Linear Integral Equations Questions and Answers

Q: For a homogeneous Fredholm integral equation with symmetric kernel :\[ x(t) = \lambda \int_{a}^{b} k(t, s)\, x(s)\, ds \]consider the following statements.(I) The eigen values are not real.(II) The eigen functions corresponding to distinct eigen values are orthogonal on \([a, b]\). Then :

Only (II) is true.For a real symmetric kernel \(k(t, s) = k(s, t)\), the Fredholm theory parallels that of a symmetric matrix:The eigenvalues are real. Multiply \(x = \lambda K x\) by \(\bar{x}\) and integrate; symmetry makes \(\int\!\int k\, x \bar{x}\) real, so \(\lambda\) is real. Statement (I), which says they are not real, is false.Eigenfunctions for distinct eigenvalues are orthogonal: if \(x_{1} = \lambda_{1} K x_{1}\) and \(x_{2} = \lambda_{2} K x_{2}\), symmetry gives \((\lambda_{1} - \lambda_{2})\int x_{1} x_{2} = 0\), so \(\int_{a}^{b} x_{1} x_{2} = 0\). Statement (II) is true.So 'only (I)', 'both' and 'both false' are ruled out.

Q: The solution of the integral equation\[ x(t) = t + \int_{0}^{t} (s - t)\, x(s)\, ds \]with the help of resolvent kernel is :

The solution is \(x(t) = \sin t\).The kernel is \(s - t = -(t - s)\). The resolvent route gives the same result as differentiating, which is quicker:\[ \begin{aligned} x(t) &= t - \int_{0}^{t} (t - s)\, x(s)\, ds \\ x'(t) &= 1 - \int_{0}^{t} x(s)\, ds \\ x''(t) &= -x(t) \end{aligned} \]with \(x(0) = 0\) and \(x'(0) = 1\). The solution is \(\sin t\). (The resolvent kernel of \(-(t - s)\) is \(-\sin(t - s)\), and \(x = t - \int_{0}^{t}\sin(t - s)\, s\, ds = \sin t\).)\(\cos t\), \(e^{t}\) and 1 all fail \(x(0) = 0\), which the equation forces at \(t = 0\) because the integral vanishes there.

Q: Eigen values of the homogeneous Fredholm integral equation\[ x(t) = \lambda \int_{0}^{1} (1 - 3ts)\, x(s)\, ds \]are :

The eigenvalues are 2 and -2.The kernel \(1 - 3ts\) is separable, so \(x(t) = \lambda(c_{1} - 3c_{2}t)\) with \(c_{1} = \int_{0}^{1} x\) and \(c_{2} = \int_{0}^{1} s\,x\). Substituting back:\[ \begin{aligned} c_{1} &= \lambda\left(c_{1} - \tfrac{3}{2}c_{2}\right) \\ c_{2} &= \lambda\left(\tfrac{1}{2}c_{1} - c_{2}\right) \end{aligned} \]A nonzero solution needs the determinant of the homogeneous system to vanish:\[ \begin{aligned} (1 - \lambda)(1 + \lambda) + \tfrac{3}{4}\lambda^{2} &= 0 \\ 1 - \tfrac{1}{4}\lambda^{2} &= 0 \\ \lambda &= \pm 2 \end{aligned} \]The pairs (1, 2), (2, -1) and (1, -2) do not satisfy \(\lambda^{2} = 4\).

Q: The following integral equation\[ x(t)=\sin t+\lambda\int_{0}^{2\pi}\sin(t+s)\,x(s)\,ds \]is :

It is a Fredholm integral equation of the second kind.The limits of integration are fixed (0 to \(2\pi\)), which makes it Fredholm; a Volterra equation has a variable upper limit such as t. The unknown x(t) also appears outside the integral, which makes it of the second kind:\[ x(t) = f(t) + \lambda\int_{a}^{b} K(t, s)\, x(s)\, ds \]In an equation of the first kind the unknown appears only under the integral sign, \(f(t) = \lambda\int K(t, s)x(s)\, ds\).

