Partial Differential Equations MCQ Quiz in मराठी - Objective Question with Answer for Partial Differential Equations - मोफत PDF डाउनलोड करा

Concept:

An equation in the form of z = px + qy + f(p, q) is analogous to Clairaut’s equation.

Its complete solution will be z = ax + by + f(a, b) where a and b are arbitrary constants.

Calculation:

Given Equation is 

Writing in general notation,

(pq – p – q)(z – px – qy) = pq

⇒ 

Now it is in the form of Clairaut’s equation, hence the general solution will be

Concept:

In standard notation, we consider x, y as independent variables and z as the dependent variables.

The order of a PDE is the order of the highest derivative occurring in it and the degree of a PDE is the degree of the highest derivative appearing in it.

Calculation:

Given PDE is t² + r² – xp³ – yq³ = 0

The highest derivative is t or r ⇒ order = 2

The power of the highest derivative is 2 (power of t or r) ⇒ degree = 2

Concept:

Wave equation:

It is a second-order linear partial differential equation for the description of waves (like mechanical waves).

The Partial Differential equation is given as, 

For One-Dimensional equation,

where, A = α2, B = 0, C = -1

Put all the values in equation (1)

∴ 0 - 4 (α2)(-1)

4α2 > 0.

So, this is a one-dimensional wave equation.

Additional Information

having A = α2, B = 0, C = 0

Put all the values in equation (1), we get 

0 - 4(α2)(0) = 0, therefore it shows parabolic function.

So, this is a one-dimensional heat equation.

having A = 1, B = 0, C = 1

Put all the values in equation (1), we get 

0 - 4(1)(1) = -4, therefore it shows elliptical function.

So, this is a two-dimensional heat equation.

Concept:

f(z, p, q) = 0 form

Calculation:

Given PDE is p³ + q³ = qz;

Let’s substitute p = dz/du and q = a dz/du;

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

Substituting u = x + ay,

⇒ 4z (1 + a³) = a (x + ay + b)²;

Concept:

A partial differentiation equation of the form z = px + qy + f(p,q) is known as Clairaut's Equation.

For such equations, the solution is given by 

z = ax + by + f(a,b) where a, b are arbitrary constants.

since 

Calculation:

Given PDE is yz = pxy + qy² + pqy

Dividing with y on both sides, we get

⇒ z = px + qy + pq

Which is in the form z = px + qy + f(p,q)

Then the general solution will be

z = ax + by + ab where a, b are arbitrary constants

Concept:

The equation below represents the general form of second-order partial differential equation in to variables:

If B2 - 4AC > 0, the equation is hyperbolic.

If B2 - 4AC

If B2 - 4AC = 0, the equation is parabola.

Calculation:

Given:

\(4\frac{\partial^2\phi}{\partial x^2}\;+\;8\frac{\partial^2\phi}{\partial x\partial y}\;+\;C\frac{\partial^2\phi}{\partial y^2}\;+\;4\phi=0\)

Comparing with standard form A = 4, B = 8, C = ?, and F = 4.

For the equation to be parabolic,

B2 - 4AC = 0

∴ (8)2 - 4 × 4 × C = 0

∴ 64 - 16C = 0

∴ 64 = 16C

∴ C = 4 is the required condition for the equation to be parabolic.

Explanation:

     ...(1)

Since u = sin (ax – y) is the given solution, so it should satisfy equation (1).

     ...(2)

     ...(3)

Putting (2) and (3) in the equation (1)

∴ a² = 9

a = ± 3

As per the given condition, here magnitude is asked so 3 will be the correct answer.

Explanation:

Jacobian of (y₁, y₂, y₃) w.r.t. (x₁, x₂, x­₃) is equal to

= -1(1 - 1) - 1(-1 - 1) + 1(1 + 1)

= 0 + 2 + 2

= 4

Concept:

Equation containing p, q, z only.

Let it be f(z, p, q) = 0;

(i) Assume u = x + ay and substitute p = dz/du and q = a dz/du

(ii) Solve the resulting ODE in z and u

(iii) Replace u by x + ay

Calculation:

Given q² = 4z²p²(1 – p²);

Assume u = x + ay and substitute p = dz/du and q = a dz/du;

⇒  

⇒ 4z² = a² + (2x + 2ay + b)² ;

Concept:

The general form of 2^nd order linear partial differential equation is given by

The above equation is said to be parabolic, elliptic, and hyperbolic based on the following,

1. Parabolic     = B² – 4AC = 0
2. Elliptic          = B² – 4AC
3. Hyperbolic   = B² – 4AC > 0

Calculation:

Given equation,

Compare the given equation with the following general equation

Hence we get,

A = 1, B = 3, C = 1

Put value of A,B, and C in the following equation

B² – 4AC

∴ B² – 4AC = 3² – (4× 1 × 1) = 5

We can see B² – 4AC is greater than Zero, i.e

B² – 4AC > 0

Hence the given partial differential equation is Hyperbolic