The general solution of y 2 d x + ( x 2 − x y + y 2 ) d y = 0 is [EAMCET 2003] Options: A . &nbsp t a n − 1 ( x y ) + l o g y + c = 0 B . &nbsp 2 t a n − 1 ( x y ) + l o g x + c = 0 C . &nbsp l o g ( y + √ x 2 + y 2 ) + l o g y + c = 0 D . &nbsp s i n h − 1 ( x y ) + l o g y + c = 0

Answer: Option A : A d x d y + x 2 − x y + y 2 y 2 = 0 d x d y + ( x y ) 2 − ( x y ) + 1 = 0 Put v = x / y ⇒ x = v y ⇒ d x d y = v + y d v d y v + y d v d y + v 2 − v + 1 = 0 ⇒ d v v 2 + 1 + d y y = 0 ⇒ ∫ d v v 2 + 1 + ∫ d y y = 0 ⇒ t a n − 1 ( v ) + l o g y + C = 0 ⇒ t a n − 1 ( x / y ) + l o g y + c = 0.

The general solution of y2 dx+(x2−xy+y2)dy=0 is [EAMCET 200

Discussion Forum : Differential Equations

Question -

The general solution of $y^{2} d x + (x^{2} - x y + y^{2}) d y = 0$ is

[EAMCET 2003]

Options:

A .

t a n^{- 1} (\frac{x}{y}) + l o g y + c = 0

B .

2 t a n^{- 1} (\frac{x}{y}) + l o g x + c = 0

C .

l o g (y + \sqrt{x^{2} + y^{2}}) + l o g y + c = 0

D .

s i n h^{- 1} (\frac{x}{y}) + l o g y + c = 0

Answer: Option A
:
A

$\frac{d x}{d y} + \frac{x^{2} - x y + y^{2}}{y^{2}} = 0 \frac{d x}{d y} + {(\frac{x}{y})}^{2} - (\frac{x}{y}) + 1 = 0$
Put $v = x / y \Rightarrow x = v y \Rightarrow \frac{d x}{d y} = v + y \frac{d v}{d y} v + y \frac{d v}{d y} + v^{2} - v + 1 = 0 \Rightarrow \frac{d v}{v^{2} + 1} + \frac{d y}{y} = 0 \Rightarrow \int \frac{d v}{v^{2} + 1} + \int \frac{d y}{y} = 0 \Rightarrow t a n^{- 1} (v) + l o g y + C = 0 \Rightarrow t a n^{- 1} (x / y) + l o g y + c = 0.$

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