Partial Differential Equations MCQ Quiz in தமிழ் - Objective Question with Answer for Partial Differential Equations - இலவச PDF ஐப் பதிவிறக்கவும்

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 $\left(\frac{\partial z}{\partial x}.\frac{\partial z}{\partial y}-\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}\right)\left(z-x\frac{\partial z}{\partial x}-y\frac{\partial z}{\partial y}\right)=\frac{\partial z}{\partial x}.\frac{\partial z}{\partial y}$

Writing in general notation,

(pq – p – q)(z – px – qy) = pq

⇒ $z=px+qy+\frac{pq}{pq-p-q}$

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

$z=ax+by+\frac{ab}{ab-a-b}$

Concept:

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

$p=\frac{\partial z}{\partial x},q=\frac{\partial z}{\partial y}$

$r=\frac{{\partial }^{2}z}{\partial x^2},s=\frac{{\partial }^{2}z}{\partial x\partial y},t=\frac{{\partial }^{2}z}{\partial y^2}$

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, 

$A\frac{{\partial }^{2}u}{\partial x^2}+B\frac{{\partial }^{2}u}{\partial x\partial y}+C\frac{{\partial }^{2}u}{\partial y^2}+D\frac{\partial u}{\partial x}+E\frac{\partial u}{\partial y}=F$

For One-Dimensional equation,

${\alpha }^2\frac{{\partial }^{2}y}{\partial x^2}=\frac{{\partial }^{2}y}{\partial t^2}$

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

$\frac{\partial y}{\partial t}={\alpha }^2\frac{{\partial }^{2}y}{\partial x^2}$

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.

$\frac{{\partial }^{2}y}{\partial t^2}=-\frac{{\partial }^{2}y}{\partial x^2}$

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;

⇒ ${\left(\frac{dz}{du}\right)}^3+{\left(a\frac{dz}{du}\right)}^3=a\left(\frac{dz}{du}\right)z$

⇒ ${\left(\frac{dz}{du}\right)}^2\left(1+a^3\right)=az$

⇒ $\frac{dz}{du}=\sqrt{\frac{a}{1+a^3}}\sqrt{z}$

⇒ $\sqrt{\frac{1+a^3}{a}}\frac{dz}{\sqrt{z}}=du$

⇒ $\sqrt{\frac{1+a^3}{a}}2\sqrt{z}=u+b$

⇒ $\frac{4\left(1+a^3\right)}{a}z={\left(u+b\right)}^2$

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 $\frac{\partial z}{\partial x}=a=p;\frac{\partial z}{\partial y}=b=q;$

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:

$\mathrm{A}\frac{{\partial }^2\mathrm{z}}{\partial {\mathrm{x}}^2}+\mathrm{B}\frac{{\partial }^2\mathrm{z}}{\partial \mathrm{x}\partial \mathrm{y}}+\mathrm{C}\frac{{\partial }^2\mathrm{z}}{\partial {\mathrm{y}}^2}+\mathrm{D}\frac{\partial \mathrm{z}}{\partial \mathrm{x}}+\mathrm{E}\frac{\partial \mathrm{z}}{\partial \mathrm{y}}+\mathrm{F}\mathrm{z}=\mathrm{G}$

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

If B2 - 4AC < 0, the equation is elliptic.

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

Calculation:

Given:

$4\frac{{\partial }^2\varphi }{\partial x^2}+8\frac{{\partial }^2\varphi }{\partial x\partial y}+C\frac{{\partial }^2\varphi }{\partial y^2}+4\varphi =0$

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:

${\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}=9{\displaystyle \frac{{\partial }^{2}u}{\partial y^2}}$     ...(1)

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

$\frac{\partial u}{\partial x}=a\cos \left(ax-y\right)$

$\frac{{\partial }^{2}u}{\partial x^2}=-a^2\sin \left(ax-y\right)$     ...(2)

$\frac{\partial u}{\partial y}=-\cos \left(ax-y\right)$

$\frac{{\partial }^{2}u}{\partial y^2}=-\sin \left(ax-y\right)$     ...(3)

$\frac{{\partial }^{2}u}{\partial x^2}=9\frac{{\partial }^{2}u}{\partial y^2}$

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

${\displaystyle \frac{{\partial }^{2}u}{\partial x^2}}=9{\displaystyle \frac{{\partial }^{2}u}{\partial y^2}}$

$-a^2\sin (ax-y)=9\times -\sin \left(ax-y\right)$

∴ 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

$\frac{\partial \left(y_1,y_2,y_3\right)}{\partial \left(x_1,x_2,x_3\right)}=\left|\begin{array}{ccc}\frac{\partial y_1}{\partial x_1}& \frac{\partial y_1}{\partial x_2}& \frac{\partial y_1}{\partial x_3}\\ \frac{\partial y_2}{\partial x_1}& \frac{\partial y_2}{\partial x_2}& \frac{\partial y_2}{\partial x_3}\\ \frac{\partial y_3}{\partial x_1}& \frac{\partial y_3}{\partial x_2}& \frac{\partial y_3}{\partial x_3}\end{array}\right|$

$=\left|\begin{array}{ccc}\frac{-x_2\cdot x_3}{x_1^2}& \frac{x_3}{x_1}& \frac{x_2}{x_1}\\ \frac{x_3}{x_2}& \frac{-x_3\cdot x_1}{x_2^2}& \frac{x_1}{x_1}\\ \frac{x_2}{x_3}& \frac{x_1}{x_3}& \frac{-x_1\cdot x_2}{x_3^2}\end{array}\right|$

$=\frac{1}{x_1^2\cdot x_2^2\cdot x_3^2}\left|\begin{array}{ccc}-x_2{x}_3& x_1{x}_3& x_1{x}_2\\ x_2{x}_3& -x_1{x}_3& x_1{x}_2\\ x_2{x}_3& x_1{x}_3& -x_1{x}_2\end{array}\right|$

$=\frac{x_1^2{x}_2^2{x}_3^2}{x_1^2{x}_2^2{x}_3^2}\left|\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right|$

= -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;

⇒ $a^2{\left(\frac{dz}{du}\right)}^2=4z^2{\left(\frac{dz}{du}\right)}^2\left(1-{\left(\frac{dz}{du}\right)}^2\right)$ 

$\Rightarrow 1-\frac{a^2}{4z^2}={\left(\frac{dz}{du}\right)}^2$

$\Rightarrow \frac{2z}{\sqrt{4z^2-a^2}}dz=du$

$\Rightarrow \frac{1}{2}\sqrt{4z^2-a^2}=u+c$

⇒ 4z² = a² + (2x + 2ay + b)² ;

Concept:

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

$A\frac{{\partial }^{2}U}{\partial X^2}+B\frac{{\partial }^{2}U}{\partial X\partial Y}+C\frac{{\partial }^{2}U}{\partial Y^2}+f\left(X,Y,Z\frac{\partial U}{\partial X},\frac{\partial U}{\partial Y}\right)=0$

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

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

Calculation:

Given equation,

$\frac{{\partial }^{2}P}{\partial X^2}+\frac{{\partial }^{2}P}{\partial Y^2}+3\frac{{\partial }^{2}P}{\partial X\partial Y}+2\frac{\partial P}{\partial X}-\frac{\partial P}{\partial Y}=0$

Compare the given equation with the following general equation

$A\frac{{\partial }^{2}U}{\partial X^2}+B\frac{{\partial }^{2}U}{\partial X\partial Y}+C\frac{{\partial }^{2}U}{\partial Y^2}+f\left(X,Y,Z\frac{\partial U}{\partial X},\frac{\partial U}{\partial Y}\right)=0$

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