Linear Integral Equations Questions and Answers

Q: The resolvent kernel \(R(x, t; \lambda)\) for this integral equation is

The resolvent kernel is \(\frac{xe^{t}}{1 - \lambda}\).The kernel \(K(x, t) = xe^{t}\) is separable. Its iterated kernels are all equal to it, because\[ \begin{aligned} K_{2}(x, t) &= \int_{0}^{1} xe^{s}\, se^{t}\, ds \\ &= xe^{t}\int_{0}^{1} se^{s}\, ds = xe^{t}\cdot 1. \end{aligned} \]The resolvent is \(R = \sum_{n \ge 1}\lambda^{n-1}K_{n}\), which here is \(xe^{t}(1 + \lambda + \lambda^{2} + \cdots)\), so\[ R(x, t; \lambda) = \frac{xe^{t}}{1 - \lambda} \]for \(|\lambda| \lt 1\).The key fact is \(\int_{0}^{1}se^{s}\,ds = [se^{s} - e^{s}]_{0}^{1} = 1\). Because every iterated kernel equals \(K\), the series is geometric with ratio \(\lambda\), not \(-\lambda\) or \(\pm\lambda^{2}\), so the denominators \(1 + \lambda\), \(1 + \lambda^{2}\) and \(1 - \lambda^{2}\) cannot arise.

Q: The solution of this integral equation is

The solution is \(u(x) = \frac{x}{1 - \lambda}\).The integral is a constant times \(x\): write \(u(x) = x(1 + \lambda C)\) with \(C = \int_{0}^{1}e^{t}u(t)\,dt\). Substituting back:\[ \begin{aligned} C &= (1 + \lambda C)\int_{0}^{1}te^{t}\,dt = 1 + \lambda C \\ C &= \frac{1}{1 - \lambda} \\ u(x) &= x\left(1 + \frac{\lambda}{1 - \lambda}\right) = \frac{x}{1 - \lambda} \end{aligned} \]The same follows from the resolvent of Q.50: \(u = x + \lambda\int_{0}^{1}\frac{xe^{t}}{1-\lambda}t\,dt\), which is \(x + \frac{\lambda x}{1-\lambda}\). \(\frac{x+1}{1-\lambda}\) adds a constant the equation cannot produce, since every term is a multiple of \(x\); \(x^{2}\) cannot appear for the same reason; and \(\frac{x}{1+\lambda^{2}}\) does not satisfy the equation.

Q: The function \(\phi(x)\) satisfying the integral equation\[ \int_{0}^{x} e^{x - \xi} \phi(\xi)\, d\xi = \frac{x^{2}}{2} \]is

The solution is \(\phi(x) = x - \frac{x^{2}}{2}\).The left side is the convolution of \(e^{x}\) with \(\phi\), so take Laplace transforms:\[ \begin{aligned} \frac{1}{s - 1}\,\Phi(s) &= \frac{1}{s^{3}} \\ \Phi(s) &= \frac{s - 1}{s^{3}} = \frac{1}{s^{2}} - \frac{1}{s^{3}} \\ \phi(x) &= x - \frac{x^{2}}{2} \end{aligned} \]Check by differentiating the equation: \(\phi(x) + \int_{0}^{x} e^{x-\xi}\phi(\xi)\,d\xi = x\), so \(\phi(x) = x - x^{2}/2\) directly. \(x^{2}/2\) mistakes the right side for the answer, and \(x + x^{2}/2\) and \(1 + x^{2}/2\) have the wrong sign or fail at \(x = 0\), where \(\phi\) must satisfy \(\phi(0) = 0\).

Q: The integral equation :\[ e^{t} + 2\int_{0}^{1} e^{(t - s)} x(s) = 0 \]is a :

It is a linear Fredholm integral equation of first kind.The integration runs over fixed limits, 0 to 1, which makes it a Fredholm equation (a Volterra equation has a variable limit such as \(t\)). The unknown \(x(s)\) appears only inside the integral, linearly, with no \(x(t)\) standing outside it, which is the mark of an equation of the first kind.Rewritten, it is \(\int_{0}^{1} e^{t-s}x(s)\,ds = -\tfrac{1}{2}e^{t}\), of the form \(\int K(t, s)x(s)\,ds = f(t)\). A second-kind equation would read \(x(t) = f(t) + \lambda\int K x\), which is not the case here.

