Linear Integral Equations MCQ Quiz in বাংলা - Objective Question with Answer for Linear Integral Equations - বিনামূল্যে ডাউনলোড করুন [PDF]
Last updated on Apr 23, 2025
Latest Linear Integral Equations MCQ Objective Questions
Top Linear Integral Equations MCQ Objective Questions
Linear Integral Equations Question 1:
Given the homogenous integral equation \(\rm y(x)=\frac{1}{e^2-1} \int_0^1 e^{x+t} y(t)\)dt with degenerate kernel then
Answer (Detailed Solution Below)
Linear Integral Equations Question 1 Detailed Solution
Calculations:
\(\rm y(x)=\frac{1}{e^2-1} \int_0^1 e^{x+t} y(t) \)dt
\(\rm y(x)=\frac{e^x}{e^2-1} \int_0^1 e^{t} y(t) \)dt
y(x) = \(\frac{e^x}{e^2-1} c_1\)...(i)
where
c1 = \(\int_0^1 e^{t} y(t) dt\)
c1 = \(\int_0^1 e^{t}\cdot\frac{c_1e^t}{e^2-1} dt\)
c1 = \(\frac{c_1}{e^2-1}[\frac{e^{2t}}{2}]_0^1\)
c1 = \(\frac{c_1}{e^2-1}\cdot\frac{e^{2}-1}{2}\)
c1 = \(\frac{c_1}{2}\)
\(\frac{c_1}{2}\) = 0
c1 = 0
Substituting value of c1 in (i) we get
y = 0
hence given integral equation has only trivial solution.
(2) correct
Linear Integral Equations Question 2:
The integral equation
g(x) y(x) = \(\rm f(x)+\lambda \int_α^β k(x, t) y(t) d t\) with f(x), g(x) and k(x, t) as known functions, α and β as known constants and λ as a known parameter is a
Answer (Detailed Solution Below)
Linear Integral Equations Question 2 Detailed Solution
Explanation:
g(x) y(x) = \(\rm f(x)+\lambda \int_α^β k(x, t) y(t) d t\) is general form of Fredholm type integral
From the definition of Linear Fredholm integral equation we can say that (2) is correct
Linear Integral Equations Question 3:
The integral equation
\(\phi(x)=1+\frac{2}{\pi}\int_{0}^{\pi}(\cos^{2}\,x)\,\phi(t)d t\)
has
Answer (Detailed Solution Below)
Linear Integral Equations Question 3 Detailed Solution
Explanation:
Given integral equation
\(ϕ(x)=1+\frac{2}{π}\int_{0}^{π}(\cos^{2}\,x)\,ϕ(t)d t\)
⇒ \(ϕ(x)=1+\frac{2}{π}(\cos^{2}\,x)\int_{0}^{π}ϕ(t)d t\)...(i)
Let \(\int_0^{π}ϕ(t)dt=c\)....(ii)
then (i) becomes
\(ϕ(x)=1+\frac{2}{π}(\cos^{2}\,x)c\)
So \(ϕ(t)=1+\frac{2c}{π}(\cos^{2}t)\)
Substituting this expression in (ii) we get
\(\int_0^{π}(1+\frac{2c}{π}\cos^2t)dt=c\\\int_0^{π}(1+\frac{c}{π}(1+\cos2t)dt=c\\ [t+\frac{c}{π}(t+\frac{\sin2t}{2}]_0^{π}=c\\ π+c=c\)
⇒ π ≠ 0, which is a contradiction.
Hence integral equation has no solution.
Linear Integral Equations Question 4:
Let ϕ be the solution of
ϕ(x) = 1 − 2x − 4x2 + \(\rm \int^x_0\)[3 + 6(x − t) − 4(x − t)2]ϕ(t)dt.
