If the solution curve, of the differential equation \(\frac{dy}{dx}=\frac{x+y-2}{x-y}\) passing through the point \((2,1)\) is \(\tan^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{\beta} \log_{e}\left(\alpha+\left(\frac{y-1}{x-1}\right)^{2}\right)=\log_{e}|x-1|\), then \(5\beta+\alpha\) is equal to

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Differential Equations

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)

tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{β} lo g_{e} (α + (\frac{y - 1}{x - 1})^{2}) = lo g_{e} ∣ x - 1∣

, then

5 β + α

is equal to

Answer: 11

\frac{d y}{d x} = \frac{x + y - 2}{x - y}

. Shift the origin,

x = X + h

y = Y + k

, choosing

h, k

so the constants vanish:

h + k - 2 = 0

and

h - k = 0

, giving

h = k = 1

.

Then

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

, which is homogeneous. Let

Y = v X

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

Separating variables:

\frac{1 - v}{1 + v ^{2}} d v = \frac{d X}{X} \Rightarrow tan^{- 1} v - \frac{1}{2} lo g_{e} (1 + v^{2}) = lo g_{e} ∣ X ∣ + C .

At the point

(2, 1)

X = 1, Y = 0, v = 0

, so

0 - 0 = 0 + C \Rightarrow C = 0

.

With

v = \frac{Y}{X} = \frac{y - 1}{x - 1}

and

X = x - 1

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

Comparing with the given form,

β = 2

and

α = 1

5 β + α = 10 + 1 = 11.

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Integer answer