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

Explanation:

Comparing $\mathrm{y}\frac{\partial \mathrm{z}}{\partial \mathrm{x}}-\mathrm{x}\frac{\partial \mathrm{z}}{\partial \mathrm{y}}=0$ with Pp+Qq=R we get P = y and Q=-x

Given x0(s) = cos(s), y0(s) = sin(s), z0(s) = 1

So P(x0,y0,z0) = sin(s), Q(x0,y0,z0) = - cos(s), $\frac{d{x}_0}{ds}$ = - sin(s), $\frac{d{y}_0}{ds}$ = cos(s)

Now $\frac{P(x_0,y_0,z_0)}{d{x}_0/ds}$ = -1 =$\frac{Q(x_0,y_0,z_0)}{d{y}_0/ds}$

So it may have no solution or infinitely many solutions

Using Lagrange characteristic equation we get

$\frac{dx}{y}=\frac{dx}{-x}=\frac{dz}{0}$

taking first two we get

x²+y² = a

and from 3rd one we get

z = b

Using given initial solutions we get

sin2s + cos2s = 1 ⇒  a = 1 and 1 = b

So we get 

x2 + y2 = 1, z = 1 ⇒ z = x2 + y2 is a solution

But we can see that z = (x2 + y2)ⁿ is also a solution for all n ∈ N as it satisfies the partial differential equation as well as the initial solution.

Hence given problem has infinitely many solutions.

Option (4) is correct

Explanation:

∵ Z_z = p, Z_y = q

So z = $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}+\frac{\partial z}{\partial x}\cdot \frac{\partial z}{\partial y}$

⇒ z = px + qy + pq ....(i) which is in Clairaut's equation

So it's complete integral is

z = ax + by + ab .... (ii)

Option (1) is correct & option (2) is incorrect.

Now for (3) & (4): Given x = 0 & z = y²

Let y = t ⇒ z = t²

∴ by (ii),  z = ax + by + ab ⇒ t² = a(0) + b(t) + ab

⇒ t² - bt - ab = 0 ....(iii)

Now we have to find value of t,

so differentiating (iii) with respect to t, we get

2t - b = 0  ⇒ t = b/2

Now put t = b/2 in equation (iii) we have

${\left(\frac{b}{2}\right)}^2-b.\frac{b}{2}-$ab = 0

$\frac{b^2}{4}+ab=0$

 ⇒ a = - b/4

Now put value of a in equation (ii)

$z=(-\frac{b}{4})x+by-\frac{b^2}{4}$ ....(iv)

Now to eliminate b, differentiating (iv) w.r.t. b & find value in terms of x & y.

$\Rightarrow 0=-\frac{x}{4}+y-b/2$

⇒ b = 2y - x/2

Now putting this b in equation (iv) and simplifying we get

z = - $\frac{1}{4}$(2y - x/2)x + (2y - x/2)y - $\frac{1}{4}$(2y - x/2)²

z = (2y - x/2)(y - x/4) -$\frac{1}{4}$(2y - x/2)2 

z = 2(y - x/4)² - $\frac{1}{2}$(y - x/4)2

z = $\frac{3}{2}$(y - x/4)2 

z = $\frac{3}{2}$ ${(\frac{x}{4}-y)}^2$

The particular solution passing through x = 0 and z = y2 is ${(\frac{x}{4}-y)}^2$

Option (3) is correct and option (4) is incorrect

Explanation:

Here, (i) $\frac{{\partial }^{2}u}{\partial x^2}+2\frac{{\partial }^{2}u}{\partial x\partial y}+(1-\mathrm{sgn}(y))\frac{{\partial }^{2}u}{\partial y^2}=0$

So A = 1, B = 2, C = 1 - sgn(y)

$\because \mathrm{sgn}(y)=\{\begin{array}{c}1:y>0\\ 0;y=0\\ -1:y<0\end{array}\Rightarrow$ $1-\mathrm{sgn}(y)=\{\begin{array}{ll}0& :y>0\\ 1& :y=0\\ 2& :y<0\end{array}$

Hence 

B2 - 4AC = $4-\{\begin{array}{ll}0,& y>0\\ 4,& y<0\\ 8,& y<0\end{array}$

 = $\{\begin{array}{ll}4,& \text{ if }y>0\Rightarrow \text{ Hyperbolic }\\ 0,& \text{ if }y=0\Rightarrow \text{ parabolic }\\ -4,& \text{ if }y<0\Rightarrow \text{ Elliptic }\end{array}$

Option (1) is incorrect, option (2) is correct.

