no intersection. True.

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · One or more correct options