If the solution curve of the differential equation dydx=x+y−4x−y passes through the point (3,2) and (p+2,3), p>0 then

dydx=(x−2)+(y−2)(x−2)−(y−2) Put x−2=h,y−2=kdkdh=h+kh−k put k=vh⇒ v+hdvdh=1+v1−v ⇒ tan−1(y−2x−2)=12ln(1+(y−2)2(x−2)2)+ln(x−2)+c it passes through (3, 2)⇒ tan−10=12ln1+ln1+c⇒c=0 it also passes through (p + 2, 3)tan−1(1p)=12ln(1+1p2)+lnp ⇒ 2tan−1(1p)=ln(1+p2)

If the solution curve of the differential equation dydx=x+y−4x−y passes through the point (3,2) and (p+2,3), p>0 then

dydx=(x−2)+(y−2)(x−2)−(y−2) Put x−2=h,y−2=kdkdh=h+kh−k put k=vh⇒ v+hdvdh=1+v1−v ⇒ tan−1(y−2x−2)=12ln(1+(y−2)2(x−2)2)+ln(x−2)+c it passes through (3, 2)⇒ tan−10=12ln1+ln1+c⇒c=0 it also passes through (p + 2, 3)tan−1(1p)=12ln(1+1p2)+lnp ⇒ 2tan−1(1p)=ln(1+p2)

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If the solution curve of the differential equation $\frac{d y}{d x} = \frac{x + y - 4}{x - y}$ passes through the point $(3, 2)$ and $(p + 2, 3)$ , $p > 0$ then

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$\tan^{- 1} (\frac{1}{p}) = \log_{e} (p^{2} + 1)$

$2 \tan^{- 1} (\frac{1}{p}) = \log_{e} (p^{2} + 1)$

$2 \tan^{- 1} (\frac{1}{p}) = \log_{e} (\frac{p^{2} + 1}{p})$

$2 \tan^{- 1} (\frac{1}{p + 1}) = \log_{e} (p^{2} + 2 p + 2)$

answer is A.

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Detailed Solution

$\frac{d y}{d x} = \frac{(x - 2) + (y - 2)}{(x - 2) - (y - 2)}$ Put $x - 2 = h, y - 2 = k$
$\frac{d k}{d h} = \frac{h + k}{h - k}$ put $k = v h \Rightarrow v + h \frac{d v}{d h} = \frac{1 + v}{1 - v}$

$\Rightarrow \tan^{- 1} (\frac{y - 2}{x - 2}) = \frac{1}{2} \ln (1 + \frac{{(y - 2)}^{2}}{{(x - 2)}^{2}}) + \ln (x - 2) + c$
it passes through (3, 2)
$\Rightarrow \tan^{- 1} 0 = \frac{1}{2} \ln 1 + \ln 1 + c \Rightarrow c = 0$
it also passes through (p + 2, 3)
$\tan^{- 1} (\frac{1}{p}) = \frac{1}{2} \ln (1 + \frac{1}{p^{2}}) + \ln p$

$\Rightarrow 2 \tan^{- 1} (\frac{1}{p}) = \ln (1 + p^{2})$