Then ϕ(1) is equal to
Answer (Detailed Solution Below)
Linear Integral Equations Question 4 Detailed Solution
Concept:
Libnitz integral rule: \(\frac{\partial }{\partial x}\int_{a(x)}^{b(x)}f(x,t)dt =\int_{a(x)}^{b(x)}\frac{\partial f}{\partial x}dt \) + f(b(x), x)\(\frac{db}{dx}\) - f(a(x), x)\(\frac{da}{dx}\)
Explanation:
ϕ(x) = 1 − 2x − 4x2 + \(\rm \int^x_0\)[3 + 6(x − t) − 4(x − t)2]ϕ(t)dt ...(i)
Differentiating both side
ϕ'(x) = - 2 − 8x + \(\rm \int^x_0\)[6 − 8(x − t)]ϕ(t)dt + 3ϕ(x) (Using Libnitz integral rule) ...(ii)
Differentiating again we get
⇒ ϕ''(x) = - 8 - 8\(\rm \int^x_0\)ϕ(t)dt + 3ϕ'(x) + 6ϕ(x) (Again using Libnitz rule) ...(iii)
Again differentiating using Libnitz rule
⇒ ϕ'''(x) = - 8ϕ(x) + 3ϕ''(x) + 6ϕ'(x) ....(iv)
⇒ ϕ'''(x) - 3ϕ''(x) - 6ϕ'(x) + 8ϕ(x) = 0....(v)
Now from (i), (ii), (iii), (iv) we get the boundary conditions as
ϕ(0) = 1...(vi)
ϕ'(0) = - 2 + 3ϕ(0) = - 2 + 3 = 1...(vii)
ϕ''(0) = -8+3+6 =1...(viii) and
ϕ'''(0) = - 8ϕ(0) + 3 ϕ''(0) + 6ϕ'(0) = -8 + 3 + 6 = 1 ...(ix)
Now auxillary equation of (v) is
m3 - 3m2 - 6m +8 = 0
⇒ (m - 1)(m2 - 2m - 8) = 0
⇒ (m - 1)(m + 2)(m - 4) = 0
⇒ m = 1, -2, 4
Hence ϕ(x) = aex + be-2x + ce4x
So ϕ'(x) = aex - 2be-2x + 4ce4x
ϕ''(x) = aex + 4be-2x + 16ce4x
ϕ'''(x) = aex - 8be-2x + 64ce4x
Using boundary conditions we get
a + b + c = 1
a - 2b + 4c = 1
a + 4b +16c = 1
a - 8b + 64c = 1
Subtracting first two equations we get
3b - 3c = 0⇒ b=c
and subtracting and 3rd equations we get
6b + 12c = 0 ⇒ b + 2c = 0 ⇒ 3c = 0 (as b = c) ⇒ c = 0
Hence b=0
so putting c=b=0 in the first equation we get a=1
Hence solution is
ϕ(x) = ex
So ϕ(1) = e
Hence option (3) is correct
Linear Integral Equations Question 5:
For the Fredholm integral equation
\(y(s)=\lambda \int_0^1 e^s e^t y(t) d t\)
Which of the following statements are true?
Answer (Detailed Solution Below)
Linear Integral Equations Question 5 Detailed Solution
Explanation:
\(y(s)=λ \int_0^1 e^s e^t y(t) d t\)
⇒ y(s) = \(λ e^s\int_0^1 e^t y(t) d t\)
⇒ y(s) = \(λ ce^s\) .....(i) where
c = \(\int_0^1 e^t y(t) d t\)
⇒ c = \(\int_0^1 e^tλ c e^td t\)....(ii)
⇒ c = \(λ c[\frac{e^{2t}}{2}]_0^1\)
⇒ c = \(λ c(\frac{e^2-1}{2})\)
If c ≠ 0 then
⇒ \(λ (\frac{e^2-1}{2})\) = 1
⇒ λ = \( \frac{2}{e^2-1}\)
So non-trivial solution for \(\int_0^1 e^t y(t) d t\) ≠ 0 and c ≠ 0
Option (4) is correct
and if \(\int_0^1 e^t y(t) d t\) = 0 then by (ii) c = 0
Hence y(x) = 0 trivial solution.
Option (2) is correct
Linear Integral Equations Question 6:
Let K(x, t) = t - x be the kernel of a Volterra integral equation and λ = 1. Then which of the following is the resolvent kernel
Answer (Detailed Solution Below)
Linear Integral Equations Question 6 Detailed Solution
Concept:
Let the Volterra integral equation of 2nd kind be
y(x) = f(x) + λ\(\displaystyle \int_0^x K(x,t)y(t) d t\) 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) = \(\displaystyle \int_t^x K(x,s)K_{n-1}(s, t)ds\)
and Resolvent kernel is
R(x, t; λ) = \(\sum_{n=1}^{\infty}λ^{n-1}K_{n}(x, t)\)
Explanation:
Given K(x, t) = t - x
K1(x, t) = K(x, t) = t - x
K2(x, t) = \(\displaystyle \int_t^x (s-x)(t-s)ds\)
= \(\displaystyle \int_t^x (st-xt-s^2+xs)ds\)
= \(\left[{s^2t\over 2}-xts-{s^3\over 3}+{xs^2\over 2}\right]_t^x\)
