Linear Integral Equations MCQ Quiz - Objective Question with Answer for Linear Integral Equations - Download Free PDF

Concept:

Let the Volterra integral equation of 2nd kind be

y(x) = f(x) + λ where K(x,t) is continuous function on 0 ≤ x ≤ a, 0 ≤ t ≤ x

and f(x) is continuous on 0 ≤ x ≤ a.

then iterative kernel is given by

K1(x, t) = K(x, t)

Kn(x, t) = 

and Resolvent kernel is

R(x, t; λ) = 

Explanation:

Given K(x, t) = t - x

K1(x, t) = K(x, t) = t - x

K2(x, t) = 

           =

          = 

          = 

        = 

       = 

       = 

Similarly we get

K3(x, t) =  

Continuing this process we get

Kn(x, t) = 

So, Resolvent kernel is

R(x, t; 1) =  = sin(t - x)

Option (3) is true

Explanation:

y''(x) + λy(x) = x with y(0) = 0 and y'(1) = 0

y''(x) = x - λy(x)

Integrating both sides from 0 to x

y'(x) - y'(0) =  - λ

y'(x) = c +  - λ ....(i) where c = y'(0)

Again integrating from 0 to x 

y(x) - y(0) = cx + 

y(x) - 0 = cx + 

y(x) = cx + 

Using 

y'(1) = 0 in (1) we get

0 = c + 

c = 

Hence

y(x) =  + 

y(x) = 

y(x) = 

y(x) = 

where k(x, t) = t\\x, x

Option (1) is true.

Explanation:

⇒ 

⇒ y(x) = x + xc

⇒ y(x) = (1 + c)x....(i)

where

c = 

⇒ c =  (by using (i))

⇒ c = 

⇒ 2c = 1 + c

⇒ c = 1

Putting in (i) we get

y(x) = 2x

Hence y(2) = 4

Option (1) is true.

Explanation:

y(x) = 1 + x + y(t)dt

Here the kernel k(x, t) = e^x-t is the function of the difference of x - t only. So the Volterra integral equation can be written as

y(x) = 1 + x + k(x)*y(x)

Taking Laplace transformation we get

Y(s) =  where L{y(x)} = Y(s)

⇒ 

⇒ Y(s).  = 

⇒ Y(s) =  × 

⇒ Y(s) = 

Using partial sum

 = 

Hence taking inverse Laplace transformation we get

y(x) =

Option (3) is true.

Concept:

Leibnitz rule of differentiation:

Let F(x, t) and  be continuous function of both x and t and let a'(x), b'(x) be continuous functions then

Explanation:

x³ = 

Using Leibnitz rule of differentiation we get

3x² =   + e^{x - x}y(x).1 - 0

⇒ 3x² = x³ + y(x)

y(x) =  3x2  - x³  

Hence y(2) = 12 - 8 =  4

Option (1) is true. 

Concept:

Leibnitz rule for differentiation:

 =  

Explanation:

(1): 

K(x, t) = x\end{cases}\) 

So 

y(x) = 2 K(x, t) y(t) dt

⇒ y(x) = 2 t(1 - x)ydt + 2 x(1 - t)ydt....(i)

⇒ y' = 2(-t)y(t)dt + x(1 - x)y(x) - 0 + 21(1-t)y(t)dt + 0 -x(1 - x)y(x).1

⇒ y' = 2(-t)y(t)dt + 2(1 - t)y(t)dt

⇒ y'' = 2[0 - xy(x).1 - 0] + 2[0 + 0 -(1 - x)y(x)] 

⇒ y'' = -2y(x)

⇒ y''(x) + 2y(x) = 0

 Also by (i)

y(0) = 2 t(1 - 0)ydt + 2 0(1 - t)ydt = 0 and  

y(1) = 2 t(1 - 1)ydt + 2 1(1 - t)ydt = 0

hence  satisfies

Therefore option (1) is correct

Explanation

Given:

y(s) = s + 2 

let t² y(t) dt = c₁    -- (i)

 t y(t) dt = c₂    -- (ii)