Q: The initial value problem corresponding to the integral equation :\[ x(t) = x - \cos x + \int_{0}^{x} (x - t)\, x(t)\, dt \]is :

The initial value problem is \(x'' - x = \cos t\), \(x(0) = -1\), \(x'(0) = 1\).The stem uses \(x\) both for the unknown function and the variable; the options show it means \(x(t) = t - \cos t + \int_{0}^{t}(t - s)x(s)\,ds\). Put \(t = 0\): \(x(0) = 0 - 1 + 0 = -1\). Differentiate using Leibniz's rule (the integrand vanishes at \(s = t\)):\[ \begin{aligned} x'(t) &= 1 + \sin t + \int_{0}^{t} x(s)\,ds \\ x''(t) &= \cos t + x(t) \end{aligned} \]So \(x'(0) = 1\) and \(x'' - x = \cos t\). Options A and B have the wrong forcing term and \(x(0) = 1\), which contradicts \(x(0) = -1\), and option D adds an \(x'\) term that the single integral cannot produce.

Q: The solution of the singular integral equation :\[ \pi t = \int_{0}^{t} \frac{1}{\sqrt{t - s}}\, x(s)\, ds \]is :

The solution is \(2\sqrt{x}\), that is \(x(t) = 2\sqrt{t}\) (the options write the variable as \(x\)).This is Abel's equation \(\int_{0}^{t}\dfrac{x(s)}{\sqrt{t - s}}ds = f(t)\), whose solution is\[ \begin{aligned} x(t) &= \frac{1}{\pi}\frac{d}{dt} \\ &\quad \int_{0}^{t}\frac{f(s)}{\sqrt{t - s}}ds \end{aligned} \] With \(f(s) = \pi s\), using the Beta integral \(\int_{0}^{t}s(t - s)^{-1/2}ds = \tfrac{4}{3}t^{3/2}\):\[ \begin{aligned} x(t) &= \frac{d}{dt}\left(\frac{4}{3}t^{3/2}\right) \\ &= 2t^{1/2} \end{aligned} \]Check: \(\int_{0}^{t}\dfrac{2\sqrt{s}}{\sqrt{t - s}}ds = 2 \cdot \tfrac{\pi}{2}t = \pi t\). The constant \(2\pi\) and the multiples with \(\pi\) do not satisfy the equation.

Q: If \(L^{-1}\{F(s)\} = f(t)\), then\(L^{-1}\{F^{(n)}(s)\} =\) is :where \(L^{-1}\) be the inverse Laplace transform

\(L^{-1}\{F^{(n)}(s)\} =\) \((-1)^{n}t^{n}f(t)\).Differentiate \(F(s) = \int_{0}^{\infty} e^{-st}f(t)\,dt\) under the integral sign: each derivative in \(s\) brings down a factor \(-t\), so\[ F^{(n)}(s) = \int_{0}^{\infty} e^{-st}(-t)^{n}f(t)\,dt \]That is, \(L\{(-1)^{n}t^{n}f(t)\} = F^{(n)}(s)\), and taking inverses gives the answer. For example \(L\{1\} = 1/s\) and \(\tfrac{d}{ds}(1/s) = -1/s^{2} = L\{-t\}\).Option A misses the sign, option C misses the factor \(t^{n}\), and option D confuses differentiation in \(s\) with differentiation in \(t\).

