Solution of the differential equation xy(dy/dx)=(1 + y2/1 + x2)(1 + x + x2) , given that x=1,y=0 is

Question

Solution of the differential equation xy(dy/dx)=(1 + y2/1 + x2)(1 + x + x2) , given that x=1,y=0 is (A) ℓ n√1 + y2=ℓ nx+tan- 1x-(π /2) (B) ln((1 + y2/x2))=2(tan)- 1x-(π /2) (C) ℓ n((1 + y2/x2))=(π /4)-2(tan)- 1x (D) ℓ n((1 + y2/x2))=(tan)- 1x-(π /4)

Tardigrade Admin · Accepted Answer

∫ ( ydy /1+ y 2)=∫ (1+ x + x 2/ x (1+ x 2)) dx (1/2) ℓ n (1+ y 2)= tan -1 x +ℓ nx + c ⇒ ℓ n ((1+ y 2/ x 2))=2 tan -1 x + c

Q. Solution of the differential equation $x y \frac{d y}{d x} = \frac{1 + y ^{2}}{1 + x ^{2}} (1 + x + x^{2})$ , given that $x = 1, y = 0$ is

$ℓ n 1 + y^{2} = ℓ n x + t a n^{- 1} x - \frac{π}{2}$

$l n (\frac{1 + y ^{2}}{x ^{2}}) = 2 (t an)^{- 1} x - \frac{π}{2}$

$ℓ n (\frac{1 + y ^{2}}{x ^{2}}) = \frac{π}{4} - 2 (t an)^{- 1} x$

$ℓ n (\frac{1 + y ^{2}}{x ^{2}}) = (t an)^{- 1} x - \frac{π}{4}$

$\int \frac{y d y}{1 + y ^{2}} = \int \frac{1 + x + x ^{2}}{x ( 1 + x ^{2} )} d x$
$\frac{1}{2} ℓ n (1 + y^{2}) = tan^{- 1} x + ℓ n x + c$
$\Rightarrow ℓ n (\frac{1 + y ^{2}}{x ^{2}}) = 2 tan^{- 1} x + c$

Q. Solution of the differential equation xydxdy​=1+x21+y2​(1+x+x2) , given that x=1,y=0 is

ℓn1+y2​=ℓnx+tan−1x−2π​

ln(x21+y2​)=2(tan)−1x−2π​

ℓn(x21+y2​)=4π​−2(tan)−1x

ℓn(x21+y2​)=(tan)−1x−4π​