Q: The nontrivial solution of the integral equation\[ x(t) = \lambda \int_{0}^{1} e^{t + s}\, x(s)\, ds \]is :

The nontrivial solution is et (up to a constant factor).The kernel \(e^{t + s} = e^{t}e^{s}\) is separable. Write\[ \begin{aligned} x(t) &= \lambda e^{t}\int_{0}^{1} e^{s}x(s)\, ds \\ &= c\lambda e^{t} \end{aligned} \]so every solution is a multiple of \(e^{t}\). Substituting \(x = Ce^{t}\) back, \(c = \int_{0}^{1} e^{2s}C\, ds\), which allows non-zero C exactly for the eigenvalue \(\lambda = \frac{2}{e^{2} - 1}\).The forms t, \(t^{2}\) and \(e^{t^{2}}\) are not multiples of \(e^{t}\), so they cannot satisfy the equation.

Q: The Volterra integral equation\[ x(t) = t + \int_{0}^{t} (s - t)\, x(s)\, ds \]is equivalent to the initial value problem :

The equivalent problem is x'' + x = 0, x(0) = 0, x'(0) = 1.Differentiate \(x(t) = t + \int_{0}^{t}(s - t)x(s)\, ds\) using Leibniz's rule:\[ \begin{aligned} x'(t) &= 1 + (t - t)x(t) - \int_{0}^{t} x(s)\, ds \\ &= 1 - \int_{0}^{t} x(s)\, ds \\ x''(t) &= -x(t) \end{aligned} \]At \(t = 0\) the integrals vanish, so \(x(0) = 0\) and \(x'(0) = 1\). The solution is \(x = \sin t\).The options with \(x'' - x\) have the wrong sign, those with \(x'\) in place of x differentiate wrongly, and \(x(0) = 1,\ x'(0) = 0\) swaps the initial values.

Q: For a homogeneous Fredholm integral equation with separable kernel :\[ x(t) = \lambda \int_{a}^{b} k(t, s)\, x(s)\, ds, \]consider the following statements :(I) If Fredholm determinant \(D(\lambda) \neq 0\), then integral equation has only trivial solution i.e. \(x(t) = 0\).(II) If Fredholm determinant \(D(\lambda) = 0\), then integral equation has infinitely many solutions.Then :

Both (I) and (II) are true.With a separable kernel \(k(t, s) = \sum a_{i}(t)b_{i}(s)\), the homogeneous equation reduces to a homogeneous linear system \((I - \lambda A)c = 0\) for the coefficients, and \(D(\lambda) = \det(I - \lambda A)\).If \(D(\lambda) \neq 0\), the system has only the zero solution, so \(x(t) = 0\) is the only solution: (I) is true.If \(D(\lambda) = 0\), the system has a non-zero solution c, giving a non-zero x(t). Every scalar multiple of it is also a solution, so there are infinitely many: (II) is true.

Q: The Neumann series solution of the Volterra integral equation\[ x(t) = 1 + t + \int_{0}^{t} (t - s)\, x(s)\, ds \]is :

The solution is x(t) = et.Differentiate twice:\[ \begin{aligned} x'(t) &= 1 + \int_{0}^{t} x(s)\, ds \\ x''(t) &= x(t) \end{aligned} \]with \(x(0) = 1\) and \(x'(0) = 1\), whose solution is \(e^{t}\). The Neumann series gives the same: starting from \(x_{0} = 1 + t\), each iteration adds the next terms \(t^{2}/2!,\ t^{3}/3!, \ldots\), building up \(\sum t^{n}/n! = e^{t}\).\(\cos t\) and \(\sin t\) solve \(x'' = -x\), not \(x'' = x\), and \(\log t\) is not even defined at t = 0.

