Let $y=y(x)$ be the solution of the differential equation $d y=e^{\alpha x+y} d x ; \alpha \in N$. If $y\left(\log _{e} 2\right)=\log _{e} 2$ and $y(0)=\log _{e}\left(\frac{1}{2}\right)$, then the value of $\alpha$ is equal to ______

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

d y = e^{α x + y} d x; α \in N

. If

y (lo g_{e} 2) = lo g_{e} 2

and

y (0) = lo g_{e} (\frac{1}{2})

, then the value of

α

is equal to ______

Answer: 2

Separating the variables in

\frac{d y}{d x} = e^{α x} e^{y}

gives

e^{- y} d y = e^{α x} d x

, and integrating,

- e^{- y} = \frac{e ^{α x}}{α} + C

x = 0

y = lo g_{e} \frac{1}{2}

, so

e^{- y} = 2

and

- 2 = \frac{1}{α} + C

. At

x = lo g_{e} 2

y = lo g_{e} 2

, so

e^{- y} = \frac{1}{2}

and

- \frac{1}{2} = \frac{2 ^{α}}{α} + C

. Subtracting the first from the second,

\frac{3}{2} = \frac{2 ^{α} - 1}{α} ⟹ 2^{α + 1} = 3 α + 2

For

α = 1

the two sides are

4

and

5

; for

α = 2

both are

8

; for

α \geq 3

the left side already exceeds the right and grows faster. The only natural number satisfying the equation is

α = 2

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Integer answer