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\(\displaystyle \int_0^x K(x,t)y(t) d t\) 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

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.

Explanation:

\(y(x)=x+ \displaystyle \int_0^1 x y(t) d t\)

⇒ \(y(x)=x+ x\displaystyle \int_0^1 y(t) d t\)

⇒ y(x) = x + xc

⇒ y(x) = (1 + c)x....(i)

where

c = \(\displaystyle \int_0^1 y(t) d t\)

⇒ c = \(\displaystyle \int_0^1 (1+c)t d t\) (by using (i))

⇒ c = \((1+c).\frac12\)

⇒ 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 + \(\int_0^xe^{x-t}\)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) = \(\frac1s +\frac1{s^2}+\frac1{s-1}Y(s)\) where L{y(x)} = Y(s)

⇒ \((1-\frac1{s-1})Y(s)={s+1\over s^2}\)

⇒ Y(s). \(s-2\over s-1\) = \({s+1\over s^2}\)

⇒ Y(s) = \({s+1\over s^2}\) × \(s-1\over s-2\)

⇒ Y(s) = \(s^2-1\over s^2(s-2)\)

Using partial sum

\(s^2-1\over s^2(s-2)\) = \(\frac14.\frac1s+\frac12.\frac1{s^2}+\frac34.\frac1{s-2}\)

Hence taking inverse Laplace transformation we get

y(x) = \(\frac14+\frac x2+\frac{3e^x}4\)

Option (3) is true.

Concept:

Leibnitz rule of differentiation:

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

\({d\over dx}\int_{a(x)}^{b(x)}F(x, t)dt=\int_{a(x)}^{b(x)}{\partial F\over \partial x}dt+F(x, a(x))a'(x)-F(x, b(x))b'(x)\)  

Explanation:

x³ = \(\int_0^x e^{x-t}y(t)dt\)

Using Leibnitz rule of differentiation we get

3x² = \(\int_0^x e^{x-t}y(t)dt \)  + 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:

\(\frac{\partial }{\partial x}\int_{a(x)}^{b(x)}f(x,t)dt\) = \(\int_{a(x)}^{b(x)}\frac{\partial f}{\partial x}dt+f(b(x),x)\frac{\partial b}{\partial x}-f(a(x),x)\frac{\partial a}{\partial x}\) 

Explanation:

(1): 

K(x, t) = \(\begin{cases} \rm t(1-x) & \text { for } \rm tx\end{cases}\) 

So 

y(x) = 2\(\displaystyle\int_0^1\) K(x, t) y(t) dt

⇒ y(x) = 2 \(\displaystyle\int_0^x\)t(1 - x)ydt + 2 \(\displaystyle\int_x^1\)x(1 - t)ydt....(i)

⇒ y' = 2\(\displaystyle\int_0^x\)(-t)y(t)dt + x(1 - x)y(x) - 0 + 2\(\displaystyle\int_x^1\)1(1-t)y(t)dt + 0 -x(1 - x)y(x).1

⇒ y' = 2\(\displaystyle\int_0^x\)(-t)y(t)dt + 2\(\displaystyle\int_x^1\)(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 \(\displaystyle\int_0^0\)t(1 - 0)ydt + 2 \(\displaystyle\int_0^1\)0(1 - t)ydt = 0 and  

y(1) = 2 \(\displaystyle\int_0^1\)t(1 - 1)ydt + 2 \(\displaystyle\int_1^1\)1(1 - t)ydt = 0

hence \(\rm \left\{\begin{aligned}\rm y^{\prime \prime}(x)+2 y(x) & =0 \quad \text { for } \rm x ∈(0,1), \\ \rm y(0) & =\rm y(1)=0 .\end{aligned}\right.\) satisfies

Therefore option (1) is correct

Explanation

Given:

y(s) = s + 2 \(∫_0^1 s t^2 y(t) d t+2 ∫_0^1 s^2 t y(t) d t\)

let \(\int_0^1\)t² y(t) dt = c₁    -- (i)

\(\int_0^1\) t y(t) dt = c₂    -- (ii)

⇒ y(s) = s + 2sc₁ + 2s² c₂ ---- (iii)