Q: For the homogeneous integral equation\[ \begin{aligned} x(t) = \lambda \int_{0}^{1} &\left(s\sqrt{t} - t\sqrt{s}\right) \\ &\times x(s)\, ds, \end{aligned} \]consider the following statements.(I) This equation does not have real eigenvalues.(II) The eigenvalues of this equation are 1 and −1.Then :

Only (I) is true: the equation has no real eigenvalues.The kernel is separable: \(x(t) = \lambda(\sqrt{t}\,A - t\,B)\), with \(A = \int_{0}^{1} s\,x(s)\,ds\) and \(B = \int_{0}^{1} \sqrt{s}\,x(s)\,ds\). Substituting back:\[ \begin{aligned} A &= \lambda\left(\tfrac{2}{5}A - \tfrac{1}{3}B\right) \\ B &= \lambda\left(\tfrac{1}{2}A - \tfrac{2}{5}B\right) \end{aligned} \]A non-zero solution needs the determinant of this system to vanish:\[ \begin{aligned} &\left(1 - \tfrac{2\lambda}{5}\right)\left(1 + \tfrac{2\lambda}{5}\right) + \tfrac{\lambda^{2}}{6} \\ &= 1 + \tfrac{\lambda^{2}}{150} = 0 \end{aligned} \]so \(\lambda^{2} = -150\) and there is no real eigenvalue. This fits the kernel being skew-symmetric (\(K(t, s) = -K(s, t)\)). In particular 1 and \(-1\) are not eigenvalues, so (II) is false.

21.

The initial value problem

\[ x''(t) - x(t) = 0,\ x(0) = 1,\ x'(0) = 1 \]

is equivalent to the integral equation :

A.\(x(t) = 1 + \int_{0}^{t} (t + s)^{2} x(s)\, ds\)
B.\(x(t) = t + \int_{0}^{t} (t + s)\, x(s)\, ds\)
C.\(x(t) = 1 + \int_{0}^{t} (t - s)^{2}\, x(s)\, ds\)
D.\(x(t) = 1 + t + \int_{0}^{t} (t - s)\, x(s)\, ds\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The equivalent integral equation is \(x(t) = 1 + t + \int_{0}^{t} (t - s)\, x(s)\, ds\).

Integrate \(x'' = x\) twice from 0, using the initial conditions:

\[ \begin{aligned} x'(t) &= 1 + \int_{0}^{t} x(s)\, ds \\ x(t) &= 1 + t + \int_{0}^{t}\!\int_{0}^{\tau} x(s)\, ds\, d\tau \\ &= 1 + t + \int_{0}^{t} (t - s)\, x(s)\, ds \end{aligned} \]

The last step is the standard formula that turns a repeated integral into a single one with kernel \(t - s\).

Check: \(x = e^{t}\) solves the initial value problem, and \(1 + t + \int_{0}^{t}(t - s)e^{s} ds = 1 + t + (e^{t} - 1 - t) = e^{t}\). The other options use \((t + s)\) or squared kernels, or drop the \(t\) that comes from \(x'(0) = 1\).

Discuss in Forum

22.

For a homogeneous Fredholm integral equation with symmetric kernel :

\[ x(t) = \lambda \int_{a}^{b} k(t, s)\, x(s)\, ds \]

consider the following statements.

(I) The eigen values are not real.
(II) The eigen functions corresponding to distinct eigen values are orthogonal on \([a, b]\). 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 - 2023, SET Exam - 2023,

Answer: B

Only (II) is true.

For a real symmetric kernel \(k(t, s) = k(s, t)\), the Fredholm theory parallels that of a symmetric matrix:

The eigenvalues are real. Multiply \(x = \lambda K x\) by \(\bar{x}\) and integrate; symmetry makes \(\int\!\int k\, x \bar{x}\) real, so \(\lambda\) is real. Statement (I), which says they are not real, is false.
Eigenfunctions for distinct eigenvalues are orthogonal: if \(x_{1} = \lambda_{1} K x_{1}\) and \(x_{2} = \lambda_{2} K x_{2}\), symmetry gives \((\lambda_{1} - \lambda_{2})\int x_{1} x_{2} = 0\), so \(\int_{a}^{b} x_{1} x_{2} = 0\). Statement (II) is true.