Q: The solution of the integral equation :\[ x(t) = e^{-t} + \int_{0}^{t} \sin(t - \tau) x(\tau) d\tau \]by using Laplace transform method is :

The solution is \(x(t) = 2e^{-t} + t - 1\).The integral is the convolution \(\sin t * x(t)\). Taking Laplace transforms, with \(L\{\sin t\} = 1/(s^{2} + 1)\):\[ \begin{aligned} X &= \frac{1}{s + 1} + \frac{X}{s^{2} + 1} \\ X\frac{s^{2}}{s^{2} + 1} &= \frac{1}{s + 1} \\ X &= \frac{s^{2} + 1}{s^{2}(s + 1)} \end{aligned} \]Partial fractions give \(X = -\tfrac{1}{s} + \tfrac{1}{s^{2}} + \tfrac{2}{s + 1}\), so \(x(t) = -1 + t + 2e^{-t}\). Check: \(x(0) = 1\), which matches \(e^{0} + 0\). The options with \(e^{-2t}\) have the wrong exponential, and \(e^{-t} + 2t - 1\) has \(x'(0) = 1\), while differentiating the equation gives \(x'(0) = -1\).

Q: If \(\tilde{F}(s)\) is the Fourier transform of \(f(t)\), then the Fourier transform of :\[ f(t)\cos at \]is :

The transform is \(\tfrac{1}{2}[\tilde{F}(s + a) + \tilde{F}(s - a)]\).Write \(\cos at = \tfrac{1}{2}(e^{iat} + e^{-iat})\). Multiplying \(f(t)\) by \(e^{\pm iat}\) shifts its Fourier transform by \(a\) in one direction or the other (the modulation theorem). The sum of the two shifts, halved, is the transform of \(f(t)\cos at\). With either sign convention for the transform, the two shifts are \(s + a\) and \(s - a\), so the formula is the same.Options A and B are the transforms of \(f(t)e^{\pm iat}\), one exponential only. The difference in option D belongs to \(f(t)\sin at\), up to a factor \(\pm i\).

Q: Consider the following statements :(I) Every initial value problem can be reduced to a Fredholm integral equation.(II) Every boundary value problem can be reduced to a Volterra integral equation.Then :

Both (I) and (II) are wrong.The pairing in the statements is reversed. An initial value problem, with all conditions at one point, integrates to a Volterra equation, whose upper limit is the variable \(x\). A boundary value problem, with conditions at both ends, leads to a Fredholm equation over the fixed interval \([a, b]\), using the Green's function.For example, \(y'' + y = 0\), \(y(0) = 0\), \(y'(0) = 1\) becomes \(y(x) = x - \int_{0}^{x}(x - t)y(t)\,dt\), a Volterra equation. So neither statement is correct, and the options calling either correct are ruled out.

Direction for [ Question No: 1 To 2 ] :

Consider the Fredholm integral equation \(u(x) = x + \lambda\int_{0}^{1} x e^{t} u(t)\, dt\).

The resolvent kernel \(R(x, t; \lambda)\) for this integral equation is

A.\(\frac{xe^{t}}{1 - \lambda}\)
B.\(\frac{\lambda xe^{t}}{1 + \lambda}\)
C.\(\frac{xe^{t}}{1 + \lambda^{2}}\)
D.\(\frac{xe^{t}}{1 - \lambda^{2}}\)

Exam Year(s)

GATE Mathematics - 2012,

Answer: A

The resolvent kernel is \(\frac{xe^{t}}{1 - \lambda}\).

The kernel \(K(x, t) = xe^{t}\) is separable. Its iterated kernels are all equal to it, because

\[ \begin{aligned} K_{2}(x, t) &= \int_{0}^{1} xe^{s}\, se^{t}\, ds \\ &= xe^{t}\int_{0}^{1} se^{s}\, ds = xe^{t}\cdot 1. \end{aligned} \]

The resolvent is \(R = \sum_{n \ge 1}\lambda^{n-1}K_{n}\), which here is \(xe^{t}(1 + \lambda + \lambda^{2} + \cdots)\), so

\[ R(x, t; \lambda) = \frac{xe^{t}}{1 - \lambda} \]

for \(|\lambda| \lt 1\).

