A solution curve of the differential equation given by x2-x y-4 x-2 y-4 d y-d x-y2-0 passes through 1- 3The equation of the tangent to the curve at 1-3 isA- 2 x-y-5-0B- x-2 y-5-0C- x-y-4-0D- x-y-2-0

The correct option is B x 2y + 5 = 0Given equation can be written as 1/(x+2)^2dx/dy=1/y^2+1/y(x+2), put 1/x+2=t ⇒dt/dy+t/y= 1/y^2, IF = y ty=∫ 1/ydy = C lo ...

A solution cu...

Question

A solution curve of the differential equation given by $(x^{2} + x y + 4 x + 2 y + 4) \frac{d y}{d x} - y^{2} = 0$ passes through (1, 3)

The equation of the tangent to the curve at (1, 3) is

2x + y - 5 = 0

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x - 2y + 5 = 0

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x - y + 2 = 0

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x + y - 4 = 0

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Solution

The correct option is B x - 2y + 5 = 0
Given equation can be written as
$\frac{1}{(x + 2)^{2}} \frac{d x}{d y} = \frac{1}{y^{2}} + \frac{1}{y (x + 2)}, p u t \frac{1}{x + 2} = t \Rightarrow \frac{d t}{d y} + \frac{t}{y} = \frac{- 1}{y^{2}}, I F = y t y = \int \frac{- 1}{y} d y = C - l o g y \Rightarrow \frac{y}{x + 2} = C - l o g y, I t i s P . T . (1, 3) \Rightarrow c = 1 + l o g 3 \Rightarrow y = (x + 2) [1 + l o g 3 - l o g y]$

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