So 'only (I)', 'both' and 'both false' are ruled out.

Discuss in Forum

23.

The solution of the integral equation

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

with the help of resolvent kernel is :

A.\(x(t) = \cos t\)
B.\(x(t) = \sin t\)
C.\(x(t) = e^{t}\)
D.\(x(t) = 1\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The solution is \(x(t) = \sin t\).

The kernel is \(s - t = -(t - s)\). The resolvent route gives the same result as differentiating, which is quicker:

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

with \(x(0) = 0\) and \(x'(0) = 1\). The solution is \(\sin t\). (The resolvent kernel of \(-(t - s)\) is \(-\sin(t - s)\), and \(x = t - \int_{0}^{t}\sin(t - s)\, s\, ds = \sin t\).)

\(\cos t\), \(e^{t}\) and 1 all fail \(x(0) = 0\), which the equation forces at \(t = 0\) because the integral vanishes there.

Discuss in Forum

24.

Eigen values of the homogeneous Fredholm integral equation

\[ x(t) = \lambda \int_{0}^{1} (1 - 3ts)\, x(s)\, ds \]

are :

A.1, 2
B.2, -2
C.2, -1
D.1, -2

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The eigenvalues are 2 and -2.

The kernel \(1 - 3ts\) is separable, so \(x(t) = \lambda(c_{1} - 3c_{2}t)\) with \(c_{1} = \int_{0}^{1} x\) and \(c_{2} = \int_{0}^{1} s\,x\). Substituting back:

\[ \begin{aligned} c_{1} &= \lambda\left(c_{1} - \tfrac{3}{2}c_{2}\right) \\ c_{2} &= \lambda\left(\tfrac{1}{2}c_{1} - c_{2}\right) \end{aligned} \]

A nonzero solution needs the determinant of the homogeneous system to vanish:

\[ \begin{aligned} (1 - \lambda)(1 + \lambda) + \tfrac{3}{4}\lambda^{2} &= 0 \\ 1 - \tfrac{1}{4}\lambda^{2} &= 0 \\ \lambda &= \pm 2 \end{aligned} \]

The pairs (1, 2), (2, -1) and (1, -2) do not satisfy \(\lambda^{2} = 4\).

Discuss in Forum

25.

The following integral equation

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

is :

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

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

It is a Fredholm integral equation of the second kind.

The limits of integration are fixed (0 to \(2\pi\)), which makes it Fredholm; a Volterra equation has a variable upper limit such as t. The unknown x(t) also appears outside the integral, which makes it of the second kind:

\[ x(t) = f(t) + \lambda\int_{a}^{b} K(t, s)\, x(s)\, ds \]

In an equation of the first kind the unknown appears only under the integral sign, \(f(t) = \lambda\int K(t, s)x(s)\, ds\).

Discuss in Forum

26.

The nontrivial solution of the integral equation

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

is :

A.\(t\)
B.\(t^{2}\)
C.\(e^{t^{2}}\)
D.\(e^{t}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The nontrivial solution is e^t (up to a constant factor).

The kernel \(e^{t + s} = e^{t}e^{s}\) is separable. Write

\[ \begin{aligned} x(t) &= \lambda e^{t}\int_{0}^{1} e^{s}x(s)\, ds \\ &= c\lambda e^{t} \end{aligned} \]

so every solution is a multiple of \(e^{t}\). Substituting \(x = Ce^{t}\) back, \(c = \int_{0}^{1} e^{2s}C\, ds\), which allows non-zero C exactly for the eigenvalue \(\lambda = \frac{2}{e^{2} - 1}\).

The forms t, \(t^{2}\) and \(e^{t^{2}}\) are not multiples of \(e^{t}\), so they cannot satisfy the equation.

Discuss in Forum

27.

The Volterra integral equation

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

is equivalent to the initial value problem :

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

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The equivalent problem is x'' + x = 0, x(0) = 0, x'(0) = 1.

