Higher Order Linear Differential Equations Questions and Answers

Q: If a transformation \(y = uv\) transforms the given differential equation \(f(x)y'' - 4f'(x)y' + g(x)y = 0\) into the equation of the form \(v'' + h(x)v = 0\), then \(u\) must be

The multiplier is \(u = f^{2}\).Write the equation as \(y'' + Py' + Qy = 0\) with \(P = -4f'/f\). Substituting \(y = uv\) gives\[ uv'' + (2u' + Pu)v' + (u'' + Pu' + Qu)v = 0, \]and the \(v'\) term disappears only if \(2u' + Pu = 0\), i.e. \(u = e^{-\frac{1}{2}\int P\,dx}\). Here\[ -\frac{1}{2}\int P\,dx = 2\int \frac{f'}{f}\,dx = 2\ln f, \]so \(u = f^{2}\). The reciprocal \(1/f^{2}\) comes from a sign slip in the exponent, and \(xf\) and \(1/2f\) do not remove the first-derivative term.

Q: Let \(f(x)\) and \(xf(x)\) be the particular solutions of a differential equation\[ y'' + R(x)y' + S(x)y = 0. \]Then the solution of the differential equation \(y'' + R(x)y' + S(x)y = f(x)\) is

The solution is \(y = \left(\frac{x^{2}}{2} + \alpha x + \beta\right)f(x)\).Try \(y = vf\). Substituting and using that \(f\) solves the homogeneous equation leaves\[ fv'' + (2f' + Rf)v' = f. \]Because \(xf\) is also a solution, \(v = x\) must make the left side vanish in the homogeneous case, which gives \(2f' + Rf = 0\). The equation therefore reduces to \(fv'' = f\), i.e. \(v'' = 1\), so\[ v = \frac{x^{2}}{2} + \alpha x + \beta. \]The \(\alpha x + \beta\) part reproduces the general homogeneous solution \(\alpha xf + \beta f\), and \(\frac{x^{2}}{2}f\) is the particular solution. \(-\frac{x^{2}}{2}\) gives \(v'' = -1\), the wrong sign; \(-x^{2}\) and \(x^{3}\) give \(v'' = -2\) and \(6x\), not 1.

Q: Let \(\phi\) and \(\psi\) be two linearly independent solutions of the ordinary differential equation\[ y'' + (2 - \cos x)\,y = 0, \quad x \in \mathbb{R}. \]Let \(\alpha, \beta \in \mathbb{R}\) be such that \(\alpha \lt \beta\), \(\phi(\alpha) = \phi(\beta) = 0\) and \(\phi(x) \ne 0\) for all \(x \in (\alpha, \beta)\).Consider the following statements:P: \(\phi'(\alpha)\phi'(\beta) \gt 0\).Q: \(\phi(x)\psi(x) \ne 0\) for all \(x \in (\alpha, \beta)\).Then

Both P and Q are FALSE.P: \(\phi\) has consecutive zeros at \(\alpha\) and \(\beta\) and keeps one sign between them; say \(\phi \gt 0\) on \((\alpha, \beta)\). Then \(\phi'(\alpha) \ge 0\) and \(\phi'(\beta) \le 0\), and neither can be zero, since a solution with \(\phi = \phi' = 0\) at a point is identically zero. So \(\phi'(\alpha)\phi'(\beta) \lt 0\), and P is false.Q: by the Sturm separation theorem, between two consecutive zeros of \(\phi\) the linearly independent solution \(\psi\) has exactly one zero. At that point \(\phi\psi = 0\), so Q is false.

