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

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"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"

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

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

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