By (ii), c2 = \(\int_0^1\)t[t + 2c1 t + 2c2t2]dt

=\(\int_0^1\)(t2 + 2t2 c1 + 2c2 t3)dt

c2​ = \(\frac{t^3}{3}\left[1+\frac{2}{c_1}\right]+\left.\frac{c_2 t^4}{2}\right|_0 ^1=\left(\frac{1}{3}+\frac{2}{3} c_1\right)+\frac{c_2}{2}\)

\(⇒ \frac{3 C_2}{2}=1+\frac{2}{C_1} \)

⇒ 4C1​ - 3c2 = -2     - (iv)

\(C_1 =∫_0^1 t^2\left[t+2 c_1 t+2 t^2 c_2\right] d t \)

\(=∫_0^1\left[t^3\left(1+2 c_1\right)+2 t^4 c_2\right] d t\)

\(c_1 =\left[\left(1+2 c_1\right) \frac{t^4}{4}+\frac{2 c_2 t^5}{5}\right]_0^1=\frac{1+2 c_1}{4}+\frac{2 c_2}{5}\)

⇒ 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:

\(y(x)=λ \displaystyle \int_0^1 x^2 e^{x+t} y(t) d t\)

⇒ \(y(x)=λ x^2 e^{x}\displaystyle \int_0^1 e^{t} y(t) d t\)

⇒ y(x) = λx²e^xc....(i)

where c = \(\displaystyle \int_0^1 e^{t} y(t) d t\)....(ii) 

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

c = \(\displaystyle \int_0^1 e^{t} λ t^2e^tc\, d t\)

⇒ c = λc\(\displaystyle \int_0^1 t^2e^{2t}\, d t\)

⇒ c = λc \(\left\{\left[ t^2{e^{2t}\over 2}\right]_0^1-\displaystyle \int_0^12t{e^{2t}\over2}\, dt\right\}\)

⇒ c = λc \(\left\{{e^{2}\over 2}-\left[{t e^{2t}\over 2}\right]_0^1+\left[{e^{2t}\over4}\right]_0^1\right\}\)

⇒ c = λc\(\left\{{e^{2}\over 2}-{e^2\over 2}+{e^{2}-1\over4}\right\}\)

⇒ c = λc\(e^{2}-1\over4\)

⇒ c - λc\(e^{2}-1\over4\) = 0

Since c ≠ 0

⇒ 1 - λ\(e^{2}-1\over4\) = 0

⇒ λ = \(\frac{4}{e^2-1}\)

(3) is correct

Explanation:

\(u(x)-λ \int_0^1 \log \left(\frac{1}{t}\right)^p u(t) d t=1\)

⇒ \(u(x)=1+λ \int_0^1 \log \left(t^{-p}\right) u(t) d t\)

⇒ u(x) = 1 + λ\(\int_0^1-p \log \left(t^{}\right) u(t) d t\)

⇒ u(x) = 1 - pλc....(i) where

c = \(\int_0^1 \log \left(t^{}\right) u(t) d t\)....(ii)

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

c = \(\int_0^1 \log \left(t^{}\right) (1 - pλc) d t\)

⇒ c = (1 - pλc) \(\int_0^1 \log \left(t^{}\right) d t\)

⇒ c = (1 - pλc) \([t\log t-t]_0^1\)

⇒ c = (1 - pλc) (0 - 1)

⇒ c = -1 + pλc 

⇒ c(pλ - 1) = 1

⇒ c = \(\frac{1}{p\lambda-1}\)

Putting this value of c in (i) we get

u(x) = 1 - \(\frac{p\lambda}{p\lambda-1}\) = \(\frac{1}{1-p\lambda}\)

(3) is correct

Explanation:

y(x) = 1 + x + \(\int_0^xe^{x-t}\)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) = \(\frac1s +\frac1{s^2}+\frac1{s-1}Y(s)\) where L{y(x)} = Y(s)

⇒ \((1-\frac1{s-1})Y(s)={s+1\over s^2}\)

⇒ Y(s). \(s-2\over s-1\) = \({s+1\over s^2}\)

⇒ Y(s) = \({s+1\over s^2}\) × \(s-1\over s-2\)

⇒ Y(s) = \(s^2-1\over s^2(s-2)\)

Using partial sum

\(s^2-1\over s^2(s-2)\) = \(\frac14.\frac1s+\frac12.\frac1{s^2}+\frac34.\frac1{s-2}\)

Hence taking inverse Laplace transformation we get

y(x) = \(\frac14+\frac x2+\frac{3e^x}4\)

Option (3) is true.

