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Mathematics Question on Area under Simple Curves

If the solution curve of the differential equation dydx=x+y−2x−y\frac{dy}{dx} = \frac{x + y - 2}{x - y} passes through the points (2,1)(2, 1) and (k + 1, 2), k > 0, then

A

2tan⁡−1(1k)=log⁡e(k2+1)2\tan^{-1}\left(\frac{1}{k}\right) = \log_e(k^2 + 1)

B

tan⁡−1(1k)=log⁡e(k2+1)\tan^{-1}\left(\frac{1}{k}\right) = \log_e(k^2 + 1)

C

2tan⁡−1(1k+1)=log⁡e(k2+2k+2)2\tan^{-1}\left(\frac{1}{k+1}\right) = \log_e(k^2 + 2k + 2)

D

2tan⁡−1(1k)=log⁡e(k2+1k2)2\tan^{-1}\left(\frac{1}{k}\right) = \log_e\left(k^2 + \frac{1}{k^2}\right)

Answer

2tan⁡−1(1k)=log⁡e(k2+1)2\tan^{-1}\left(\frac{1}{k}\right) = \log_e(k^2 + 1)

Explanation

Solution

dydx=x−1+y−1x−1−y+1\frac{dy}{dx} = \frac{x-1 + y-1}{x-1 - y+1}
Let x–1=X,y–1=Yx – 1 = X, y – 1 = Y

dYdX=X+YX−Y\frac{dY}{dX}=\frac{X+Y}{X−Y}

LetY=tXLet Y=tX

dYdX=t+XdtdX\frac{dY}{dX} = t + X\frac{dt}{dX}

t+XdtdX=1+t1−tt + X\frac{dt}{dX} = 1 + \frac{t}{1 - t}

XdtdX=1+t1−t−t=1+t21−tX\frac{dt}{dX} = 1 + \frac{t}{1 - t} - t = 1 + \frac{t^2}{1 - t}

∫1−t1+t2 dt=∫dXX\int \frac{1 - t}{1 + t^2} \, dt = \int \frac{dX}{X}

tan⁡−1(t−1)−12ln⁡(1+t2)=ln⁡∣X∣+C\tan^{-1}(t - 1) - \frac{1}{2}\ln(1 + t^2) = \ln|X| + C

tan⁡−1(y−1x−1)−12ln⁡(1+(y−1x−1)2)=ln⁡∣x−1∣+C\tan^{-1}\left(\frac{y-1}{x-1}\right) - \frac{1}{2}\ln\left(1+\left(\frac{y-1}{x-1}\right)^2\right) = \ln|x-1| + C
Curve passes through (2,1)(2, 1)
0–0=0+c⇒c=00–0=0+c⇒c=0
If (k\+1,2)(k \+ 1, 2) also satisfies the curve
tan⁡−1(1k)−12ln⁡(1+k2k2)=ln⁡k\tan^{-1}\left(\frac{1}{k}\right) - \frac{1}{2}\ln\left(\frac{1 + k^2}{k^2}\right) = \ln k
2tan⁡−1(1k)=ln⁡(1+k2)2\tan^{-1}\left(\frac{1}{k}\right) = \ln(1+k^2)
So, the correct option is (A): 2tan⁡−1(1k)=ln⁡(k2+1)2\tan^{-1}\left(\frac{1}{k}\right) = \ln(k^2 + 1)