Question

If $\dfrac{{dy}}{{dx}} = {\left( {{e^y} - x} \right)^{ - 1}}$ where $y\left( 0 \right) = 0$then $y$ is expressed explicitly as
(A) $0.5{\log _e}\left( {1 + {x^2}} \right)$
(B) ${\log _e}\left( {1 + {x^2}} \right)$
(C) ${\log _e}\left( {x + \sqrt {1 + {x^2}} } \right)$
(D) ${\log _e}\left( {x + \sqrt {1 - {x^2}} } \right)$

Answer 1

Hint: First of all identify the type of differential equation. Here the given differential equation is in the form of $\dfrac{{dx}}{{dy}} + Px = Q$, so it is linear differential equation of first order whose solution can be find in several steps:
(i) Convert the given differential equation in the form of $\dfrac{{dx}}{{dy}} + Px = Q$ to find out the values of $P$ and $Q$.
(ii) Now find integrating factor (I.F.) by the formula; $I.F. = {e^{\int {Pdy} }}$.
(iii) Now multiplying the both sides of equation obtained in step 1 by I.F., so that the LHS becomes the derivative of some function of $x$ and $y$.
(iv) Now integrate the both sides of equation obtained in step 3 with respect to $y$ and then find the value of $C$.

Complete step-by-step answer:
Given, $\dfrac{{dy}}{{dx}} = {\left( {{e^y} - x} \right)^{ - 1}}$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\left( {{e^y} - x} \right)}}$
$ \Rightarrow \dfrac{{dx}}{{dy}} = \left( {{e^y} - x} \right)$
$ \Rightarrow \dfrac{{dx}}{{dy}} + x = {e^y}$ …. (1)
It is of the form of first order differential equation, i.e., $\dfrac{{dx}}{{dy}} + Px = Q$, where $P = 1,Q = {e^y}$.
The solution of a linear differential equation can be find by following steps:
Step 1: First of all, find the Integrating Factor (I.F.), which is given by,
$I.F. = {e^{\int {Pdy} }}$
Put $P = 1$,
$I.F. = {e^{\int {1 \cdot dy} }}$
$I.F. = {e^y}$
Step 2: Now multiplying the both sides of equation (1) by I.F., so that the LHS becomes the derivative of some function of $x$ and $y$.
${e^y}\dfrac{{dx}}{{dy}} + {e^y}x = {e^y}{e^y}$
$ \Rightarrow {e^y}\dfrac{{dx}}{{dy}} + x{e^y} = {e^{2y}}$ ….. (2)
Now, using product rule of derivative, i.e., $\dfrac{d}{{dx}}\left[ {f\left( x \right) \cdot g\left( x \right)} \right] = f\left( x \right)\dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] + g\left( x \right)\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right]$ to find $\dfrac{d}{{dy}}\left( {{e^y} \cdot x} \right)$.
$\therefore $$\dfrac{d}{{dy}}\left( {{e^y} \cdot x} \right) = {e^y} \cdot \dfrac{{dx}}{{dy}} + x{e^y}$
Now, equation (2) becomes
$ \Rightarrow \dfrac{d}{{dy}}\left( {{e^y} \cdot x} \right) = {e^{2y}}$ [Put ${e^y} \cdot \dfrac{{dx}}{{dy}} + x{e^y} = \dfrac{d}{{dy}}\left( {{e^y} \cdot x} \right)$]
Step 3: Now integrating both sides of equation (3) with respect to $y$, we get
${e^y} \cdot x = \int {{e^{2y}}} + C$, where $C$ is the constant of integration.
$ \Rightarrow {e^y} \cdot x = \dfrac{{{e^{2y}}}}{2} + C$ …. (3)
We have, $y\left( 0 \right) = 0$. Therefore, put $x = 0$ and $y = 0$ in equation (3) to find out the value of $C$.
$\therefore {e^0} \cdot 0 = \dfrac{{{e^{2\left( 0 \right)}}}}{2} + C$
$ \Rightarrow 0 = \dfrac{1}{2} + C$
$ \Rightarrow C = \dfrac{{ - 1}}{2}$
Put the value of $C$ in equation (3),
$\therefore {e^y} \cdot x = \dfrac{{{e^{2y}}}}{2} - \dfrac{1}{2}$
$ \Rightarrow {e^y} \cdot x = \dfrac{{{e^{2y}} - 1}}{2}$
$ \Rightarrow 2{e^y} \cdot x = {e^{2y}} - 1$
$ \Rightarrow {e^{2y}} - 2x{e^y} - 1 = 0$
Step 4: Now find the value of $y$ by substituting ${e^y} = t$,
$\therefore {t^2} - 2xt - 1 = 0$
Now find the value of $t$ by using the quadratic formula, i.e.,
$t = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$\therefore t = \dfrac{{ - \left( { - 2x} \right) \pm \sqrt {{{\left( { - 2x} \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}}$
$ \Rightarrow t = \dfrac{{2x \pm \sqrt {4{x^2} + 4} }}{2}$
\[ \Rightarrow t = \dfrac{{2x \pm \sqrt {4\left( {{x^2} + 1} \right)} }}{2}\]
\[ \Rightarrow t = \dfrac{{2x \pm 2\sqrt {\left( {{x^2} + 1} \right)} }}{2}\]
\[ \Rightarrow t = \dfrac{{2\left[ {x \pm \sqrt {\left( {{x^2} + 1} \right)} } \right]}}{2}\]
\[ \Rightarrow t = x \pm \sqrt {\left( {{x^2} + 1} \right)} \]
\[\therefore {e^y} = x \pm \sqrt {\left( {{x^2} + 1} \right)} \] $\left( {\because {e^y} = t} \right)$
\[ \Rightarrow y = {\log _e}\left[ {x \pm \sqrt {\left( {{x^2} + 1} \right)} } \right]\]
Log value cannot be so we have to take positive value
$\Rightarrow y = $${\log _e}\left( {x + \sqrt {1 + {x^2}} } \right)$
Hence, option (C) is the correct answer.

Note: Linear differential equation of first order is of two types: one is the form of $\dfrac{{dx}}{{dy}} + Px = Q$ and second is the form of $\dfrac{{dy}}{{dx}} + Py = Q$. It may be noted that integrating factors for the above two equations are different. For $\dfrac{{dx}}{{dy}} + Px = Q$; $I.F. = {e^{\int {Pdy} }}$ and for $\dfrac{{dy}}{{dx}} + Py = Q$; $I.F. = {e^{\int {Pdx} }}$.