For (ii) $y\frac{{\partial }^{2}u}{\partial x^2}+x\frac{{\partial }^{2}u}{\partial y^2}=0$ 

A = y, B = 0, C = x

∴  B² - 4AC = 0 - 4yx = -4xy

B² - 4AC will lie in different quadrant for different x and y

In quadrant I and III, x and y are of same sign ⇒ -4xy < 0 so PDE is elliptic in I and III

While in quadrant II and IV, x and y are of opposite sign ⇒ -4xy > 0

⇒ B² - 4AC > 0 so PDE is hyperbolic in II and IV

Option (3) is correct option (4) is incorrect

Explanation:

 General form of a 2nd order Homogenous Linear PDE is 

Auxx + Buxy + Cuyy + Dux + Euy + Fu + G = 0 

So, x2uxy + 3y2u = 0 is also Linear ( A = 0, B = x2, C =0, D = 0, E = 0, F = 3y2, G = 0 )

Option 1 is not correct:

Let u(x, y) = X(x)Y(y) be a solution of the given p.d.e 

  x2uxy + 3y2u = 0, with the initial condition u(x,0) = e1/x

Then, x² $X^{\prime }{Y}^{\prime }$ + 3y2u = 0

⇒ x² $X^{\prime }{Y}^{\prime }$  = - 3y2u

⇒ x² $\frac{X^{\prime }}{X}=-3y^2\frac{Y}{Y^{\prime }}=k$  

⇒ $\frac{X^{\prime }}{X}=\frac{k}{x^2}$ and $-3y^2\frac{Y}{Y^{\prime }}=k$

⇒ $\ln X=-\frac{k}{x}+C_1$ and $-\frac{3y^2}{k}=\frac{Y^{\prime }}{Y}$

⇒ $X=C_1{e}^{-\frac{k}{x}}$ and $Y=C_2{e}^{-\frac{y^3}{k}}$

⇒ $u(x,y)=C_1{e}^{\frac{-k}{x}}\cdot C_2{e}^{\frac{-y^3}{k}}=C{e}^{-\frac{k}{x}}\cdot e^{-\frac{y^3}{k}}$

where $C=C_1{C}_2$ 

Now, using u(x,0) = e1/x 

C = 1, k = -1 

$\therefore u(x,y)=e^{\frac{1}{x}}\cdot e^{y^3}$

Hence $u(1,1)=e^1\cdot e^{1^3}=e^2$

Option 2 is correct and option 3 is incorrect.

The method of separation is variables is utilized.

Option 4 is correct: 

The correct answers are Options 2 and 4.

Concept:

The non-linear PDE of the form

f(x, y, z, p, q) = 0 satisfy Charpit equation

$\frac{dx}{f_p}=\frac{dy}{f_q}=\frac{dz}{p{f}_p+q{f}_q}$$=\frac{dp}{-(f_x+p{f}_z)}=\frac{dq}{-(f_y+q{f}_z)}$

Explanation:

Here f(x, y, z, p, q) = (p2 + q2)y - qz 

Using Charpit formula

$\frac{dx}{2py}=\frac{dy}{2qy-z}$$=\frac{dz}{2p^{2}y+2q^{2}y-qz}$$=\frac{dp}{0+pq}=\frac{dq}{-p^2-q^2+q^2}$

⇒ $\frac{dx}{2py}=\frac{dy}{2qy-z}$$=\frac{dz}{2p^{2}y+2q^{2}y-qz}$$=\frac{dp}{pq}=\frac{dq}{-p^2}$

Taking

$\frac{dp}{pq}=\frac{dq}{-p^2}$

⇒ pdp + qdq = 0

Integrating

p² + q² = a²....(i)

Putting in the given equation

a²y = qz

⇒ q = $\frac{a^{2}y}{z}$

Putting in (i)

p = $\sqrt{a^2-q^2}$ = $\frac{a}{z}\sqrt{z^2-a^2{y}^2}$

Putting p and q in

dz = pdx + qdy

⇒ dz = $\frac{a}{z}\sqrt{z^2-a^2{y}^2}$dx + $\frac{a^{2}y}{z}$dy

⇒ $\frac{zdz-a^{2}ydy}{\sqrt{z^2-a^2{y}^2}}$ = a dx

Integrating

$\sqrt{z^2-a^2{y}^2}$ = ax + b

Squaring both sides

z² = a²y² + (ax+b)²

Both (1) and (2) are correct.

