Your Answer : (Not Answered)

Correct Answer : A

The particular integral is \(-\frac{x}{2}\cos(x - y)\).

For \(\sin(x - y)\), replacing \(D_{x}^{2}\) by \(-1\) and \(D_{y}^{2}\) by \(-1\) makes the denominator \(-1 - (-1) = 0\), the case of failure. The rule is then to multiply by \(x\) and differentiate the denominator with respect to \(D_{x}\):

\[ \begin{aligned} \frac{1}{D_{x}^{2} - D_{y}^{2}}\sin(x - y) &= x\cdot\frac{1}{2D_{x}}\sin(x - y) \\ &= \frac{x}{2}\int \sin(x - y)\,dx \\ &= -\frac{x}{2}\cos(x - y). \end{aligned} \]

Check: with \(u = -\frac{x}{2}\cos(x - y)\), \(u_{xx} - u_{yy} = \sin(x - y)\). The option that adds \(\sin(x - y)\) differs only by a term the operator annihilates, so it is not what the inverse operator produces; the particular integral carries no complementary-function part. \(-\frac{x}{2}\sin(x - y) + \cos(x - y)\) gives \(-\cos(x - y)\) under the operator, and \(\frac{3x}{2}\sin(x - y)\) also produces a cosine term, so neither satisfies the equation.

Your Answer : (Not Answered)

Correct Answer : C

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

Write \(y'' = -y\). Integrating twice from 0 and using \(y(0) = y'(0) = 0\):

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

The first step is the standard formula for a repeated integral. Option A has the sign wrong, which would correspond to \(y'' = y\), and options B and D carry an extra factor \(t\) that the double integration does not produce. (Of course the only solution here is \(y \equiv 0\), but the question asks for the equivalent equation.)

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct Answer : D

There are infinitely many extremals if \(\lambda = -1\).

The Euler-Lagrange equation for \(F = y'^{2} - y^{2}\) is \(y'' + y = 0\), with solutions \(y = A\cos x + B\sin x\). The condition \(y(0) = 1\) gives \(A = 1\). Then

\[ y(\pi) = \cos\pi + B\sin\pi = -1 \]

whatever \(B\) is. So if \(\lambda = -1\), every \(y = \cos x + B\sin x\) is an extremal, infinitely many; if \(\lambda \neq -1\), there is none.

In particular there is no extremal at all when \(\lambda = 1\), so A and B fail, and C fails because the extremal is not unique.

Your Answer : (Not Answered)

Correct Answer : C

The shortest distance is \(\dfrac{19\sqrt{2}}{8}\).

The distance from \((x, x^{2})\) to the line \(x - y - 5 = 0\) is \(|x - x^{2} - 5|/\sqrt{2}\). Since \(x^{2} - x + 5 \gt 0\) always, the line and parabola do not meet, and we minimise \(x^{2} - x + 5\):

\[ \begin{aligned} x &= \tfrac{1}{2}: \ \tfrac{1}{4} - \tfrac{1}{2} + 5 = \tfrac{19}{4} \\ d &= \frac{19/4}{\sqrt{2}} = \frac{19\sqrt{2}}{8} \end{aligned} \]

(At \(x = 1/2\) the tangent slope \(2x = 1\) is parallel to the line.) The distance is not 0 because the curves do not intersect, and 5 and \(\sqrt{5}\) do not come from the minimisation.

Your Answer : (Not Answered)

Correct Answer : A

The equation is \(y\dfrac{d^{2}y}{dx^{2}} - \left(\dfrac{dy}{dx}\right)^{2} = 0\).

From \(y = ae^{bx}\): \(y' = by\) and \(y'' = b^{2}y\). Eliminating \(b\), \(b = y'/y\), so

\[ \begin{aligned} y'' &= \frac{(y')^{2}}{y}, \\ \text{that is,} \quad yy'' - (y')^{2} &= 0 \end{aligned} \]

Both constants disappear, as required for a family with two parameters and a second-order equation.

The other options contain \(x\) explicitly, but the family \(ae^{bx}\) is unchanged by shifts in \(x\) (a shift only changes \(a\)), so its differential equation cannot involve \(x\).

Your Answer : (Not Answered)

Correct Answer : A

The keyed answer is both (I) and (II) are true.

(I): if \(c_{1}x + c_{2}|x| = 0\) for all \(x\), then \(x = 1\) gives \(c_{1} + c_{2} = 0\) and \(x = -1\) gives \(-c_{1} + c_{2} = 0\), so \(c_{1} = c_{2} = 0\). The set is linearly independent.

(II): for \(x \gt 0\), \(|x| = x\) and \(W = x \cdot 1 - x \cdot 1 = 0\); for \(x \lt 0\), \(|x| = -x\) and \(W = x(-1) - (-x)(1) = 0\). At \(x = 0\), \(|x|\) is not differentiable, so the Wronskian vanishes wherever it is defined, which is the textbook reading of (II).

This is the standard example showing that a zero Wronskian does not imply dependence when the functions are not solutions of one linear ODE.

Your Answer : (Not Answered)

Correct Answer : C

The solution is \(z = \phi_{1}(x + y) + \phi_{2}(x - y)\).

This is the wave equation \(z_{xx} = z_{yy}\). Its auxiliary equation \(m^{2} - 1 = 0\) has distinct roots \(m = \pm 1\), so the general solution is d'Alembert's: an arbitrary function of \(x + y\) plus an arbitrary function of \(x - y\).