Differentiate \(x(t) = t + \int_{0}^{t}(s - t)x(s)\, ds\) using Leibniz's rule:

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

At \(t = 0\) the integrals vanish, so \(x(0) = 0\) and \(x'(0) = 1\). The solution is \(x = \sin t\).

The options with \(x'' - x\) have the wrong sign, those with \(x'\) in place of x differentiate wrongly, and \(x(0) = 1,\ x'(0) = 0\) swaps the initial values.

Discuss in Forum

28.

For a homogeneous Fredholm integral equation with separable kernel :

\[ x(t) = \lambda \int_{a}^{b} k(t, s)\, x(s)\, ds, \]

consider the following statements :

(I) If Fredholm determinant \(D(\lambda) \neq 0\), then integral equation has only trivial solution i.e. \(x(t) = 0\).
(II) If Fredholm determinant \(D(\lambda) = 0\), then integral equation has infinitely many solutions.

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 - 2024, SET Exam - 2024,

Answer: C

Both (I) and (II) are true.

With a separable kernel \(k(t, s) = \sum a_{i}(t)b_{i}(s)\), the homogeneous equation reduces to a homogeneous linear system \((I - \lambda A)c = 0\) for the coefficients, and \(D(\lambda) = \det(I - \lambda A)\).

If \(D(\lambda) \neq 0\), the system has only the zero solution, so \(x(t) = 0\) is the only solution: (I) is true.

If \(D(\lambda) = 0\), the system has a non-zero solution c, giving a non-zero x(t). Every scalar multiple of it is also a solution, so there are infinitely many: (II) is true.

Discuss in Forum

29.

The Neumann series solution of the Volterra integral equation

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

is :

A.\(x(t) = \log t\)
B.\(x(t) = \cos t\)
C.\(x(t) = \sin t\)
D.\(x(t) = e^{t}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The solution is x(t) = e^t.

Differentiate twice:

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

with \(x(0) = 1\) and \(x'(0) = 1\), whose solution is \(e^{t}\). The Neumann series gives the same: starting from \(x_{0} = 1 + t\), each iteration adds the next terms \(t^{2}/2!,\ t^{3}/3!, \ldots\), building up \(\sum t^{n}/n! = e^{t}\).

\(\cos t\) and \(\sin t\) solve \(x'' = -x\), not \(x'' = x\), and \(\log t\) is not even defined at t = 0.

Discuss in Forum

30.

For the homogeneous integral equation

\[ \begin{aligned} x(t) = \lambda \int_{0}^{1} &\left(s\sqrt{t} - t\sqrt{s}\right) \\ &\times x(s)\, ds, \end{aligned} \]

consider the following statements.

(I) This equation does not have real eigenvalues.
(II) The eigenvalues of this equation are 1 and −1.

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: A

Only (I) is true: the equation has no real eigenvalues.

The kernel is separable: \(x(t) = \lambda(\sqrt{t}\,A - t\,B)\), with \(A = \int_{0}^{1} s\,x(s)\,ds\) and \(B = \int_{0}^{1} \sqrt{s}\,x(s)\,ds\). Substituting back:

\[ \begin{aligned} A &= \lambda\left(\tfrac{2}{5}A - \tfrac{1}{3}B\right) \\ B &= \lambda\left(\tfrac{1}{2}A - \tfrac{2}{5}B\right) \end{aligned} \]

A non-zero solution needs the determinant of this system to vanish:

\[ \begin{aligned} &\left(1 - \tfrac{2\lambda}{5}\right)\left(1 + \tfrac{2\lambda}{5}\right) + \tfrac{\lambda^{2}}{6} \\ &= 1 + \tfrac{\lambda^{2}}{150} = 0 \end{aligned} \]

so \(\lambda^{2} = -150\) and there is no real eigenvalue. This fits the kernel being skew-symmetric (\(K(t, s) = -K(s, t)\)). In particular 1 and \(-1\) are not eigenvalues, so (II) is false.

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