Hence option (3) is true.

Concept:

Let Pp + Qq = R be a PDE where P, Q, R are functions of x, y, z then by Lagrange's method

$\frac{dx}{P}$ = $\frac{dy}{Q}$ = $\frac{du}{R}$

Explanation:

Given 

ux - uuy = 0, x, y ∈ ℝ,

u(0, y) = 2y, y ∈ ℝ.

Using Lagrange's method

$\frac{dx}{P}$ = $\frac{dy}{Q}$ = $\frac{du}{R}$

⇒ $\frac{dx}{1}$ = $\frac{dy}{-u}$ = $\frac{du}{0}$

Taking last two ratio

u = 0

⇒ u = c1...(i)

and putting u = c1 we get from first two terms

$\frac{dx}{1}$ = $\frac{dy}{-c_1}$  

⇒ dy + c1dx = 0

Integrating we get

⇒ y + c1x = c2

⇒ y + ux = c2...(ii)

 From (i) and (ii) we get the general solution as

u = ϕ(y + ux)

Using u(0, y) = 2y we get

2y = ϕ(y) ⇒ ϕ(y + ux) = 2(y + ux)

Hence solution is

u = 2(y + ux) 

⇒  u(1 - 2x) = 2y ⇒ u = $\frac{2y}{1-2x}$

Therefore u(1, 2) = $\frac{4}{1-2}$ = - 4

Option (2) is correct

Explanation:

Given 

$u_t=u_{xx},x\in \mathbb{R},t>0$ with the initial condition

$u(x,0)=\sin (4x)+x+1,x\in \mathbb{R}$ 

which is a heat equation for infinite domain.

So solution is

u(x, t) = $\frac{1}{\sqrt{4\pi ct}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{(x-y{)}^2}{4c^{2}t}}f(y)dy$

Here f(x) = sin 4x + x + 1 and c = 1 then

u(x, t) = $\frac{1}{\sqrt{4\pi t}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{(x-y{)}^2}{4t}}(\sin 4y+y+1)dy$....(i)

(1): $\mathrm{u}(\frac{\pi }{8},1)$ = $\frac{1}{\sqrt{4\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{(\frac{\pi }{8}-y{)}^2}{4}}(\sin 4y+y+1)dy$

Let $\frac{\pi }{8}-y=u\Rightarrow dy=-du$ so

$\mathrm{u}(\frac{\pi }{8},1)$ = $\frac{1}{\sqrt{4\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}(\sin 4(\frac{\pi }{8}-u)+(\frac{\pi }{8}-u)+1)du$

$\mathrm{u}(\frac{\pi }{8},1)$ = $\frac{1}{\sqrt{4\pi }}[{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}\cos 4udu+{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}(\frac{\pi }{8}-u)+{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}du]$......(ii)

and $\mathrm{u}(-\frac{\pi }{8},1)$ = $\frac{1}{\sqrt{4\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{(-\frac{\pi }{8}-y{)}^2}{4}}(\sin 4y+y+1)dy$

Let $\frac{\pi }{8}+y=u\Rightarrow dy=du$ so

$\mathrm{u}(-\frac{\pi }{8},1)$ = $\frac{1}{\sqrt{4\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}(\sin 4(-\frac{\pi }{8}+u)+(-\frac{\pi }{8}+u)+1)du$

$\mathrm{u}(-\frac{\pi }{8},1)$ = $\frac{1}{\sqrt{4\pi }}[-{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}\cos 4udu-{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}(\frac{\pi }{8}-u)+{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}du]$.....(iii)

adding (ii) and (iii) we get

$\mathrm{u}(\frac{\pi }{8},1)+\mathrm{u}(-\frac{\pi }{8},1)$ = $\frac{2}{\sqrt{4\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}du$

                                 = $\frac{1}{\sqrt{\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\frac{u^2}{4}}du$

                               = $\frac{1}{\sqrt{\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-p^2}2dp$ (let u = 2p then du = 2dp)

                              = $\frac{1}{\sqrt{\pi }}.2\sqrt{\pi }(\because {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-p^2}dp=\sqrt{\pi })$

                             = 2

(1) is true.

From (ii) and (iii) we can see that (2), (3), (4) are false.