= \(\left[{x^2t\over 2}-x^2t-{x^3\over 3}+{x^3\over 2}\right]-\left[{t^3\over 2}-xt^2-{t^3\over 3}+{xt^2\over 2}\right]\)
= \(\left[-{x^2t\over 2}+{x^3\over 6}\right]-\left[{t^3\over 6}-{xt^2\over 2}\right]\)
= \(x^3-t^3-3x^2t+3xt^2\over6\)
= \(-{(t-x)^3\over 3!}\)
Similarly we get
K3(x, t) = \({(t-x)^5\over 5!}\)
Continuing this process we get
Kn(x, t) = \({(-1)^n(t-x)^{2n+1}\over (2n+1)!}\)
So, Resolvent kernel is
R(x, t; 1) = \(\sum_{n=1}^{\infty}1^{n-1}.{(-1)^n(t-x)^{2n+1}\over (2n+1)!}\) = sin(t - x)
Option (3) is true
Linear Integral Equations Question 7:
The kernel corresponding to the boundary value problem
y''(x) + λy(x) = x with y(0) = 0 and y'(1) = 0 is
Answer (Detailed Solution Below)
Linear Integral Equations Question 7 Detailed Solution
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) = \(x^2\over 2\) - λ\(\int_0^xy(t)dt\)
y'(x) = c + \(x^2\over 2\) - λ\(\int_0^xy(t)dt\) ....(i) where c = y'(0)
Again integrating from 0 to x
y(x) - y(0) = cx + \({x^3\over 6}-\lambda\int_0^x\int_0^xy(t)dt\)
y(x) - 0 = cx + \({x^3\over 6}-\lambda\int_0^x(x-t)y(t)dt\)
y(x) = cx + \({x^3\over 6}-\lambda\int_0^x(x-t)y(t)dt\)
Using
y'(1) = 0 in (1) we get
0 = c + \(\frac12-\lambda\int_0^1y(t)dt\)
c = \(-\frac12+\lambda\int_0^1y(t)dt\)
Hence
y(x) = \(-\frac x2+\lambda\int_0^1xy(t)dt\) + \({x^3\over 6}-\lambda\int_0^x(x-t)y(t)dt\)
y(x) = \(\frac16(x^3-3x)+\lambda[\int_0^xxy(t)dt+\int_x^1 xy(t)dt-\int_0^x(x-t)y(t)dt]\)
y(x) = \(\frac16(x^3-3x)+\lambda[\int_0^xty(t)dt+\int_x^1 xy(t)dt]\)
y(x) = \(\frac16(x^3-3x)+\lambda\int_0^1k(x,t)y(t)dt\)
where k(x, t) = \(\begin{cases}t, x>t\\x, x
Option (1) is true.
Linear Integral Equations Question 8:
The value of λ for which the integral equation
\(y(x)=\lambda \displaystyle \int_0^1 x e^{x+2t} y(t) d t\)
has a non-zero solution, is
Answer (Detailed Solution Below)
Linear Integral Equations Question 8 Detailed Solution
Explanation:
\(y(x)=\lambda \displaystyle \int_0^1 x e^{x+2t} y(t) d t\)
⇒ \(y(x)=λ x e^{x}\displaystyle \int_0^1 e^{2t} y(t) d t\)
⇒ y(x) = λxexc....(i)
where c = \(\displaystyle \int_0^1 e^{2t} y(t) d t\)....(ii)
Putting the expression of y(x) from (i) in (ii) we get
c = \(\displaystyle \int_0^1 e^{t} λ te^{2t}c\, d t\)
⇒ c = λc\(\displaystyle \int_0^1 te^{3t}\, d t\)
⇒ c = λc \(\left\{\left[ t{e^{3t}\over 3}\right]_0^1-\displaystyle \int_0^1{e^{3t}\over3}\, dt\right\}\)
⇒ c = λc \(\left\{{e^{3}\over 3}-\left[{ e^{3t}\over 9}\right]_0^1\right\}\)
⇒ c = λc\(\left\{{e^{3}\over 3}-{{e^3-1}\over 9}\right\}\)
⇒ c = λc\(2e^{3}+1\over9\)
⇒ 1 = λ\(2e^{3}+1\over9\) (since c ≠ 0)
⇒ λ = \(\frac{9}{2e^3+1}\)
(2) is correct
Linear Integral Equations Question 9:
for λ ∈ ℝ such that |λ| < \(\frac{5}{32}\), let R(x, t, λ) and u denote the resolvent kernel and the solution, respectively, of the Fredholm integral equation
\(\rm u(x)=x+\frac{\lambda}{2}\int_{-2}^2(xt+x^2t^2)u(t)dt\)
Then which of the following statements are true?
Answer (Detailed Solution Below)
Linear Integral Equations Question 9 Detailed Solution
Concept:
The resolvent kernel, denoted as \(R(x,t,\lambda) \)
For second-kind Fredholm integral equations, the resolvent kernel helps us express the solution in terms of a known series.
