Linear Integral Equations Questions and Answers

Q: The solution of the integral equation\[ \frac{\pi t}{2}=\int_{0}^{t}\frac{1}{\sqrt{t-s}}\,x(s)\,ds \]is :

The solution is \(x(t) = \sqrt{t}\).This is Abel's integral equation. Test \(x(s) = \sqrt{s}\) with the substitution \(s = tu\):\[ \begin{aligned} \int_{0}^{t} \frac{\sqrt{s}}{\sqrt{t - s}}\,ds &= t\int_{0}^{1} u^{1/2}(1 - u)^{-1/2}\,du \\ &= t\,B\left(\tfrac{3}{2}, \tfrac{1}{2}\right) = t \cdot \frac{\pi}{2} \end{aligned} \]since\[ \begin{aligned} B\left(\tfrac{3}{2}, \tfrac{1}{2}\right) &= \frac{\Gamma(3/2)\Gamma(1/2)}{\Gamma(2)} \\ &= \frac{\sqrt{\pi}}{2} \cdot \sqrt{\pi} = \frac{\pi}{2} \end{aligned} \]That matches the left-hand side \(\frac{\pi t}{2}\).By the same computation, \(x = 1\) gives \(2\sqrt{t}\), \(x = t\) gives \(\frac{4}{3}t^{3/2}\), and \(x = \frac{1}{\sqrt{t}}\) gives the constant \(\pi\); none equals \(\frac{\pi t}{2}\).

Q: The boundary value problem\[ y'' + \lambda y = x, \; y(0) = 0, \; y'(1) = 0 \]is equivalent to the Fredholm integral equation.where \(k(x, t) = \begin{cases} t; & 0 \lt t \lt x \\ x; & x \lt t \lt 1 \end{cases}\)

The equivalent equation is \(y(x) = \frac{1}{6}(x^{3} - 3x) + \lambda\int_{0}^{1} k(x, t)\, y(t)\, dt\).Write the problem as \(y'' = x - \lambda y\) with \(y(0) = 0\), \(y'(1) = 0\). For these boundary conditions the Green's function of \(y'' = g\) is \(-\min(x, t) = -k(x, t)\), so\[ \begin{aligned} y(x) = &-\int_{0}^{1} k(x, t)\,t\,dt \\ &+ \lambda\int_{0}^{1} k(x, t)\,y(t)\,dt \end{aligned} \]The first integral is\[ \begin{aligned} \int_{0}^{x} t^{2}\,dt + x\int_{x}^{1} t\,dt &= \frac{x^{3}}{3} + \frac{x(1 - x^{2})}{2} \\ &= \frac{x}{2} - \frac{x^{3}}{6} \end{aligned} \]so the free term is \(\frac{x^{3}}{6} - \frac{x}{2} = \frac{1}{6}(x^{3} - 3x)\). Check: it vanishes at 0 and its derivative \(\frac{x^{2} - 1}{2}\) vanishes at 1. The other free terms fail these boundary checks, and their kernels carry extra factors.

Q: Eigenvalues of the Fredholm integral equation,\[ x(t) = 1 + \lambda \int_{0}^{\pi} \cos(t + s)\, x(s)\, ds \]are :

The eigenvalues are \(2/\pi\) and \(-2/\pi\).Expand the kernel: \(\cos(t + s) = \cos t\cos s - \sin t\sin s\). For the homogeneous equation, \(x(t) = \lambda(A\cos t - B\sin t)\) with \(A = \int_{0}^{\pi} \cos s\, x(s)\,ds\) and \(B = \int_{0}^{\pi} \sin s\, x(s)\,ds\).On \([0, \pi]\), \(\int \cos^{2} = \int \sin^{2} = \frac{\pi}{2}\) and \(\int \sin s\cos s\,ds = 0\), so\[ A = \lambda\frac{\pi}{2}A, \qquad B = -\lambda\frac{\pi}{2}B \]A non-trivial solution needs \(\lambda = \frac{2}{\pi}\) (with \(x = \cos t\)) or \(\lambda = -\frac{2}{\pi}\) (with \(x = \sin t\)).The pairs \(\pi, -\pi\), \(1, -1\) and \(\sqrt{2}, -\sqrt{2}\) do not satisfy \(\lambda\frac{\pi}{2} = \pm 1\), so they are not eigenvalues.