Concept:

Let quasi-linear partial differential equation of order one is of the form Pp + Qq = R, where P, Q and R are functions of x, y, z (called Lagrange equation). then Lagrange’s auxiliary equations to solve this PDE is

$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}$

Explanation:

Given PDE is

$x^2\frac{\partial z}{\partial x}+y^2\frac{\partial z}{\partial y}=(x+y)z$

Lagrange’s auxiliary equations are

$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dz}{(x+y)z}$

Taking 1st two terms

$\frac{dx}{x^2}=\frac{dy}{y^2}$

⇒ $\frac{dy}{y^2}-\frac{dx}{x^2}$ = 0

Integrating both sides

$-\frac{1}{y}+\frac{1}{x}$ = c₁....(i)

Also, $\frac{\frac{dx}{x}+\frac{dy}{y}-\frac{dz}{z}}{x+y-x-y}=\frac{dx}{x^2}$

⇒ $\frac{dx}{x}+\frac{dy}{y}-\frac{dz}{z}$ = 0

Integrating both sides

ln x + ln y - ln z = ln c₂

⇒ $\frac{xy}{z}$ = c₂ or, $\frac{z}{xy}$ = c2.....(ii)

From (i) and (ii) general solution is

$\frac{z}{xy}$ = $f(\frac{1}{x}-\frac{1}{y})$

z = xy$f(\frac{1}{x}-\frac{1}{y})$ for an arbitrary differentiable function f

Option (4) is true and (3) is false

Also if we take

$\frac{d(x-y)}{x^2-y^2}=\frac{dz}{(x+y)z}$

⇒ $\frac{d(x-y)}{x-y}=\frac{dz}{z}$

Integrating both sides

ln(x - y) - ln z = lnc₃

⇒ $\frac{x-y}{z}$ = c₃....(iii)

Taking (ii) and (iii) we get

$F(\frac{xy}{z},\frac{x-y}{z})$ = 0, for an arbitrary differentiable function F

Option (1) is true.

Taking (i) and (iii) we get the general solution as 

$F(\frac{x-y}{z},\frac{1}{x}-\frac{1}{y})$ = 0, for an arbitrary differentiable function F

Option (2) is true.

Concept:

The PDE of the Pp + Qq = R is solved by using Lagrange's auxiliary equation

$\frac{dx}{P}=\frac{dy}{Q}=\frac{du}{R}$

Explanation:

Given PDE is 

$\begin{array}{c}2u_x+3u_y=5\\ u=1\text{ on the line }3x-2y=0\end{array}\}$

Using Lagrange's auxiliary equation

$\frac{dx}{2}=\frac{dy}{3}=\frac{du}{5}$

Taking $\frac{dx}{2}=\frac{dy}{3}$

⇒ 3dx - 2dy = 0

Integrating both sides

3x - 2y = c1...(i)

Using 

$\frac{dx}{2}=\frac{du}{5}$

⇒ 2du - 5dx = 0

Integrating both sides

⇒ 2u - 5x = c2 ....(ii)

From (i) and (ii) general solution is

2u - 5x = ϕ(3x - 2y)....(iii)

Given u = 1 on the line 3x - 2y = 0 ⇒ 2y = 3x

(iii) ⇒ 2 - 5x = ϕ(3x - 3x)

So, ϕ(0) = 2 - 5x, which is not possible

Therefore 

PDE has no solution

 Option (4) is correct

Concept:

If the Pde is second-order in two variables x and y, it should be in the form: Auxx+2Buxy+Cuyy+Dux+Euy+F=0 where A,B,C,D,E, and F are functions of x and y.

Now, Examine the discriminant

If B2- 4AC>0 Then Pde is hyperbolic

If B2- 4AC=0 Then Pde is parabolic

If B2- 4AC<0 Then Pde is Elliptic

Explanation:

Here, A = P(x,y), B = $e^{x^2}.e^{y^2}$, C = Q(x,y)

⇒  B² - 4AC = $(e^{x^2+y^2}{)}^2$ - 4PQ

Let P(x,y) = Q(x,y) = 1

⇒ B² - 4AC = $(e^{x^2+y^2}{)}^2$ - 4

Let, $(e^{x^2+y^2}{)}^2$ - 4 = 0 ⇒ $(e^{x^2+y^2}{)}^2$ = 4 ⇒ 2(x²+y²) = ln4 

⇒ (x2+y2) = ln2 

Now, If (x2+y2) > ln2, then Pde is hyperbolic

If (x2+y2) = ln2, then Pde is parabola

If (x2+y2) < ln2, then Pde is elliptic 

Here, Treating R = ln2

Hence, Option (2) is true