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

Engineering

Mathematics

Linear Differential Equation

Question

Let y = f(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:

1 + log_e 2

2 + log_e 2

log_e 2

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e^y = t

$e^{y} \frac{dy}{dx} = \frac{dt}{dx}$

$\frac{dt}{dx} - t = e^{x}$

$IF = e^{\int - 1 . dx} = e^{- x}$

$t (e^{- x}) = \int e^{x} . e^{- x} dx$

e^y–x = x + c

Put x = 0, y = 0 then c = 1

e^y–x = x + 1

y = x + ln(x + 1)

at x = 1, y = 1 + ln(2)