Q: The resolvent kernel for Volterra integral equation with kernel :\[ k(t, s) = a^{t-s}, \quad a \gt 0 \]is :

The resolvent kernel is \(R(t, s; \lambda) = a^{t-s}\, e^{\lambda(t-s)}\).The iterated kernels of \(k(t, s) = a^{t-s}\) are\[ k_{n}(t, s) = a^{t-s}\,\frac{(t - s)^{n-1}}{(n - 1)!} \](each iteration integrates a constant over \([s, t]\)). Summing the Neumann series,\[ \begin{aligned} R &= \sum_{n \ge 1} \lambda^{n-1} k_{n} \\ &= a^{t-s} \sum_{m \ge 0} \frac{[\lambda(t - s)]^{m}}{m!} = a^{t-s} e^{\lambda(t-s)} \end{aligned} \]Options B and D contain \(a^{t+s}\), which is not a function of \(t - s\), although every iterated kernel of a difference kernel is. Option A puts \(\lambda\) in the exponent of a instead of e.

Q: For the integral equation\[ x(t)=\frac{1}{2}\int_{t}^{1}(t\sqrt{s}-s\sqrt{t})\,x(s)\,ds \]consider the following statements. The above integral equation is :(I) Fredholm integral equation of second kind.(II) Volterra integral equation of first kind.Then :

Both (I) and (II) are false.The upper limit is 1 but the lower limit is the variable t, so the range of integration depends on t: that makes it a Volterra-type equation, not a Fredholm equation (which has fixed limits). So (I) is false.The unknown x appears outside the integral as well as inside it, which makes the equation of the second kind. A first-kind equation has the unknown only under the integral sign. So (II) is false as well; the equation is a Volterra equation of the second kind.

Q: Eigen values of the homogeneous Fredholm integral equation\[ x(t) = \lambda \int_{0}^{2\pi} \sin(t + s)\, x(s)\, ds \]are :

The eigenvalues are 1/π and −1/π.Expand \(\sin(t + s) = \sin t\cos s + \cos t\sin s\), so \(x(t) = \lambda(c_{1}\sin t + c_{2}\cos t)\) with \(c_{1} = \int_{0}^{2\pi}\cos s\, x(s)\, ds\) and \(c_{2} = \int_{0}^{2\pi}\sin s\, x(s)\, ds\). Using \(\int_{0}^{2\pi}\sin^{2} = \int_{0}^{2\pi}\cos^{2} = \pi\) and \(\int_{0}^{2\pi}\sin\cos = 0\):\[ \begin{aligned} c_{1} &= \lambda\pi c_{2} \\ c_{2} &= \lambda\pi c_{1} \\ c_{1} &= \lambda^{2}\pi^{2}c_{1} \end{aligned} \]A non-zero solution needs \(\lambda^{2}\pi^{2} = 1\), so \(\lambda = \pm 1/\pi\). The values \(\pm\pi\) are the reciprocals (the eigenvalues of the integral operator itself), and 1 or \(\pi\) do not satisfy the condition.

31.

The solution of the integral equation

\[ \frac{\pi t}{2}=\int_{0}^{t}\frac{1}{\sqrt{t-s}}\,x(s)\,ds \]

is :

A.\(x(t) = \frac{1}{\sqrt{t}}\)
B.\(x(t) = \sqrt{t}\)
C.\(x(t) = 1\)
D.\(x(t) = t\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The solution is \(x(t) = \sqrt{t}\).

This is Abel's integral equation. Test \(x(s) = \sqrt{s}\) with the substitution \(s = tu\):

\[ \begin{aligned} \int_{0}^{t} \frac{\sqrt{s}}{\sqrt{t - s}}\,ds &= t\int_{0}^{1} u^{1/2}(1 - u)^{-1/2}\,du \\ &= t\,B\left(\tfrac{3}{2}, \tfrac{1}{2}\right) = t \cdot \frac{\pi}{2} \end{aligned} \]

since

\[ \begin{aligned} B\left(\tfrac{3}{2}, \tfrac{1}{2}\right) &= \frac{\Gamma(3/2)\Gamma(1/2)}{\Gamma(2)} \\ &= \frac{\sqrt{\pi}}{2} \cdot \sqrt{\pi} = \frac{\pi}{2} \end{aligned} \]

That matches the left-hand side \(\frac{\pi t}{2}\).

