The solution curve of the differential equation. (1+ e - x )(1+ y 2) ( dy / dx )= y 2, which passes through the point (0,1), is :

Question

The solution curve of the differential equation. (1+ e - x )(1+ y 2) ( dy / dx )= y 2, which passes through the point (0,1), is : (A) y2=1+y log e((1+ex/2)) (B) y2+1=y( log e((1+ex/2))+2) (C) y2=1+y log e((1+e-x/2)) (D) y2+1=y( log e((1+e-x/2))+2)

Tardigrade Admin · Accepted Answer

(1+ e -x)(1+ y 2) ( dy / dx )= y 2 ⇒ (1+y-2) d y=((ex/1+ex)) d x ⇒ (y-(1/y))= ln (1+ex)+c ∴ It passes through (0,1) ⇒ c=- ln 2 ⇒ y2=1+y ln ((1+ex/2))

Q. The solution curve of the differential equation. $(1 + e^{- x}) (1 + y^{2}) \frac{d y}{d x} = y^{2},$ which passes through the point $(0, 1),$ is :

$y^{2} = 1 + y lo g_{e} (\frac{1 + e ^{x}}{2})$

$y^{2} + 1 = y (lo g_{e} (\frac{1 + e ^{x}}{2}) + 2)$

$y^{2} = 1 + y lo g_{e} (\frac{1 + e ^{- x}}{2})$

$y^{2} + 1 = y (lo g_{e} (\frac{1 + e ^{- x}}{2}) + 2)$

$(1 + e^{- x}) (1 + y^{2}) \frac{d y}{d x} = y^{2}$
$\Rightarrow (1 + y^{- 2}) d y = (\frac{e ^{x}}{1 + e ^{x}}) d x$
$\Rightarrow (y - \frac{1}{y}) = ln (1 + e^{x}) + c$
$∴$ It passes through $(0, 1)$
$\Rightarrow c = - ln 2$
$\Rightarrow y^{2} = 1 + y ln (\frac{1 + e ^{x}}{2})$

Q. The solution curve of the differential equation. (1+e−x)(1+y2)dxdy​=y2, which passes through the point (0,1), is :

y2=1+yloge​(21+ex​)

y2+1=y(loge​(21+ex​)+2)

y2=1+yloge​(21+e−x​)

y2+1=y(loge​(21+e−x​)+2)