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, isOption 4 : one initial conditions and two boundary conditions

**Calculation:**

**Given:**

\(\frac{{\partial v}}{{\partial y}} = \frac{{{\partial ^2}v}}{{\partial {x^2}}};\) y ≥ 0; x_{1} ≤ x ≤ x_{2}

∵ 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 = (c_{1} cos px + c_{2} 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 = 1

2) 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.

Consider the differential equation given below:

\(\dfrac{dy}{dx} + \dfrac{x}{1-x^2} y= x\sqrt{y}\)

The integrating factor of the differential equation is:

Option 1 : (1 - x^{2})^{-1/4}

__Concept__**:**

\(\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)\)

The solution of a linear differential equation of a general form shown above is:

y(I.F) = ∫Q(x) (IF) dx + C

Where:

IF = Integrating factor calculated as:

\(I.F = {e^{\smallint Pdx}}\)

__Calculation__**:**

Given:

\(\frac{{dy}}{{dx}} + \frac{x}{{1 - {x^2}}}y = x\sqrt y \)

The above differential equation is not in a general form. Converting it first in the general form of a linear differential equation, we divide the equation by √y to get:

\(\frac{1}{{\sqrt y }}\frac{{dy}}{{dx}} + \frac{x}{{1 - {x^2}}}\;\sqrt y = x\)

Let √y = u

\(\left[ {\frac{1}{{2\sqrt y }}\;\;\frac{{dy}}{{dx}}} \right] = \frac{{du}}{{dx}}\)

\(\frac{1}{{\sqrt y }}\;\frac{{dy}}{{dx}} = 2\frac{{du}}{{dx}}\)

\(2\frac{{du}}{{dx}} + \frac{x}{{1 - {x^2}}}u = x\)

\(\frac{{du}}{{dx}} + \frac{1}{2}\left( {\frac{x}{{1 - {x^2}}}} \right)u = \frac{x}{2}\)

\(\frac{{du}}{{dx}} + P\left( x \right)u = Q\left( x \right)\)

\(IF = {e^{\smallint Pdx}}\)

\(IF = {e^{\smallint \frac{x}{{2\left( {1 - {x^2}} \right)}}dx}}\)

Let 1 – x2 = t

-2xdx = dt

\(xdt = \frac{{ - dt}}{2}\)

\(IF = {e^{\smallint - \frac{1}{t}\;\frac{{dt}}{4}}}\)

\( = {e^{\frac{{ - 1}}{4}\ln t}}\)

\( = {e^{\ln {t^{\frac{{ - 1}}{4}}}}}\)

\( = {t^{\frac{{ - 1}}{4}}}\)

\(IF = {\left( {1 - {x^2}} \right)^{\frac{{ - 1}}{4}}}\)

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, isOption 4 : one initial conditions and two boundary conditions

**Calculation:**

**Given:**

\(\frac{{\partial v}}{{\partial y}} = \frac{{{\partial ^2}v}}{{\partial {x^2}}};\) y ≥ 0; x_{1} ≤ x ≤ x_{2}

∵ 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 = (c_{1} cos px + c_{2} 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 = 1

2) 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.

Consider the differential equation given below:

\(\dfrac{dy}{dx} + \dfrac{x}{1-x^2} y= x\sqrt{y}\)

The integrating factor of the differential equation is:

Option 1 : (1 - x^{2})^{-1/4}

__Concept__**:**

\(\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)\)

The solution of a linear differential equation of a general form shown above is:

y(I.F) = ∫Q(x) (IF) dx + C

Where:

IF = Integrating factor calculated as:

\(I.F = {e^{\smallint Pdx}}\)

__Calculation__**:**

Given:

\(\frac{{dy}}{{dx}} + \frac{x}{{1 - {x^2}}}y = x\sqrt y \)

The above differential equation is not in a general form. Converting it first in the general form of a linear differential equation, we divide the equation by √y to get:

\(\frac{1}{{\sqrt y }}\frac{{dy}}{{dx}} + \frac{x}{{1 - {x^2}}}\;\sqrt y = x\)

Let √y = u

\(\left[ {\frac{1}{{2\sqrt y }}\;\;\frac{{dy}}{{dx}}} \right] = \frac{{du}}{{dx}}\)

\(\frac{1}{{\sqrt y }}\;\frac{{dy}}{{dx}} = 2\frac{{du}}{{dx}}\)

\(2\frac{{du}}{{dx}} + \frac{x}{{1 - {x^2}}}u = x\)

\(\frac{{du}}{{dx}} + \frac{1}{2}\left( {\frac{x}{{1 - {x^2}}}} \right)u = \frac{x}{2}\)

\(\frac{{du}}{{dx}} + P\left( x \right)u = Q\left( x \right)\)

\(IF = {e^{\smallint Pdx}}\)

\(IF = {e^{\smallint \frac{x}{{2\left( {1 - {x^2}} \right)}}dx}}\)

Let 1 – x2 = t

-2xdx = dt

\(xdt = \frac{{ - dt}}{2}\)

\(IF = {e^{\smallint - \frac{1}{t}\;\frac{{dt}}{4}}}\)

\( = {e^{\frac{{ - 1}}{4}\ln t}}\)

\( = {e^{\ln {t^{\frac{{ - 1}}{4}}}}}\)

\( = {t^{\frac{{ - 1}}{4}}}\)

\(IF = {\left( {1 - {x^2}} \right)^{\frac{{ - 1}}{4}}}\)

Option 3 : \(\frac{{{\rm{\delta c}}}}{{{\rm{\delta t}}}} = {{\rm{D}}_{\rm{x}}}\frac{{{{\rm{\delta }}^2}{\rm{c}}}}{{{\rm{\delta }}{{\rm{x}}^2}}}\)

The one dimensional diffusion equation can be written as

\(\frac{{{\rm{\delta c}}}}{{{\rm{\delta t}}}} = {{\rm{D}}_{\rm{x}}}\frac{{{{\rm{\delta }}^2}{\rm{c}}}}{{{\rm{\delta }}{{\rm{x}}^2}}}\)