If y = y(x) is the solution of the differential equation, [e^y( dy dx - 1 ) = e^x] such th…

Question

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:

Accepted Answer

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 = ex

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 = loge2

$$ \Rightarrow $$ y = 1 + loge2

Solutions · 1