Given, $${e^y}\left( {{{dy} \over {dx}} - 1} \right) = {e^x}$$

$$ \Rightarrow $$ $${e^y}{{dy} \over {dx}} - {e^y} = {e^x}$$ ....(1)

Let $${e^y} = t$$ $$ \Rightarrow $$ $${e^y}{{dy} \over {dx}} = {{dt} \over {dx}}$$

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

So here p = -1 and q = e^x

We know, IF = $${e^{\int {pdx} }}$$

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

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

$$ \Rightarrow $$ t.$${e^{ - x}}$$ = x + c

Putting value of t, we get

$${e^y}{e^{ - x}}$$ = x + c

$$ \Rightarrow $$ $${e^{y - x}} = x + c$$ .....(2)

Given y(0) = 0 means y = 0 when x = 0.

Putting in equation (2), we get

e⁰ = 0 + c

$$ \Rightarrow $$ c = 1

$$ \therefore $$ $${e^{y - x}} = x + 1$$ ....(3)

Now we have to find y(1), which means when x = 1 find the value of y in the above equation.

Putting x = 1 in equation (3)

$${e^{y - 1}} = 1 + 1$$

$$ \Rightarrow $$ y - 1 = log_e2

$$ \Rightarrow $$ y = 1 + log_e2