The key fact is \(\int_{0}^{1}se^{s}\,ds = [se^{s} - e^{s}]_{0}^{1} = 1\). Because every iterated kernel equals \(K\), the series is geometric with ratio \(\lambda\), not \(-\lambda\) or \(\pm\lambda^{2}\), so the denominators \(1 + \lambda\), \(1 + \lambda^{2}\) and \(1 - \lambda^{2}\) cannot arise.

Discuss in Forum

The solution of this integral equation is

A.\(\frac{x + 1}{1 - \lambda}\)
B.\(\frac{x^{2}}{1 - \lambda^{2}}\)
C.\(\frac{x}{1 + \lambda^{2}}\)
D.\(\frac{x}{1 - \lambda}\)

Exam Year(s)

GATE Mathematics - 2012,

Answer: D

The solution is \(u(x) = \frac{x}{1 - \lambda}\).

The integral is a constant times \(x\): write \(u(x) = x(1 + \lambda C)\) with \(C = \int_{0}^{1}e^{t}u(t)\,dt\). Substituting back:

\[ \begin{aligned} C &= (1 + \lambda C)\int_{0}^{1}te^{t}\,dt = 1 + \lambda C \\ C &= \frac{1}{1 - \lambda} \\ u(x) &= x\left(1 + \frac{\lambda}{1 - \lambda}\right) = \frac{x}{1 - \lambda} \end{aligned} \]

The same follows from the resolvent of Q.50: \(u = x + \lambda\int_{0}^{1}\frac{xe^{t}}{1-\lambda}t\,dt\), which is \(x + \frac{\lambda x}{1-\lambda}\). \(\frac{x+1}{1-\lambda}\) adds a constant the equation cannot produce, since every term is a multiple of \(x\); \(x^{2}\) cannot appear for the same reason; and \(\frac{x}{1+\lambda^{2}}\) does not satisfy the equation.

Discuss in Forum

The function \(\phi(x)\) satisfying the integral equation

\[ \int_{0}^{x} e^{x - \xi} \phi(\xi)\, d\xi = \frac{x^{2}}{2} \]

A.\(\frac{x^{2}}{2}\)
B.\(x + \frac{x^{2}}{2}\)
C.\(x - \frac{x^{2}}{2}\)
D.\(1 + \frac{x^{2}}{2}\)

Exam Year(s)

GATE Mathematics - 2012,

Answer: C

The solution is \(\phi(x) = x - \frac{x^{2}}{2}\).

The left side is the convolution of \(e^{x}\) with \(\phi\), so take Laplace transforms:

\[ \begin{aligned} \frac{1}{s - 1}\,\Phi(s) &= \frac{1}{s^{3}} \\ \Phi(s) &= \frac{s - 1}{s^{3}} = \frac{1}{s^{2}} - \frac{1}{s^{3}} \\ \phi(x) &= x - \frac{x^{2}}{2} \end{aligned} \]

Check by differentiating the equation: \(\phi(x) + \int_{0}^{x} e^{x-\xi}\phi(\xi)\,d\xi = x\), so \(\phi(x) = x - x^{2}/2\) directly. \(x^{2}/2\) mistakes the right side for the answer, and \(x + x^{2}/2\) and \(1 + x^{2}/2\) have the wrong sign or fail at \(x = 0\), where \(\phi\) must satisfy \(\phi(0) = 0\).

Discuss in Forum

The integral equation :

\[ e^{t} + 2\int_{0}^{1} e^{(t - s)} x(s) = 0 \]

is a :

A.linear Fredholm integral equation of second kind
B.linear Volterra integral equation of second kind
C.linear Fredholm integral equation of first kind
D.linear Volterra integral equation of first kind

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

It is a linear Fredholm integral equation of first kind.

The integration runs over fixed limits, 0 to 1, which makes it a Fredholm equation (a Volterra equation has a variable limit such as \(t\)). The unknown \(x(s)\) appears only inside the integral, linearly, with no \(x(t)\) standing outside it, which is the mark of an equation of the first kind.

Rewritten, it is \(\int_{0}^{1} e^{t-s}x(s)\,ds = -\tfrac{1}{2}e^{t}\), of the form \(\int K(t, s)x(s)\,ds = f(t)\). A second-kind equation would read \(x(t) = f(t) + \lambda\int K x\), which is not the case here.

