Test: Diffusion Equation - Engineering Mathematics MCQ

m = 0.5, 0.5
The solution is:
y = (C₁ + C₂x)e^0.5x
For y = 0 at x = 0
0 = (C₁ + C₂(0))e^0.5(0)
0 = C₁

C₂ = 1
y = (0 + (1)x)e^0.5x
y = xe^0.5x

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:

Calculation:
Given:

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:

Given:

∵ y ≥ 0 ⇒ It can be replaced with ‘t’.

This is a 1-D Heat equation. It measures temperature distribution in a uniform rod.
The general solution is u = f(x, t)

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:
 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.

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^∫Pdx
Calculation:
Given:

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:

Let √y = u

Concept:

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:

Calculation:

Given:

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:


-2xdx = dt