The solution of the differential equation, x y (d y/d x)=(1+y2/1+x2)(1+x+x2) given that when x=1, y=0 is

Question

The solution of the differential equation, x y (d y/d x)=(1+y2/1+x2)(1+x+x2) given that when x=1, y=0 is (A)  ln √1+y2= ln x+ tan -1 x-(π/2) (B)  ln (1+y2/x2)=2 tan -1 x-(π/2) (C)  ln ((1+ y 2/ x 2))=(π/4)-2 tan -1 x (D)  ln (1+y2/x2)= tan -1 x-(π/4)

Tardigrade Admin · Accepted Answer

∫ ( ydy /1+ y 2)=∫ (1+ x + x 2/ x (1+ x 2)) dx ; (1/2) ln (1+ y 2)= tan -1 x + ln x + C ln (1+y2/x2)=2 tan -1 x+C.

Q. The 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 when $x = 1, y = 0$ is

$ln 1 + y^{2} = ln x + tan^{- 1} x - \frac{π}{2}$

$ln \frac{1 + y ^{2}}{x ^{2}} = 2 tan^{- 1} x - \frac{π}{2}$

$ln (\frac{1 + y ^{2}}{x ^{2}}) = \frac{π}{4} - 2 tan^{- 1} x$

$ln \frac{1 + y ^{2}}{x ^{2}} = tan^{- 1} x - \frac{π}{4}$

$\int \frac{y d y}{1 + y ^{2}} = \int \frac{1 + x + x ^{2}}{x ( 1 + x ^{2} )} d x; \frac{1}{2} ln (1 + y^{2}) = tan^{- 1} x + ln x + C$
$ln \frac{1 + y ^{2}}{x ^{2}} = 2 tan^{- 1} x + C$ .

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

ln1+y2​=lnx+tan−1x−2π​

lnx21+y2​=2tan−1x−2π​

ln(x21+y2​)=4π​−2tan−1x

lnx21+y2​=tan−1x−4π​