Q: Let \(y(x)\) be the solution of the differential equation\[ x^{2}y'' + 7xy' + 9y = x^{-3}\log_{e} x, \quad x \gt 0, \]satisfying \(y(1) = 0\) and \(y'(1) = 0\). Then, the value of \(y(e)\) is equal to

The value is \(\dfrac{1}{6}e^{-3}\).This is a Cauchy-Euler equation. With \(x = e^{t}\) it becomes\[ \begin{aligned} y'' + 6y' + 9y &= te^{-3t} \\ (D + 3)^{2}y &= te^{-3t} \end{aligned} \]The root -3 is double, so try \(y_{p} = At^{3}e^{-3t}\): \((D + 3)^{2}(t^{3}e^{-3t}) = 6te^{-3t}\), giving A = 1/6. Thus\[ y = (c_{1} + c_{2}t)e^{-3t} + \tfrac{1}{6}t^{3}e^{-3t} \]x = 1 is t = 0, and the conditions give \(c_{1} = 0\) and \(c_{2} = 0\). At x = e, t = 1, so \(y(e) = \tfrac{1}{6}e^{-3}\). Missing the factor 6 from the double root leads to the other fractions.

Q: Let \(y_{1}(x)\) and \(y_{2}(x)\) be the two linearly independent solutions of the differential equation\[ (1 + x^{2})y'' - xy' + (\cos^{2} x)y = 0, \]satisfying the initial conditions\[ \begin{aligned} &y_{1}(0) = 3,\ y_{1}'(0) = -1 \ \text{and} \\ &y_{2}(0) = -5,\ y_{2}'(0) = 2. \end{aligned} \]Define\[ W(x) = \begin{vmatrix} y_{1}(x) & y_{2}(x) \\ y_{1}'(x) & y_{2}'(x) \end{vmatrix}. \]Then, the value of \(W\left(\frac{1}{2}\right)\) is

\(W(1/2) = \dfrac{\sqrt{5}}{2}\).Abel's formula: for \(y'' + p(x)y' + q(x)y = 0\), \(W(x) = W(0)\exp\left(-\int_{0}^{x} p\right)\). Dividing by \(1 + x^{2}\), \(p(x) = -x/(1 + x^{2})\), so\[ \begin{aligned} W(x) &= W(0)\exp\left(\int_{0}^{x}\frac{s}{1 + s^{2}}ds\right) = W(0)\sqrt{1 + x^{2}} \\ W(0) &= (3)(2) - (-5)(-1) = 1 \\ W(\tfrac{1}{2}) &= \sqrt{1 + \tfrac{1}{4}} = \frac{\sqrt{5}}{2} \end{aligned} \]The coefficient of y plays no part in Abel's formula. \(2/\sqrt{5}\) is the reciprocal, from a sign slip in the exponent, and \(\sqrt{5}/4\) halves the answer.

Q: Consider the differential equation\[ \frac{d^{2}y}{dx^{2}} + 3\frac{dy}{dx} + 2y = \sin(e^{x}). \]Which ONE of the following is a particular solution of this differential equation?

A particular solution is \(-e^{-2x}\sin(e^{x})\).The operator factors as \((D + 1)(D + 2)\). Put \(u = y' + 2y\); then \(u' + u = y'' + 3y' + 2y = \sin(e^{x})\). Solve the two first-order equations in turn:\[ \begin{aligned} (e^{x}u)' &= e^{x}\sin(e^{x}) \\ e^{x}u &= -\cos(e^{x}) \\ (e^{2x}y)' &= e^{2x}u = -e^{x}\cos(e^{x}) \\ e^{2x}y &= -\sin(e^{x}) \end{aligned} \]Taking both constants of integration as zero gives \(y = -e^{-2x}\sin(e^{x})\). The substitution \(w = e^{x}\), with \(dw = e^{x}dx\), is what makes each integral elementary.The other options do not satisfy the equation. Substituting \(-e^{-x}\sin(e^{x})\) gives \(e^{x}\sin(e^{x}) - 2\cos(e^{x})\) on the left, not \(\sin(e^{x})\). Substituting \(-e^{-2x}(\sin(e^{x}) + \cos(e^{x}))\) gives \(\sin(e^{x}) + \cos(e^{x})\): the added cosine term survives as an extra \(\cos(e^{x})\).

