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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Solving Linear Differential Equations of First Order

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x - 2y + 5 = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x - y + 2 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x + y - 4 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

Similar questions

Q. 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 -

Q. A solution 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:

Q. 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

Q. The equation of the common tangent to the curves

y^{2} = 8 x and x y = - 1

Q. The equation of common tangent to the curves

y^{2} = 16 x

and

x y = - 4

, is :

Join BYJU'S Learning Program

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

The correct option is B x - 2y + 5 = 0Given 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 equation of the tangent to the curve at (1, 3) is