The following partial differential equation is defined for u:u (x, y) \(\frac{{\partial u}}{{\partial y}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}};y \ge 0; \le {x_1} \le x \le {x_2}\)The set auxiliary conditions necessary to solve the equation uniquely, is

Calculation:GIVEN:\(\frac{{\PARTIAL V}}{{\partial y}} = \frac{{{\partial ^2}v}}{{\partial {x^2}}};\) y ≥ 0; x1 ≤ x ≤ x2∵ y ≥ 0 ⇒ It can be REPLACED with ‘t’.\(\therefore \frac{{\partial v}}{{\partial t}} = \frac{{{\partial ^2}v}}{{\partial {x^2}}}\)This is a 1-D Heat equation. It measures temperature distribution in a uniform rod.The general solution is u = f(x, t)u = (c1 cos px + c2 sin px) \(\left( {{c_3}{e^{ - {c^2}{p^2}t}}} \right)\)Auxiliary solutions INCLUDE both initial and boundary conditions.1) Number of initial conditions = Highest order of time derivative in partial differential = 12) The number of boundary conditions:\(\frac{{\partial v}}{{\partial t}} = \frac{{{\partial ^2}v}}{{\partial {x^2}}}\) ; To solve this partial differential equation, it needs to be integrated twice that will introduce two arbitrary constants.Hence 2 boundary conditions and 1 initial condition are required to solve this Partial differential equation.