Q: Consider the following linear second order differential equation\[ \frac{d^{2}y}{dt^{2}} + \omega^{2}y = 0, \]where \(\omega\) is a positive constant. The boundary conditions are \(\left.\frac{dy}{dt}\right|_{t=0} = 1\), and \(y(t = 0) = \frac{1}{2}\).Which of the following options gives the value of \(y(t = \frac{\pi}{2\omega})\)?

The value is \(\frac{1}{\omega}\).The general solution of \(\ddot{y} + \omega^2 y = 0\) is \(y = A\cos\omega t + B\sin\omega t\). The two conditions fix the constants:\[ \begin{aligned} y(0) &= A = \tfrac{1}{2} \\ \dot{y}(0) &= B\omega = 1, \quad B = \tfrac{1}{\omega} \end{aligned} \]At \(t = \frac{\pi}{2\omega}\), \(\omega t = \frac{\pi}{2}\), so the cosine term vanishes and the sine term is at its maximum:\[ \begin{aligned} y\left(\tfrac{\pi}{2\omega}\right) &= \tfrac{1}{2}\cos\tfrac{\pi}{2} + \tfrac{1}{\omega}\sin\tfrac{\pi}{2} \\ &= \tfrac{1}{\omega} \end{aligned} \]0 is what you get by keeping only the cosine term, and 1 by forgetting to divide the initial slope by \(\omega\).

Q: The differential equation obtained by eliminating the arbitrary constants A and \(\phi\) from :\[ y = Ae^{-\alpha t} \sin(\omega t + \phi) \]is :

The equation is \(\dfrac{d^{2}y}{dt^{2}} + 2\alpha\dfrac{dy}{dt} + (\alpha^{2} + \omega^{2})y = 0\).Write \(y = \text{Im}(Ce^{(-\alpha + i\omega)t})\) with \(C = Ae^{i\phi}\). So y is a solution of the constant-coefficient equation whose characteristic roots are \(-\alpha \pm i\omega\). Those roots satisfy\[ \begin{aligned} (m + \alpha)^{2} + \omega^{2} &= 0 \\ m^{2} + 2\alpha m + \alpha^{2} + \omega^{2} &= 0 \end{aligned} \]which corresponds to \(y'' + 2\alpha y' + (\alpha^{2} + \omega^{2})y = 0\). The two arbitrary constants A and \(\phi\) are exactly the two constants of the general solution of this second-order equation, so eliminating them gives it.The options with a factor t in the middle term have variable coefficients and do not have damped sinusoids as solutions. \(\alpha^{2} - \omega^{2}\) would correspond to real roots \(-\alpha \pm \omega\), giving exponentials rather than oscillations.

Q: Let \(\phi_{1}, \phi_{2}\) be linearly independent solutions of the linear differential equation :\[ y'' + a_{1}y' + a_{2}y = 0, \]where \(a_{1}, a_{2}\) are constants, then the Wronskian \(W(\phi_{1}, \phi_{2})\) is constant if and only if :

The Wronskian is constant if and only if \(a_{1} = 0\).By Abel's formula, the Wronskian of two solutions of \(y'' + a_{1}y' + a_{2}y = 0\) satisfies \(W' = -a_{1}W\), so\[ W(x) = W(x_{0})\, e^{-a_{1}(x - x_{0})} \]Because \(\phi_{1}, \phi_{2}\) are linearly independent, \(W(x_{0}) \ne 0\). The exponential is constant exactly when \(a_{1} = 0\).The coefficient \(a_{2}\) does not enter Abel's formula at all, so conditions on \(a_{2}\) are irrelevant; and \(a_{1} \ne 0\) makes the Wronskian grow or decay exponentially, not stay constant.

Q: An nth order linear ordinary differential equation :

The answer is has exactly \(n\) linearly independent solutions.For a homogeneous linear equation \(y^{(n)} + a_{1}(x)y^{(n-1)} + \cdots + a_{n}(x)y = 0\) with continuous coefficients, the solution set is a vector space of dimension exactly \(n\). The existence-uniqueness theorem gives \(n\) independent solutions (one for each choice of initial data \(e_{1}, \ldots, e_{n}\)), and any \(n + 1\) solutions are dependent.The paper does not say 'homogeneous', but that is the standard textbook statement it rests on. 'At most \(n\)' and 'minimum \(n\)' are each half of the truth, and 'less than \(n\)' is false.