Explanation:
The Fredholm integral equation is
\(u(x) = x + \frac{\lambda}{2} \int_{-2}^{2} \left( xt + x^2t^2 \right) u(t) \, dt\)
We are also given that \(R(x, t, \lambda)\) is the resolvent kernel, and \(\lambda \in \mathbb{R} \) with \(|\lambda| < \frac{5}{32}\) .
The function \(u(x)\) is a combination of a known term \(x \), and an integral term that depends on \(\lambda\), \(x \) and \(u(t)\) .
The integral term \(\int_{-2}^{2} \left( xt + x^2 t^2 \right) u(t) dt \) involves multiplying \(u(t)\) by a kernel function \((xt + x^2 t^2)\)
and then integrating over t from -2 to 2.
\(\int_{-2}^{2} \left( xt + x^2 t^2 \right) u(t) \, dt = x \int_{-2}^{2} t u(t) \, dt + x^2 \int_{-2}^{2} t^2 u(t) \, dt\)
Given that the non-integral part of \(u(x)\) is a polynomial in \(x \) , we can try to assume that the
solution \(u(t)\) is also a polynomial, say,
\(u(t) = a_0 + a_1 t + a_2 t^2\)
Substituting this assumed solution for \\(u(t)\) into the integral expressions, we can compute the
integrals separately \( \int_{-2}^{2} t u(t) \, dt \)
Substitute \(u(t) = a_0 + a_1 t + a_2 t^2 \):
\(\int_{-2}^{2} t \left( a_0 + a_1 t + a_2 t^2 \right) dt = a_0 \int_{-2}^{2} t dt + a_1 \int_{-2}^{2} t^2 dt + a_2 \int_{-2}^{2} t^3 dt\)
Using standard integrals, \(\int_{-2}^{2} t dt = 0, \quad \int_{-2}^{2} t^2 dt = \frac{16}{3}, \quad \int_{-2}^{2} t^3 dt = 0\)
Thus, \(\int_{-2}^{2} t u(t) \, dt = a_1 \cdot \frac{16}{3}\)
and \(\int_{-2}^{2} t^2 u(t) \, dt\)
Substitute, \(u(t) = a_0 + a_1 t + a_2 t^2\)
\(\int_{-2}^{2} t^2 \left( a_0 + a_1 t + a_2 t^2 \right) dt = a_0 \int_{-2}^{2} t^2 dt + a_1 \int_{-2}^{2} t^3 dt + a_2 \int_{-2}^{2} t^4 dt\)
Using standard integrals \(\int_{-2}^{2} t^2 dt = \frac{16}{3}, \quad \int_{-2}^{2} t^3 dt = 0, \quad \int_{-2}^{2} t^4 dt = \frac{64}{5}\)
Thus, \(\int_{-2}^{2} t^2 u(t) \, dt = a_0 \cdot \frac{16}{3} + a_2 \cdot \frac{64}{5}\)
\(u(x) = x + \frac{\lambda}{2} \left( x \cdot \frac{16}{3} \cdot a_1 + x^2 \cdot \left( a_0 \cdot \frac{16}{3} + a_2 \cdot \frac{64}{5} \right) \right)\)
Now, equate the powers of \( x \) on both sides and solve for the constants \(a_0, a_1, a_2 \).
After solving for \(a_0, a_1, a_2 \), you can evaluate the expression for \(u(x) \), particularly at \(x = 1 \) to determine \(u(1)\).
For the resolvent kernel \(R(x,t,\lambda) \), it should be determined based on the integral operator
and the values of \( \lambda \) that satisfy the equation.
Thus, Option 2) and Option 4) are correct.
Linear Integral Equations Question 10:
The integral equation \(y(x)=λ \int_{0}^{1}(4 x^2-3x) \operatorname{ty}(t) d t\), with λ as a parameter, has
Answer (Detailed Solution Below)
Linear Integral Equations Question 10 Detailed Solution
Explanation:
\(y(x)=λ \int_{0}^{1}(4 x^2-3x) \operatorname{ty}(t) d t\)
⇒ \(y(x)=λ \int_{0}^{1}(4 x^2-3x) \operatorname{ty}(t) d t\)
⇒ \(y(x)=λ (4 x^2-3x)\int_{0}^{1} t y(t) d t\)....(i)
⇒ y(x) = λ(4x2 - 3x)c ....(ii) where
c = \(\int_0^1ty(t)dt\)....(iii)
Putting the expression of ϕ(t) from (ii) in (iii) we get
c = \(\int_0^1λ c(4t^2-3t)tdt\)
⇒ c = λc \([t^4-t^3]_0^1\)
⇒ c = λc × 0
⇒ c = 0
Hence in (ii) putting c = 0, we get
y(x) = 0
Therefore we are getting trivial solution
Hence the homogeneous integral equation has no characteristic number
(3) is correct