Question

Find the singular solution of the differential equation $ y = px + p - {p^2} $ where, $ p = \dfrac{{dy}}{{dx}} $.

Answer 1

Hint: In this question, the singular solution is to be determined for the differential equation $ y = px + p - {p^2} $ for which we need to define an expression that shows the relation between $ x $ and $ y $ only with no other parameters included. To get the relation, we will use the general properties of differentiation along with algebraic identities.


Complete step by step solution:

Differentiate the equation $ y = px + p - {p^2} $ with respect to $ x $ such that $ p = \dfrac{{dy}}{{dx}} $:
\[
  \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {px + p - {p^2}} \right) \\
  p = p + x\dfrac{{dp}}{{dx}} + \dfrac{{dp}}{{dx}} - 2p\dfrac{{dp}}{{dx}} \\
  \dfrac{{dp}}{{dx}}\left( {x + 1 - 2p} \right) = 0 \\
  \dfrac{{dp}}{{dx}} = 0;\left( {x + 1 - 2p} \right) = 0 \\
  \dfrac{{dp}}{{dx}} = 0;p = \dfrac{{x + 1}}{2} \\
 \]

Now, substitute $ p = \dfrac{{x + 1}}{2} $ in the equation $ y = px + p - {p^2} $ to determine the singular solution as:

$
  y = px + p - {p^2} \\
   = \dfrac{{x(x + 1)}}{2} + \dfrac{{(x + 1)}}{2} - {\left( {\dfrac{{(x + 1)}}{2}} \right)^2} \\
   = \dfrac{{x(x + 1)}}{2} + \dfrac{{(x + 1)}}{2} - \dfrac{{{x^2} + 2x + 1}}{4} \\
   = \dfrac{{2{x^2} + 2x + 2x + 2 - ({x^2} + 2x + 1)}}{4} \\
   = \dfrac{{{x^2} + 2x + 1}}{4} \\
   = {\left( {\dfrac{{x + 1}}{2}} \right)^2} \\
 $

Hence, the singular solution of the differential equation $ y = px + p - {p^2} $ where $ p = \dfrac{{dy}}{{dx}} $ is $ y = {\left( {\dfrac{{x + 1}}{2}} \right)^2} $.

Note: This type of question can also be solved by determining an auxiliary equation with a parameter and determining the value of that parameter and at last substituting back the real parameters of the given differential equation. However, this method is quite long and includes more parameters than actual.