Let $y = y(x)$ be the solution of the differential equation $\left((x+2)e^{\left(\frac{y+1}{x+2}\right)} + (y+1)\right)dx = (x+2)dy$, $y(1) = 1$. If the domain of $y = y(x)$ is an open interval $(\alpha, \beta)$, then $|\alpha + \beta|$ is equal to ____

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

((x + 2) e^{(\frac{y + 1}{x + 2})} + (y + 1)) d x = (x + 2) d y

y (1) = 1

.

If the domain of

y = y (x)

is an open interval

(α, β)

, then

∣ α + β ∣

is equal to ____

Answer: 1

Write the equation as

\frac{d y}{d x} = e^{(y + 1) / (x + 2)} + \frac{y + 1}{x + 2}

and substitute

v = \frac{y + 1}{x + 2}, u = x + 2

. Then

\frac{d y}{d x} = v + u \frac{d v}{d u}

, so

u \frac{d v}{d u} = e^{v}

, i.e.

e^{- v} d v = \frac{d u}{u}

. Integrating gives

- e^{- v} = ln u + C

. With

y (1) = 1

(so

u = 3, v = \frac{2}{3}

C = - e^{- 2/3} - ln 3

, hence

e^{- v} = e^{- 2/3} + ln \frac{3}{x + 2}

. Reality of

v

needs this positive together with

x + 2 > 0

0 < x + 2 < 3 e^{e^{- 2/3}}

. So the domain is

(α, β) = (- 2, 3 e^{e^{- 2/3}} - 2)

and

∣ α + β ∣ = 3 e^{e^{- 2/3}} - 4 \approx 1.01

, i.e.

1

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Integer answer