By the same computation, \(x = 1\) gives \(2\sqrt{t}\), \(x = t\) gives \(\frac{4}{3}t^{3/2}\), and \(x = \frac{1}{\sqrt{t}}\) gives the constant \(\pi\); none equals \(\frac{\pi t}{2}\).

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32.

The boundary value problem

\[ y'' + \lambda y = x, \; y(0) = 0, \; y'(1) = 0 \]

is equivalent to the Fredholm integral equation.

where \(k(x, t) = \begin{cases} t; & 0 \lt t \lt x \\ x; & x \lt t \lt 1 \end{cases}\)

A.\(y(x) = \frac{1}{6}(x^{3} - 3x)\) \(+ \lambda\int_{0}^{1} k(x, t)\, y(t)\, dt\)
B.\(y(x) = \frac{1}{6}(x^{2} - 2x)\) \(+ \lambda\int_{0}^{1} x\, k(x, t)\, y(t)\, dt\)
C.\(y(x) = \frac{1}{3}(x - 3x^{2})\) \(+ \lambda\int_{0}^{1} (x - t)\, k(x, t)\, y(t)\, dt\)
D.\(y(x) = (x - 2x^{2})\) \(+ \lambda\int_{0}^{1} e^{(x-t)}\, k(x, t)\, y(t)\, dt\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The equivalent equation is \(y(x) = \frac{1}{6}(x^{3} - 3x) + \lambda\int_{0}^{1} k(x, t)\, y(t)\, dt\).

Write the problem as \(y'' = x - \lambda y\) with \(y(0) = 0\), \(y'(1) = 0\). For these boundary conditions the Green's function of \(y'' = g\) is \(-\min(x, t) = -k(x, t)\), so

\[ \begin{aligned} y(x) = &-\int_{0}^{1} k(x, t)\,t\,dt \\ &+ \lambda\int_{0}^{1} k(x, t)\,y(t)\,dt \end{aligned} \]

The first integral is

\[ \begin{aligned} \int_{0}^{x} t^{2}\,dt + x\int_{x}^{1} t\,dt &= \frac{x^{3}}{3} + \frac{x(1 - x^{2})}{2} \\ &= \frac{x}{2} - \frac{x^{3}}{6} \end{aligned} \]

so the free term is \(\frac{x^{3}}{6} - \frac{x}{2} = \frac{1}{6}(x^{3} - 3x)\). Check: it vanishes at 0 and its derivative \(\frac{x^{2} - 1}{2}\) vanishes at 1. The other free terms fail these boundary checks, and their kernels carry extra factors.

Discuss in Forum

33.

Eigenvalues of the Fredholm integral equation,

\[ x(t) = 1 + \lambda \int_{0}^{\pi} \cos(t + s)\, x(s)\, ds \]

are :

A.\(2/\pi, -2/\pi\)
B.\(\pi, -\pi\)
C.1, −1
D.\(\sqrt{2}, -\sqrt{2}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The eigenvalues are \(2/\pi\) and \(-2/\pi\).

Expand the kernel: \(\cos(t + s) = \cos t\cos s - \sin t\sin s\). For the homogeneous equation, \(x(t) = \lambda(A\cos t - B\sin t)\) with \(A = \int_{0}^{\pi} \cos s\, x(s)\,ds\) and \(B = \int_{0}^{\pi} \sin s\, x(s)\,ds\).

On \([0, \pi]\), \(\int \cos^{2} = \int \sin^{2} = \frac{\pi}{2}\) and \(\int \sin s\cos s\,ds = 0\), so

\[ A = \lambda\frac{\pi}{2}A, \qquad B = -\lambda\frac{\pi}{2}B \]

A non-trivial solution needs \(\lambda = \frac{2}{\pi}\) (with \(x = \cos t\)) or \(\lambda = -\frac{2}{\pi}\) (with \(x = \sin t\)).

