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 solution curve intersects the line y - x -2 atA- 1-3B- 2-4C- 0-2D- 3-5

The correct option is A (1, 3)Given 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 log y ⇒y/x+ ...

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Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

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 solution curve intersects the line y = x + 2 at

(1, 3)

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(2, 4)

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(0, 2)

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(3, 5)

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Solution

The correct option is A (1, 3)
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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A solution curve of the differential equation given by (x2+xy+4x+2y+4)dydx−y2=0 passes through (1, 3) The solution curve intersects the line y = x + 2 at

The correct option is A (1, 3)Given equation can be written as 1(x+2)2dxdy=1y2+1y(x+2),put1x+2=t⇒dtdy+ty=−1y2,IF=yty=∫−1ydy=C−logy⇒yx+2=C−log y,ItisP.T.(1,3)⇒c=1+log3⇒y=(x+2)[1+log3−logy]

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 solution curve intersects the line y = x + 2 at