⇒ y(s) = s + 2sc₁ + 2s² c₂ ---- (iii)

By (ii), c2 = t[t + 2c1 t + 2c2t2]dt

=(t2 + 2t2 c1 + 2c2 t3)dt

c2​ = 

⇒ 4C1​ - 3c2 = -2     - (iv)

⇒ 20c1 = 5 + 10c1 + 8c2

⇒ 10c₁ - 8c₂ = 5 - (v)

Now solving equation (iv) and (v) to find c₁ & c₂.

20c₁ - 15c₂ = - 10 

20c1 - 16c2 = 10

subtracting we get

c2 = - 20

Hence (iv) implies 

 4c1​ + 60 = -2 ⇒ 4c1 = - 62 ⇒ c1 = - 31/2 

thus y(s) = s + 2sc1 + 2s2c2

= s + (-31) s + 2s2(-20)

y(s) = s - 31s - 40s²  = -(30s + 40s²)

⇒ option (3) is correct

Explanation:

⇒

⇒ y(x) = λx²e^xc....(i)

where c = ....(ii) 

Putting the expression of y(x) from (i) in (ii) we get

c = 

⇒ c = λc

⇒ c = λc 

⇒ c = λc 

⇒ c = λc

⇒ c = λc

⇒ c - λc = 0

Since c ≠ 0

⇒ 1 - λ = 0

⇒ λ = 

(3) is correct

Explanation:

⇒ 

⇒ u(x) = 1 + λ

⇒ u(x) = 1 - pλc....(i) where

c = ....(ii)

Putting the expression of u(t) from (i) in (ii) we get

c = 

⇒ c = (1 - pλc) 

⇒ c = (1 - pλc) 

⇒ c = (1 - pλc) (0 - 1)

⇒ c = -1 + pλc 

⇒ c(pλ - 1) = 1

⇒ c = 

Putting this value of c in (i) we get

u(x) = 1 -  = 

(3) is correct

Explanation:

y(x) = 1 + x + y(t)dt

Here the kernel k(x, t) = e^x-t is the function of the difference of x - t only. So the Volterra integral equation can be written as

y(x) = 1 + x + k(x)*y(x)

Taking Laplace transformation we get

Y(s) =  where L{y(x)} = Y(s)

⇒ 

⇒ Y(s).  = 

⇒ Y(s) =  × 

⇒ Y(s) = 

Using partial sum

 = 

Hence taking inverse Laplace transformation we get

y(x) =

Option (3) is true.

Concept:

(i) Convolution theorem: The convolution  f∗g is defined as
(f∗g)(t) =

(ii) Laplace transformation of f∗g is given by L{f∗g} = F(s)G(s) where L{f} = F(s) and L{g} = G(s) 

Explanation:

⇒ y(x) = x³ + sin(x)∗y(x)

Laking Laplace transformation we get

⇒ L{y(x)} = L{x³} + L{sin(x)∗y(x)}

⇒ Y(s) =  + Y(s) (as )

⇒ Y(s)(1 - ) = 

⇒ Y(s)( ) = 

⇒ Y(s) = 

⇒ Y(s) = 

Taking inverse Laplace transformation we get

y(x) =  (as )

y(x) =  

Therefore y(1) =  = 21/20

(4) is correct

Explanation:

IE 

 xt u(t)dt

u(x) = c ...(i)

where 

c = 

c =  (using (i))

c = 

c = 1/3

Hence solution is

u(x) =  

u(x) = x

(2) correct

Explanation:

...(i)

where 

 and 

so using (i)

 = 0...(ii)

and 

 = 0...(iii)

(ii) and (iii) has solution if

 = 0

 = 0

 + λ - 1 = 0

λ² +36λ - 36 = 0

λ = 

λ = 

(1) correct

Since f1, f2 the corresponding eigen functions for the homogeneous integral equation so 

(4) correct