Question

If we have a function as ${e^y} + xy = e,\;then\;{\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)_{x = 0}}\;is - $
$\begin{align}
  &A.\;\dfrac{1}{{{e^2}}} \\
  &B.\;{e^{ - 1}} \\
  &C.\;e \\
  &D.\;None\;of\;these \\
\end{align} $

Answer 1

Hint: This question requires the knowledge of differential equations. Here, we can see that we cannot write y in terms of x directly, so we will use partial differentiation. This means that we will differentiate each term separately with respect to the two variables x and y, and then take the terms with dx and dy common, to find the derivative of this equation. Some common rules of differentiation to be used are-
$\begin{align}
  &\frac{{d\left( {u\left( x \right).v\left( x \right)} \right)}}{{dx}} = u\left( x \right)v'\left( x \right) + v\left( x \right)u'\left( x \right) \\
  &\frac{{d\left( {{e^x}} \right)}}{{dx}} = {e^x} \\
  &\frac{{d\left( k \right)}}{{dx}} = 0 \\
\end{align} $

Complete step-by-step solution -
We have been given the equation ${e^y} + xy = e$. We can see that we cannot be isolated y on one side of the equation due to the presence of an exponential term in y. So, we will differentiate the equation to find the derivative using the product rule which is-
$\frac{{d\left( {u\left( x \right).v\left( x \right)} \right)}}{{dx}} = u\left( x \right)v'\left( x \right) + v\left( x \right)u'\left( x \right)$
${e^y}\dfrac{{dy}}{{dx}} + x\frac{{dy}}{{dx}} + y = 0$
Multiply both sides by dx,
${e^y}dy + xdy + ydx = 0$
The term on the RHS was a constant so it becomes 0. So, this differential equation can be simplified as-
$\begin{align}
  &{e^y}dy + xdy = - ydx \\
  &dy\left( {{e^y} + x} \right) = - ydx \\
  &\text{Cross - multiplying both sides we get -} \\
  &\dfrac{{dy}}{{dx}} = \dfrac{{ - y}}{{{e^y} + x}} \\
\end{align} $
We will now differentiate this equation with respect to x, to get the second differential. This can be done by applying the division rule which is-
$\dfrac{{d\left( {\dfrac{{u\left( x \right)}}{{v\left( x \right)}}} \right)}}{{dx}} = \dfrac{{v\left( x \right).u'\left( x \right) - u\left( x \right)v'\left( x \right)}}{{{{\left[ {v\left( x \right)} \right]}^2}}}$
So, this equation can be differentiated as-
$\begin{align}
  &\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{\left( {{e^y} + x} \right)\left( { - \dfrac{{dy}}{{dx}}} \right) - \left( { - y} \right)\left( {{e^y}\dfrac{{dy}}{{dx}} + 1} \right)}}{{{{\left( {{e^y} + x} \right)}^2}}} \\
  &\text{Substituting the value of}\; \dfrac{{dy}}{{dx}}\; \text{we get} - \\
  &\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{\left( {{e^y} + x} \right)\left( {\dfrac{y}{{{e^y} + 1}}} \right) - \left( { - y} \right)\left( { - \dfrac{{{e^y}y}}{{{e^y} + x}} + 1} \right)}}{{{{\left( {{e^y} + x} \right)}^2}}} \\
\end{align} $

We need to find the value of this differential equation at x = 0. In substituting x = 0 in the original equation we get that-
$\begin{align}
  &{e^y} + xy = e \\
  &At\;x = 0,\; \\
  &{e^y} = {e^1} \\
  &y = 1\; \\
\end{align} $
We will now substitute x = 0 and y = 1 in the second order differential as-
$\begin{align}
&{\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)_{x = 0}} = \dfrac{{\left( {{e^1} + 0} \right)\left( {\dfrac{1}{{{e^1} + 0}}} \right) - \left( { - 1} \right)\left( {\dfrac{{ - {e^1}}}{{{e^1} + 0}} + 1} \right)}}{{{{\left( {{e^1} + 0} \right)}^2}}} \\
&{\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)_{x = 0}} = \dfrac{{\left( {\dfrac{e}{e}} \right) + \left( { - \dfrac{e}{e} + 1} \right)}}{{{e^2}}} = \dfrac{1}{{{e^2}}} \\
\end{align} $
This is the required answer, the correct option is A.

Note: These types of questions must be solved with caution, as there are chances of doing a mistake in the calculation. There is a shorter method to solve this question, which is less calculative, and does not require us to find the value of the differential. This can be done as-
$\begin{align}
  &{e^y} + xy = e \\
 & \text{Differentiating w.r.t. x} \\
  &{e^y}{y_1} + x{y_1} + y = 0\;\text{where} \; {y_1}\; \text{is the first order differential} \\
  &At\;{\text{x}} = 0\;and\;{\text{y}} = 1, \\
  &e{y_1} + 1 = 0 \\
  &{y_1} = - \dfrac{1}{e}..\left( 1 \right) \\
\end{align} $
$\begin{align}
& \text{ Differentiating the first order differential equation w.r.t. x again} \\
  &{e^y}{y_2} + {y_1}^2{e^y} + {y_1} + x{y_2} + {y_1} = 0 \\
  &At\;{\text{x}} = 0,\;{\text{y}}\; = 1\; \text{and using equation}\;\left( 1 \right), \\
  &{e^1}{y_2} + {\left( { - \dfrac{1}{e}} \right)^2}{e^1} - \dfrac{1}{e} + 0 - \dfrac{1}{e} = 0 \\
  &e{y_2} + \dfrac{1}{e} - \dfrac{1}{e} - \dfrac{1}{e} = 0 \\
  &e{y_2} = \dfrac{1}{e} \\
  &{y_2} = \dfrac{1}{{{e^2}}} \\
\end{align} $