Concept:

(i) Convolution theorem: The convolution  f∗g is defined as
(f∗g)(t) = \(∫_0^tf(u)g(t-u)du\)

(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:

\(y(x)=x^{3}+\int_{0}^{x} \operatorname{sin}(x-t) y(t) d t\)

⇒ y(x) = x³ + sin(x)∗y(x)

Laking Laplace transformation we get

⇒ L{y(x)} = L{x³} + L{sin(x)∗y(x)}

⇒ Y(s) = \(\frac6{s^4}\) + \(\frac{1}{s^2+1}\)Y(s) (as \(L\{x^n\}=\frac{n!}{s^{n+1}}\))

⇒ Y(s)(1 - \(\frac{1}{s^2+1}\)) = \(\frac6{s^4}\)

⇒ Y(s)( \(\frac{s^2}{s^2+1}\)) = \(\frac6{s^4}\)

⇒ Y(s) = \(\frac{6(s^2+1)}{s^6}\)

⇒ Y(s) = \(\frac6{s^4}+\frac6{s^6}\)

Taking inverse Laplace transformation we get

y(x) = \(x^3+\frac6{5!}x^5\) (as \(L^{-1}\{s^{n}\}=\frac{x^n}{(n-1)!}\))

y(x) = \(x^3+\frac{x^5}{20}\) 

Therefore y(1) = \(1+\frac{1}{20}\) = 21/20

(4) is correct

Explanation:

IE 

\(\rm u(x)=\frac{5 x}{6}+\frac{1}{2} \int_0^1\) xt u(t)dt

u(x) = \(\frac{5 x}{6}+\frac{x}{2}\)c ...(i)

where 

c = \(\int_0^1 tu(t)dt\)

c = \(\int_0^1 (\frac{5t^2}{6}+\frac{ct^2}{2})dt\) (using (i))

c = \(\frac{5}{18}+\frac{c}{6}\)

\(\frac{5c}{6}=\frac{5}{18}\)

c = 1/3

Hence solution is

u(x) = \(\frac{5 x}{6}+\frac{x}{6}\) 

u(x) = x

(2) correct

Explanation:

\(\phi(x)=λ \int_0^1\left(x t+2 x^2\right) \phi(t) dt\)

\(\phi(x)=λ x\int_0^1t \phi(t) dt+2λ x^2\int_0^1 \phi(t)dt\)

\(\phi(t)=λ c_1x+2λ c_2x^2\)...(i)

where 

\(c_1=\int_0^1 t\phi(t)dt\) and \(c_2=\int_0^1 \phi(t)dt\)

so using (i)

\(c_1=\int_0^1 t(λ c_1t+2λ c_2t^2)dt\)

\(c_1=\frac{λ c_1}{3}+\frac{λ c_2}{2}\)

\((1-\frac{λ}{3})c_1-\frac{λ }{2}c_2\) = 0...(ii)

and 

\(c_2=\int_0^1 (λ c_1t+2λ c_2t^2)dt\)

\(c_2=\frac{λ c_1}{2}+\frac{2λ c_2}{3}\)

\(-\frac{λ}{2}c_1+(1-\frac{2λ }{3})c_2\) = 0...(iii)

(ii) and (iii) has solution if

\(\begin{vmatrix}1-\frac{λ}{3}&-\frac{λ}{2}\\-\frac{λ}{2}&1-\frac{2λ}{3}\end{vmatrix}\) = 0

\(1-\frac{λ}{3}-\frac{2λ}{3}+\frac{2λ^2}{9}-\frac{λ^2}{4}\) = 0

\(\frac{λ^2}{36}\) + λ - 1 = 0

λ² +36λ - 36 = 0

λ = \(\frac{-36\pm\sqrt{1296+144}}{2}\)

λ = \(-18 \pm 6 \sqrt{10}\)

(1) correct

Since f1, f2 the corresponding eigen functions for the homogeneous integral equation so \(\int_0^1 f_1(x) f_2(x) d x=0\)

(4) correct