Partial Differential Equations MCQ Quiz in বাংলা - Objective Question with Answer for Partial Differential Equations - বিনামূল্যে ডাউনলোড করুন [PDF]

Explanation:

Comparing  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),  = - sin(s),  = cos(s)

Now  = -1 =

So it may have no solution or infinitely many solutions

Using Lagrange characteristic equation we get

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 = 

⇒ 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

ab = 0

 ⇒ a = - b/4

Now put value of a in equation (ii)

 ....(iv)

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

⇒ b = 2y - x/2

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

z = - (2y - x/2)x + (2y - x/2)y - (2y - x/2)²

z = (2y - x/2)(y - x/4) -(2y - x/2)2 

z = 2(y - x/4)² - (y - x/4)2

z = (y - x/4)2 

z =  

The particular solution passing through x = 0 and z = y2 is 

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

Explanation:

Here, (i) 

So A = 1, B = 2, C = 1 - sgn(y)

0 \\ 0 ; y=0 \\ -1: y 0 \\ 1 & : y=0 \\ 2 & : y

Hence 

B2 - 4AC = 0 \\ 4, & y

 = 0 \Rightarrow \text { Hyperbolic } \\ 0, & \text { if } y=0 \Rightarrow \text { parabolic } \\ -4, & \text { if } y

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

For (ii)  

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

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  + 3y2u = 0

⇒ x²   = - 3y2u

⇒ x²   

⇒  and 

⇒  and 

⇒  and 

⇒ 

where  

Now, using u(x,0) = e1/x 

C = 1, k = -1 

Hence 

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

Explanation:

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

Using Charpit formula

⇒ 

Taking

⇒ pdp + qdq = 0

Integrating

p² + q² = a²....(i)

Putting in the given equation

a²y = qz

⇒ q = 

Putting in (i)

p =  = 

Putting p and q in

dz = pdx + qdy

⇒ dz = dx + dy

⇒  = a dx

Integrating

 = 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

 =  = 

Explanation:

Given 

ux - uuy = 0, x, y ∈ ℝ,

u(0, y) = 2y, y ∈ ℝ.

Using Lagrange's method

 =  = 

⇒  =  = 

Taking last two ratio

u = 0

⇒ u = c1...(i)

and putting u = c1 we get from first two terms

 =   

⇒ 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 = 

Therefore u(1, 2) =  = - 4

Option (2) is correct

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

Explanation:

Given PDE is

Lagrange’s auxiliary equations are

Taking 1st two terms

⇒  = 0

Integrating both sides

 = c₁....(i)

Also, 

⇒  = 0

Integrating both sides

ln x + ln y - ln z = ln c₂

⇒  = c₂ or,  = c2.....(ii)

From (i) and (ii) general solution is

 = 

z = xy for an arbitrary differentiable function f

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

Also if we take

⇒ 

Integrating both sides

ln(x - y) - ln z = lnc₃

⇒  = c₃....(iii)

Taking (ii) and (iii) we get

 = 0, for an arbitrary differentiable function F

Option (1) is true.

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

 = 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

Explanation:

Given PDE is 

Using Lagrange's auxiliary equation

Taking 

⇒ 3dx - 2dy = 0

Integrating both sides

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

Using 

⇒ 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

Explanation:

Here, A = P(x,y), B = , C = Q(x,y)

⇒  B² - 4AC =  - 4PQ

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

⇒ B² - 4AC =  - 4

Let,  - 4 = 0 ⇒  = 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)

Here, Treating R = ln2

Hence, Option (2) is true

Concept:

If the Pde is second-order in two variables x and y, it should be in the form: Au_xx+2Bu_xy+Cu_yy+Du_x+Eu_y+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

Explanation:

zxx - x2y zyy = 0 

Here, A=1,B=0,C=-x²y

Now, B2- 4AC=4x2y 

As, x>0,y>0

⇒ B2- 4AC=4x2y >0

⇒ Hyperbolic 

Hence, (3) option is true