Discuss in Forum

The initial value problem corresponding to the integral equation :

\[ x(t) = x - \cos x + \int_{0}^{x} (x - t)\, x(t)\, dt \]

is :

A.\(x''(t) + x(t) = \sin t\), \(x(0) = 1\), \(x'(0) = 0\)
B.\(x''(t) + x'(t) + x(t) = \sin t\), \(x(0) = 1\), \(x'(0) = 0\)
C.\(x''(t) - x(t) = \cos t\), \(x(0) = -1\), \(x'(0) = 1\)
D.\(x''(t) - x'(t) - x(t) = \cos t\), \(x(0) = -1\), \(x'(0) = 1\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The initial value problem is \(x'' - x = \cos t\), \(x(0) = -1\), \(x'(0) = 1\).

The stem uses \(x\) both for the unknown function and the variable; the options show it means \(x(t) = t - \cos t + \int_{0}^{t}(t - s)x(s)\,ds\). Put \(t = 0\): \(x(0) = 0 - 1 + 0 = -1\). Differentiate using Leibniz's rule (the integrand vanishes at \(s = t\)):

\[ \begin{aligned} x'(t) &= 1 + \sin t + \int_{0}^{t} x(s)\,ds \\ x''(t) &= \cos t + x(t) \end{aligned} \]

So \(x'(0) = 1\) and \(x'' - x = \cos t\). Options A and B have the wrong forcing term and \(x(0) = 1\), which contradicts \(x(0) = -1\), and option D adds an \(x'\) term that the single integral cannot produce.

Discuss in Forum

The solution of the singular integral equation :

\[ \pi t = \int_{0}^{t} \frac{1}{\sqrt{t - s}}\, x(s)\, ds \]

is :

A.\(2\pi\)
B.\(2\pi\sqrt{x}\)
C.\(2\sqrt{x}\)
D.\(\pi\sqrt{x}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The solution is \(2\sqrt{x}\), that is \(x(t) = 2\sqrt{t}\) (the options write the variable as \(x\)).

This is Abel's equation \(\int_{0}^{t}\dfrac{x(s)}{\sqrt{t - s}}ds = f(t)\), whose solution is

\[ \begin{aligned} x(t) &= \frac{1}{\pi}\frac{d}{dt} \\ &\quad \int_{0}^{t}\frac{f(s)}{\sqrt{t - s}}ds \end{aligned} \]

With \(f(s) = \pi s\), using the Beta integral \(\int_{0}^{t}s(t - s)^{-1/2}ds = \tfrac{4}{3}t^{3/2}\):

\[ \begin{aligned} x(t) &= \frac{d}{dt}\left(\frac{4}{3}t^{3/2}\right) \\ &= 2t^{1/2} \end{aligned} \]

Check: \(\int_{0}^{t}\dfrac{2\sqrt{s}}{\sqrt{t - s}}ds = 2 \cdot \tfrac{\pi}{2}t = \pi t\). The constant \(2\pi\) and the multiples with \(\pi\) do not satisfy the equation.

Discuss in Forum

If \(L^{-1}\{F(s)\} = f(t)\), then

\(L^{-1}\{F^{(n)}(s)\} =\) is :

where \(L^{-1}\) be the inverse Laplace transform

A.\(t^{n} f(t)\)
B.\((-1)^{n} t^{n} f(t)\)
C.\((-1)^{n} f(t)\)
D.\((-1)^{n} f^{(n)}(t)\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

\(L^{-1}\{F^{(n)}(s)\} =\) \((-1)^{n}t^{n}f(t)\).

Differentiate \(F(s) = \int_{0}^{\infty} e^{-st}f(t)\,dt\) under the integral sign: each derivative in \(s\) brings down a factor \(-t\), so

\[ F^{(n)}(s) = \int_{0}^{\infty} e^{-st}(-t)^{n}f(t)\,dt \]

That is, \(L\{(-1)^{n}t^{n}f(t)\} = F^{(n)}(s)\), and taking inverses gives the answer. For example \(L\{1\} = 1/s\) and \(\tfrac{d}{ds}(1/s) = -1/s^{2} = L\{-t\}\).