If a transformation \(y = uv\) transforms the given differential equation \(f(x)y'' - 4f'(x)y' + g(x)y = 0\) into the equation of the form \(v'' + h(x)v = 0\), then \(u\) must be

A.\(1/f^{2}\)
B.\(xf\)
C.\(1/2f\)
D.\(f^{2}\)

Exam Year(s)

GATE Mathematics - 2012,

Answer: D

The multiplier is \(u = f^{2}\).

Write the equation as \(y'' + Py' + Qy = 0\) with \(P = -4f'/f\). Substituting \(y = uv\) gives

\[ uv'' + (2u' + Pu)v' + (u'' + Pu' + Qu)v = 0, \]

and the \(v'\) term disappears only if \(2u' + Pu = 0\), i.e. \(u = e^{-\frac{1}{2}\int P\,dx}\). Here

\[ -\frac{1}{2}\int P\,dx = 2\int \frac{f'}{f}\,dx = 2\ln f, \]

so \(u = f^{2}\). The reciprocal \(1/f^{2}\) comes from a sign slip in the exponent, and \(xf\) and \(1/2f\) do not remove the first-derivative term.

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Let \(f(x)\) and \(xf(x)\) be the particular solutions of a differential equation

\[ y'' + R(x)y' + S(x)y = 0. \]

Then the solution of the differential equation \(y'' + R(x)y' + S(x)y = f(x)\) is

A.\(y = \left(-\frac{x^{2}}{2} + \alpha x + \beta\right)f(x)\)
B.\(y = \left(\frac{x^{2}}{2} + \alpha x + \beta\right)f(x)\)
C.\(y = (-x^{2} + \alpha x + \beta)f(x)\)
D.\(y = (x^{3} + \alpha x + \beta)f(x)\)

Exam Year(s)

GATE Mathematics - 2012,

Answer: B

The solution is \(y = \left(\frac{x^{2}}{2} + \alpha x + \beta\right)f(x)\).

Try \(y = vf\). Substituting and using that \(f\) solves the homogeneous equation leaves

\[ fv'' + (2f' + Rf)v' = f. \]

Because \(xf\) is also a solution, \(v = x\) must make the left side vanish in the homogeneous case, which gives \(2f' + Rf = 0\). The equation therefore reduces to \(fv'' = f\), i.e. \(v'' = 1\), so

\[ v = \frac{x^{2}}{2} + \alpha x + \beta. \]

The \(\alpha x + \beta\) part reproduces the general homogeneous solution \(\alpha xf + \beta f\), and \(\frac{x^{2}}{2}f\) is the particular solution. \(-\frac{x^{2}}{2}\) gives \(v'' = -1\), the wrong sign; \(-x^{2}\) and \(x^{3}\) give \(v'' = -2\) and \(6x\), not 1.

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Let \(\phi\) and \(\psi\) be two linearly independent solutions of the ordinary differential equation

\[ y'' + (2 - \cos x)\,y = 0, \quad x \in \mathbb{R}. \]

Let \(\alpha, \beta \in \mathbb{R}\) be such that \(\alpha \lt \beta\), \(\phi(\alpha) = \phi(\beta) = 0\) and \(\phi(x) \ne 0\) for all \(x \in (\alpha, \beta)\).

Consider the following statements:

P: \(\phi'(\alpha)\phi'(\beta) \gt 0\).
Q: \(\phi(x)\psi(x) \ne 0\) for all \(x \in (\alpha, \beta)\).

Then

A.P is TRUE and Q is FALSE
B.P is FALSE and Q is TRUE
C.both P and Q are FALSE
D.both P and Q are TRUE

Exam Year(s)

GATE Mathematics - 2023,

Answer: C

Both P and Q are FALSE.

