If the solution curve, of the differential equation passing through the point (2, 1) is , then is equal to __________. [2024]

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Solution

Q.
If the solution curve, of the differential equation $\frac{d y}{d x} = \frac{x + y - 2}{x - y}$ passing through the point (2, 1) is $\tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{β} \log_{e} (α + {(\frac{y - 1}{x - 1})}^{2}) = \log_{e}$ $| x - 1 |$ , then $5 β + α$ is equal to __________. [2024]

Ans.
(11)

We have, $\frac{d y}{d x} = \frac{x + y - 2}{x - y}$

Substitute x = X + h and y = Y + k

$\Rightarrow \frac{d Y}{d X} = \frac{X + Y + h + k - 2}{X - Y + h - k}$

Put h + k – 2 = 0 and h – k = 0

$\Rightarrow$ h = 1 and k = 1

$∴ \frac{d Y}{d X} = \frac{X + Y}{X - Y}$

Now, put Y = vX $\Rightarrow \frac{d Y}{d X} = v + X \frac{d v}{d X}$

$∴ v + X \frac{d v}{d X} = \frac{X + v X}{X - v X} = \frac{1 + v}{1 - v}$

$\Rightarrow X \frac{d v}{d X} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v + v^{2}}{1 - v} = \frac{1 + v^{2}}{1 - v}$

$\Rightarrow \int \frac{1 - v}{1 + v^{2}} d v = \int \frac{d X}{X} \Rightarrow \int \frac{1}{1 + v^{2}} d v - \int \frac{v}{1 + v^{2}} d v = \int \frac{d X}{X}$

$\Rightarrow \tan^{- 1} v - \frac{1}{2} \log_{e} | 1 + v^{2} | = \log | X | + c$

$\Rightarrow \tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{2} \log_{e}$ $| 1 + {(\frac{y - 1}{x - 1})}^{2} |$ $= \log$ $| x - 1 |$ $+ c$ ... (i)

Equation (i) passing through (2, 1), then

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

$\Rightarrow \log_{e} 1 + c = 0 \Rightarrow c = 0$

$∴ \tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{2} \log_{e} | 1 + {(\frac{y - 1}{x - 1})}^{2} | = \log_{e} | x - 1 |$

By comparing with

$\tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{β} \log_{e}$ $| α + {(\frac{y - 1}{x - 1})}^{2} |$ $= \log_{e}$ $| x - 1 |$

$β = 2, α = 1$

$∴ 5 β + α = 5 \times 2 + 1 = 11$ .

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