The solution of differential equation 4 x y (d y/d x)=(3(1+x)2(1+y2)/(1+x2)) is (where C is constant of integration)

Question

The solution of differential equation 4 x y (d y/d x)=(3(1+x)2(1+y2)/(1+x2)) is (where C is constant of integration) (A)  ln (1+ y )= ln x +2 tan x + C (B)  ln (1+ y 2)=3 ln (1/ x )+6 tan -1 x + C (C) 2 ln (1+y2)=3 ln x+6 tan -1 x+C (D) 2 ln (1+ y 2)= ln x +2 tan -1 x + C

Tardigrade Admin · Accepted Answer

Given equation can be written as ∫ (4 y/1+y2) d y=∫ (3(1+x)2/x(1+x2)) d x ⇒ 2 ln (1+ y 2) =3 ∫((1/ x )+(2/1+ x 2)) dx =3( ln x +2 tan -1 x )+ C

Q. The solution of differential equation $4 x y \frac{d y}{d x} = \frac{3 ( 1 + x ) ^{2} ( 1 + y ^{2} )}{( 1 + x ^{2} )}$ is
(where $C$ is constant of integration)

$ln (1 + y) = ln x + 2 tan x + C$

$ln (1 + y^{2}) = 3 ln \frac{1}{x} + 6 tan^{- 1} x + C$

$2 ln (1 + y^{2}) = 3 ln x + 6 tan^{- 1} x + C$

$2 ln (1 + y^{2}) = ln x + 2 tan^{- 1} x + C$

Given equation can be written as $\int \frac{4 y}{1 + y ^{2}} d y = \int \frac{3 ( 1 + x ) ^{2}}{x ( 1 + x ^{2} )} d x$
$\Rightarrow 2 ln (1 + y^{2}) = 3 \int (\frac{1}{x} + \frac{2}{1 + x ^{2}}) d x$
$= 3 (ln x + 2 tan^{- 1} x) + C$

Q. The solution of differential equation 4xydxdy​=(1+x2)3(1+x)2(1+y2)​ is (where C is constant of integration)

ln(1+y)=lnx+2tanx+C

ln(1+y2)=3lnx1​+6tan−1x+C

2ln(1+y2)=3lnx+6tan−1x+C

2ln(1+y2)=lnx+2tan−1x+C