P: \(\phi\) has consecutive zeros at \(\alpha\) and \(\beta\) and keeps one sign between them; say \(\phi \gt 0\) on \((\alpha, \beta)\). Then \(\phi'(\alpha) \ge 0\) and \(\phi'(\beta) \le 0\), and neither can be zero, since a solution with \(\phi = \phi' = 0\) at a point is identically zero. So \(\phi'(\alpha)\phi'(\beta) \lt 0\), and P is false.

Q: by the Sturm separation theorem, between two consecutive zeros of \(\phi\) the linearly independent solution \(\psi\) has exactly one zero. At that point \(\phi\psi = 0\), so Q is false.

Discuss in Forum

Let \(y(x)\) be the solution of the differential equation

\[ x^{2}y'' + 7xy' + 9y = x^{-3}\log_{e} x, \quad x \gt 0, \]

satisfying \(y(1) = 0\) and \(y'(1) = 0\). Then, the value of \(y(e)\) is equal to

A.\(\dfrac{1}{3}e^{-3}\)
B.\(\dfrac{1}{6}e^{-3}\)
C.\(\dfrac{2}{3}e^{-3}\)
D.\(\dfrac{1}{2}e^{-3}\)

Exam Year(s)

GATE Mathematics - 2025,

Answer: B

The value is \(\dfrac{1}{6}e^{-3}\).

This is a Cauchy-Euler equation. With \(x = e^{t}\) it becomes

\[ \begin{aligned} y'' + 6y' + 9y &= te^{-3t} \\ (D + 3)^{2}y &= te^{-3t} \end{aligned} \]

The root -3 is double, so try \(y_{p} = At^{3}e^{-3t}\): \((D + 3)^{2}(t^{3}e^{-3t}) = 6te^{-3t}\), giving A = 1/6. Thus

\[ y = (c_{1} + c_{2}t)e^{-3t} + \tfrac{1}{6}t^{3}e^{-3t} \]

x = 1 is t = 0, and the conditions give \(c_{1} = 0\) and \(c_{2} = 0\). At x = e, t = 1, so \(y(e) = \tfrac{1}{6}e^{-3}\). Missing the factor 6 from the double root leads to the other fractions.

Discuss in Forum

Let \(y_{1}(x)\) and \(y_{2}(x)\) be the two linearly independent solutions of the differential equation

\[ (1 + x^{2})y'' - xy' + (\cos^{2} x)y = 0, \]

satisfying the initial conditions

\[ \begin{aligned} &y_{1}(0) = 3,\ y_{1}'(0) = -1 \ \text{and} \\ &y_{2}(0) = -5,\ y_{2}'(0) = 2. \end{aligned} \]

Define

\[ W(x) = \begin{vmatrix} y_{1}(x) & y_{2}(x) \\ y_{1}'(x) & y_{2}'(x) \end{vmatrix}. \]

Then, the value of \(W\left(\frac{1}{2}\right)\) is

A.\(\dfrac{\sqrt{5}}{4}\)
B.\(\dfrac{\sqrt{5}}{2}\)
C.\(\dfrac{2}{\sqrt{5}}\)
D.\(\dfrac{4}{\sqrt{5}}\)

Exam Year(s)

GATE Mathematics - 2025,

Answer: B

\(W(1/2) = \dfrac{\sqrt{5}}{2}\).

Abel's formula: for \(y'' + p(x)y' + q(x)y = 0\), \(W(x) = W(0)\exp\left(-\int_{0}^{x} p\right)\). Dividing by \(1 + x^{2}\), \(p(x) = -x/(1 + x^{2})\), so

\[ \begin{aligned} W(x) &= W(0)\exp\left(\int_{0}^{x}\frac{s}{1 + s^{2}}ds\right) = W(0)\sqrt{1 + x^{2}} \\ W(0) &= (3)(2) - (-5)(-1) = 1 \\ W(\tfrac{1}{2}) &= \sqrt{1 + \tfrac{1}{4}} = \frac{\sqrt{5}}{2} \end{aligned} \]

The coefficient of y plays no part in Abel's formula. \(2/\sqrt{5}\) is the reciprocal, from a sign slip in the exponent, and \(\sqrt{5}/4\) halves the answer.

