If $\mathrm{y}=\mathrm{y}(\mathrm{x})$ is the solution of the differential equation, $\mathrm{e}^{\mathrm{y}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}-1\right)=\mathrm{e}^{\mathrm{x}}$ such that $y(0)=0$, then $y(1)$ is equal to:

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Differential Equations

y = y (x)

is the solution of the differential equation,

e^{y} (\frac{dy}{dx} - 1) = e^{x}

such that

y (0) = 0

, then

y (1)

is equal to:

Answer: (A)

Rewrite

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

by setting

t = e^{y}

, so

\frac{d t}{d x} = e^{y} \frac{d y}{d x}

and the equation becomes the linear form

\frac{d t}{d x} - t = e^{x}

. With integrating factor

e^{- x}

t e^{- x} = \int e^{x} e^{- x} d x = x + c

, i.e.

e^{y - x} = x + c

. The condition

y (0) = 0

gives

c = 1

, so

e^{y - x} = x + 1

and

y = x + ln (x + 1)

; at

x = 1

y = 1 + lo g_{e} 2

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Single correct option

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