The pairs \(\pi, -\pi\), \(1, -1\) and \(\sqrt{2}, -\sqrt{2}\) do not satisfy \(\lambda\frac{\pi}{2} = \pm 1\), so they are not eigenvalues.

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34.

The resolvent kernel for Volterra integral equation with kernel :

\[ k(t, s) = a^{t-s}, \quad a \gt 0 \]

is :

A.\(R(t, s; \lambda) = a^{\lambda(t-s)} e^{(t-s)}\)
B.\(R(t, s; \lambda) = a^{t+s} e^{\lambda(t+s)}\)
C.\(R(t, s; \lambda) = a^{t-s} . e^{\lambda(t-s)}\)
D.\(R(t, s; \lambda) = a^{t+s} e^{\lambda(t-s)}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The resolvent kernel is \(R(t, s; \lambda) = a^{t-s}\, e^{\lambda(t-s)}\).

The iterated kernels of \(k(t, s) = a^{t-s}\) are

\[ k_{n}(t, s) = a^{t-s}\,\frac{(t - s)^{n-1}}{(n - 1)!} \]

(each iteration integrates a constant over \([s, t]\)). Summing the Neumann series,

\[ \begin{aligned} R &= \sum_{n \ge 1} \lambda^{n-1} k_{n} \\ &= a^{t-s} \sum_{m \ge 0} \frac{[\lambda(t - s)]^{m}}{m!} = a^{t-s} e^{\lambda(t-s)} \end{aligned} \]

Options B and D contain \(a^{t+s}\), which is not a function of \(t - s\), although every iterated kernel of a difference kernel is. Option A puts \(\lambda\) in the exponent of a instead of e.

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35.

For the integral equation

\[ x(t)=\frac{1}{2}\int_{t}^{1}(t\sqrt{s}-s\sqrt{t})\,x(s)\,ds \]

consider the following statements. The above integral equation is :

(I) Fredholm integral equation of second kind.
(II) Volterra integral equation of first kind.

Then :

A.Only (I) is true
B.Only (II) is true
C.Both (I) and (II) are true
D.Both (I) and (II) are false

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

Both (I) and (II) are false.

The upper limit is 1 but the lower limit is the variable t, so the range of integration depends on t: that makes it a Volterra-type equation, not a Fredholm equation (which has fixed limits). So (I) is false.

The unknown x appears outside the integral as well as inside it, which makes the equation of the second kind. A first-kind equation has the unknown only under the integral sign. So (II) is false as well; the equation is a Volterra equation of the second kind.

Discuss in Forum

36.

Eigen values of the homogeneous Fredholm integral equation

\[ x(t) = \lambda \int_{0}^{2\pi} \sin(t + s)\, x(s)\, ds \]

are :

A.1, −1
B.\(\pi, -\pi\)
C.\(1, \pi\)
D.\(1/\pi, -1/\pi\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The eigenvalues are 1/π and −1/π.

Expand \(\sin(t + s) = \sin t\cos s + \cos t\sin s\), so \(x(t) = \lambda(c_{1}\sin t + c_{2}\cos t)\) with \(c_{1} = \int_{0}^{2\pi}\cos s\, x(s)\, ds\) and \(c_{2} = \int_{0}^{2\pi}\sin s\, x(s)\, ds\). Using \(\int_{0}^{2\pi}\sin^{2} = \int_{0}^{2\pi}\cos^{2} = \pi\) and \(\int_{0}^{2\pi}\sin\cos = 0\):

\[ \begin{aligned} c_{1} &= \lambda\pi c_{2} \\ c_{2} &= \lambda\pi c_{1} \\ c_{1} &= \lambda^{2}\pi^{2}c_{1} \end{aligned} \]

A non-zero solution needs \(\lambda^{2}\pi^{2} = 1\), so \(\lambda = \pm 1/\pi\). The values \(\pm\pi\) are the reciprocals (the eigenvalues of the integral operator itself), and 1 or \(\pi\) do not satisfy the condition.

Linear Integral Equations Questions and Answers

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