Discuss in Forum

Consider the differential equation

\[ \frac{d^{2}y}{dx^{2}} + 3\frac{dy}{dx} + 2y = \sin(e^{x}). \]

Which ONE of the following is a particular solution of this differential equation?

A.\(-e^{-x}\sin(e^{x})\)
B.\(-e^{-2x}\sin(e^{x})\)
C.\(-e^{-x}(\sin(e^{x}) + \cos(e^{x}))\)
D.\(-e^{-2x}(\sin(e^{x}) + \cos(e^{x}))\)

Exam Year(s)

IIT JAM - 2026,

Answer: B

A particular solution is \(-e^{-2x}\sin(e^{x})\).

The operator factors as \((D + 1)(D + 2)\). Put \(u = y' + 2y\); then \(u' + u = y'' + 3y' + 2y = \sin(e^{x})\). Solve the two first-order equations in turn:

\[ \begin{aligned} (e^{x}u)' &= e^{x}\sin(e^{x}) \\ e^{x}u &= -\cos(e^{x}) \\ (e^{2x}y)' &= e^{2x}u = -e^{x}\cos(e^{x}) \\ e^{2x}y &= -\sin(e^{x}) \end{aligned} \]

Taking both constants of integration as zero gives \(y = -e^{-2x}\sin(e^{x})\). The substitution \(w = e^{x}\), with \(dw = e^{x}dx\), is what makes each integral elementary.

The other options do not satisfy the equation. Substituting \(-e^{-x}\sin(e^{x})\) gives \(e^{x}\sin(e^{x}) - 2\cos(e^{x})\) on the left, not \(\sin(e^{x})\). Substituting \(-e^{-2x}(\sin(e^{x}) + \cos(e^{x}))\) gives \(\sin(e^{x}) + \cos(e^{x})\): the added cosine term survives as an extra \(\cos(e^{x})\).

Discuss in Forum

Consider the following linear second order differential equation

\[ \frac{d^{2}y}{dt^{2}} + \omega^{2}y = 0, \]

where \(\omega\) is a positive constant. The boundary conditions are \(\left.\frac{dy}{dt}\right|_{t=0} = 1\), and \(y(t = 0) = \frac{1}{2}\).

Which of the following options gives the value of \(y(t = \frac{\pi}{2\omega})\)?

A.\(\frac{1}{\omega}\)
B.\(\frac{2\pi}{\omega}\)
C.0
D.1

Exam Year(s)

IIT JAM - 2026,

Answer: A

The value is \(\frac{1}{\omega}\).

The general solution of \(\ddot{y} + \omega^2 y = 0\) is \(y = A\cos\omega t + B\sin\omega t\). The two conditions fix the constants:

\[ \begin{aligned} y(0) &= A = \tfrac{1}{2} \\ \dot{y}(0) &= B\omega = 1, \quad B = \tfrac{1}{\omega} \end{aligned} \]

At \(t = \frac{\pi}{2\omega}\), \(\omega t = \frac{\pi}{2}\), so the cosine term vanishes and the sine term is at its maximum:

\[ \begin{aligned} y\left(\tfrac{\pi}{2\omega}\right) &= \tfrac{1}{2}\cos\tfrac{\pi}{2} + \tfrac{1}{\omega}\sin\tfrac{\pi}{2} \\ &= \tfrac{1}{\omega} \end{aligned} \]

0 is what you get by keeping only the cosine term, and 1 by forgetting to divide the initial slope by \(\omega\).