Option A misses the sign, option C misses the factor \(t^{n}\), and option D confuses differentiation in \(s\) with differentiation in \(t\).

Discuss in Forum

The solution of the integral equation :

\[ x(t) = e^{-t} + \int_{0}^{t} \sin(t - \tau) x(\tau) d\tau \]

by using Laplace transform method is :

A.\(x(t) = e^{-2t} + 2t - 1\)
B.\(x(t) = 2e^{-2t} + t - 1\)
C.\(x(t) = e^{-t} + 2t - 1\)
D.\(x(t) = 2e^{-t} + t - 1\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The solution is \(x(t) = 2e^{-t} + t - 1\).

The integral is the convolution \(\sin t * x(t)\). Taking Laplace transforms, with \(L\{\sin t\} = 1/(s^{2} + 1)\):

\[ \begin{aligned} X &= \frac{1}{s + 1} + \frac{X}{s^{2} + 1} \\ X\frac{s^{2}}{s^{2} + 1} &= \frac{1}{s + 1} \\ X &= \frac{s^{2} + 1}{s^{2}(s + 1)} \end{aligned} \]

Partial fractions give \(X = -\tfrac{1}{s} + \tfrac{1}{s^{2}} + \tfrac{2}{s + 1}\), so \(x(t) = -1 + t + 2e^{-t}\). Check: \(x(0) = 1\), which matches \(e^{0} + 0\). The options with \(e^{-2t}\) have the wrong exponential, and \(e^{-t} + 2t - 1\) has \(x'(0) = 1\), while differentiating the equation gives \(x'(0) = -1\).

Discuss in Forum

If \(\tilde{F}(s)\) is the Fourier transform of \(f(t)\), then the Fourier transform of :

\[ f(t)\cos at \]

is :

A.\(\tilde{F}(s + a)\)
B.\(\tilde{F}(s - a)\)
C.\(\dfrac{1}{2}[\tilde{F}(s + a) + \tilde{F}(s - a)]\)
D.\(\dfrac{1}{2}[\tilde{F}(s + a) - \tilde{F}(s - a)]\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The transform is \(\tfrac{1}{2}[\tilde{F}(s + a) + \tilde{F}(s - a)]\).

Write \(\cos at = \tfrac{1}{2}(e^{iat} + e^{-iat})\). Multiplying \(f(t)\) by \(e^{\pm iat}\) shifts its Fourier transform by \(a\) in one direction or the other (the modulation theorem). The sum of the two shifts, halved, is the transform of \(f(t)\cos at\). With either sign convention for the transform, the two shifts are \(s + a\) and \(s - a\), so the formula is the same.

Options A and B are the transforms of \(f(t)e^{\pm iat}\), one exponential only. The difference in option D belongs to \(f(t)\sin at\), up to a factor \(\pm i\).

Discuss in Forum

10.

Consider the following statements :

(I) Every initial value problem can be reduced to a Fredholm integral equation.
(II) Every boundary value problem can be reduced to a Volterra integral equation.

Then :

A.Only (I) is correct
B.Only (II) is correct
C.Both (I) and (II) are correct
D.Both (I) and (II) are wrong

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

Both (I) and (II) are wrong.

The pairing in the statements is reversed. An initial value problem, with all conditions at one point, integrates to a Volterra equation, whose upper limit is the variable \(x\). A boundary value problem, with conditions at both ends, leads to a Fredholm equation over the fixed interval \([a, b]\), using the Green's function.

For example, \(y'' + y = 0\), \(y(0) = 0\), \(y'(0) = 1\) becomes \(y(x) = x - \int_{0}^{x}(x - t)y(t)\,dt\), a Volterra equation. So neither statement is correct, and the options calling either correct are ruled out.

Linear Integral Equations Questions and Answers

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