Discuss in Forum

The differential equation obtained by eliminating the arbitrary constants A and \(\phi\) from :

\[ y = Ae^{-\alpha t} \sin(\omega t + \phi) \]

is :

A.\(\dfrac{d^{2}y}{dt^{2}} + 2\alpha\dfrac{dy}{dt} + (\alpha^{2} + \omega^{2})y = 0\)
B.\(\dfrac{d^{2}y}{dt^{2}} + 2t\dfrac{dy}{dt} + (\alpha^{2} - \omega^{2})y = 0\)
C.\(\dfrac{d^{2}y}{dt^{2}} + 2\alpha t\dfrac{dy}{dt}\) \(+ (\alpha^{2} + \omega^{2})y = 0\)
D.\(\dfrac{d^{2}y}{dt^{2}} + 2\alpha t\dfrac{dy}{dt}\) \(+ (\alpha^{2} - \omega^{2})y = 0\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The equation is \(\dfrac{d^{2}y}{dt^{2}} + 2\alpha\dfrac{dy}{dt} + (\alpha^{2} + \omega^{2})y = 0\).

Write \(y = \text{Im}(Ce^{(-\alpha + i\omega)t})\) with \(C = Ae^{i\phi}\). So y is a solution of the constant-coefficient equation whose characteristic roots are \(-\alpha \pm i\omega\). Those roots satisfy

\[ \begin{aligned} (m + \alpha)^{2} + \omega^{2} &= 0 \\ m^{2} + 2\alpha m + \alpha^{2} + \omega^{2} &= 0 \end{aligned} \]

which corresponds to \(y'' + 2\alpha y' + (\alpha^{2} + \omega^{2})y = 0\). The two arbitrary constants A and \(\phi\) are exactly the two constants of the general solution of this second-order equation, so eliminating them gives it.

The options with a factor t in the middle term have variable coefficients and do not have damped sinusoids as solutions. \(\alpha^{2} - \omega^{2}\) would correspond to real roots \(-\alpha \pm \omega\), giving exponentials rather than oscillations.

Discuss in Forum

Let \(\phi_{1}, \phi_{2}\) be linearly independent solutions of the linear differential equation :

\[ y'' + a_{1}y' + a_{2}y = 0, \]

where \(a_{1}, a_{2}\) are constants, then the Wronskian \(W(\phi_{1}, \phi_{2})\) is constant if and only if :

A.\(a_{1} = 0\)
B.\(a_{1} \ne 0\)
C.\(a_{2} = 0\)
D.\(a_{2} \ne 0\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The Wronskian is constant if and only if \(a_{1} = 0\).

By Abel's formula, the Wronskian of two solutions of \(y'' + a_{1}y' + a_{2}y = 0\) satisfies \(W' = -a_{1}W\), so

\[ W(x) = W(x_{0})\, e^{-a_{1}(x - x_{0})} \]

Because \(\phi_{1}, \phi_{2}\) are linearly independent, \(W(x_{0}) \ne 0\). The exponential is constant exactly when \(a_{1} = 0\).

The coefficient \(a_{2}\) does not enter Abel's formula at all, so conditions on \(a_{2}\) are irrelevant; and \(a_{1} \ne 0\) makes the Wronskian grow or decay exponentially, not stay constant.

Discuss in Forum

10.

An nth order linear ordinary differential equation :

A.has exactly \(n\) linearly independent solutions
B.has at most \(n\) linearly independent solutions
C.has less than \(n\) independent solutions
D.has minimum \(n\) linearly independent solutions

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The answer is has exactly \(n\) linearly independent solutions.

For a homogeneous linear equation \(y^{(n)} + a_{1}(x)y^{(n-1)} + \cdots + a_{n}(x)y = 0\) with continuous coefficients, the solution set is a vector space of dimension exactly \(n\). The existence-uniqueness theorem gives \(n\) independent solutions (one for each choice of initial data \(e_{1}, \ldots, e_{n}\)), and any \(n + 1\) solutions are dependent.

The paper does not say 'homogeneous', but that is the standard textbook statement it rests on. 'At most \(n\)' and 'minimum \(n\)' are each half of the truth, and 'less than \(n\)' is false.

Higher Order Linear Differential Equations Questions and Answers

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