Check: for \(z = \phi_{1}(x + y)\), \(z_{xx} = \phi_{1}''\) and \(z_{yy} = \phi_{1}''\), so they cancel; likewise for \(\phi_{2}(x - y)\). Factors of \(x\) or \(y\) in front belong to repeated roots and do not satisfy this equation.

Your Answer : (Not Answered)

Correct Answer : C

The complete integral is \(z = \phi(a)x + ay + b\).

For an equation in \(p = z_{x}\) and \(q = z_{y}\) alone, try \(z = \alpha x + \beta y + b\). Then \(p = \alpha\) and \(q = \beta\), so the equation holds whenever \(f(\alpha, \beta) = 0\). Taking \(q = a\) and solving for \(p = \phi(a)\) gives a two-parameter family \(z = \phi(a)x + ay + b\), which is the complete integral.

Option A fixes \(q = 1\), leaving only one effective parameter. Options with \(x^{2}\) give \(p = 2\phi(a)x\) or \(p = 2x\), which depend on \(x\) and do not satisfy a relation between constants \(p\) and \(q\).

Your Answer : (Not Answered)

Correct 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 \(x'(0) = -1\).

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct Answer : A

The extremal is a parabola.

For an isoperimetric problem, extremise \(\int (F + \lambda G)\,dx\) with the constraint integrand \(G = y\) and a Lagrange multiplier \(\lambda\). Here \(H = (y')^{2} + \lambda y\), and its Euler equation is

\[ \begin{aligned} \lambda - \frac{d}{dx}(2y') &= 0 \\ y'' &= \frac{\lambda}{2} \end{aligned} \]

So \(y = \tfrac{\lambda}{4}x^{2} + c_{1}x + c_{2}\), a quadratic, whose graph is a parabola. The three constants are fixed by \(y(1) = 3\), \(y(4) = 24\) and the area condition; they give \(y = x^{2} + 2x\), which indeed has \(\int_{1}^{4} y\,dx = 21 + 15 = 36\).

The circle is the answer to the classical isoperimetric problem of maximising area for a fixed arc length, where the functional is \(\int\sqrt{1 + y'^{2}}\); this functional is different.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct Answer : A

The weight function is \(e^{-3x}\).

Bring the equation to Sturm-Liouville form (p y')' + ... = \(\lambda w y\). Multiply by an integrating factor \(\mu(x)\) chosen so that \(\mu(y'' - 3y') = (\mu y')'\), which needs \(\mu' = -3\mu\), so \(\mu = e^{-3x}\):

\[ \begin{aligned} &\frac{d}{dx}\left(e^{-3x}\frac{dy}{dx}\right) \\ &= \lambda e^{-3x} y \end{aligned} \]

With p = \(e^{-3x}\) and the boundary conditions y(0) = y(\(\pi\)) = 0, eigenfunctions for distinct eigenvalues are orthogonal with weight \(w(x) = e^{-3x}\).

\(e^{3x}\) is the reciprocal, from taking the factor with the wrong sign, and \(e^{\pm 2x}\) does not turn \(y'' - 3y'\) into an exact derivative.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct Answer : A

The maximum value is 1.

With zero initial velocity, D'Alembert's formula reduces to the average of two travelling copies of the initial shape:

\[ \begin{aligned} u(x, t) &= \tfrac{1}{2}[f(x - ct) + f(x + ct)] \\ u(0, t) &= \tfrac{1}{2}[f(-ct) + f(ct)] \end{aligned} \]

For \(0 \lt t \lt 1/c\) both \(|ct|\) and \(|-ct|\) are below 1, so both terms equal 1 and \(u(0, t) = 1\). For \(t \gt 1/c\) both copies have moved past the origin and \(u(0, t) = 0\).

So the largest value is 1. The value \(\frac{1}{2}\) is what a point outside the initial pulse sees when one half-pulse passes it, which is the usual trap; \(\frac{1}{4}\) and 0 are not the maximum at the origin.

Your Answer : (Not Answered)

Correct Answer : A

The answer is \(\hat{u}(w, t) = \frac{2a}{a^2+w^2} e^{-kw^2 t}\).

Transforming the equation in x turns the second derivative into multiplication by \(-w^2\), leaving an ordinary differential equation in t:

\[ \begin{aligned} \frac{\partial \hat{u}}{\partial t} &= -k w^2 \hat{u} \\ \hat{u}(w, t) &= \hat{u}(w, 0)\, e^{-k w^2 t} \end{aligned} \]

The transform of the initial condition is

\[ \begin{aligned} \hat{u}(w, 0) &= \int_{-\infty}^{\infty} e^{-a|x|} e^{iwx} dx \\ &= \frac{2a}{a^2 + w^2}. \end{aligned} \]

The option with a in the numerator has lost the factor 2 from the two halves of the line; the one with \(e^{-ka^2 t}\) decays at the wrong rate; and the Gaussian is the solution kernel in x, not this transform.

Your Answer : (Not Answered)

Correct Answer : C

\(u\left(\frac{1}{2}, 0\right) = \) \(\frac{1}{4}\).

On the unit circle \(x^{2} + y^{2} = 1\), so the boundary data can be rewritten:

\[ 1 - 2y^{2} = (x^{2} + y^{2}) - 2y^{2} = x^{2} - y^{2} \]

The function \(x^{2} - y^{2}\) is harmonic (\(u_{xx} + u_{yy} = 2 - 2 = 0\)) and matches the boundary values, so by uniqueness of the Dirichlet problem it is the solution: \(u(x, y) = x^{2} - y^{2}\).

Then \(u\left(\frac{1}{2}, 0\right) = \frac{1}{4}\). Taking \(1 - 2y^{2}\) itself as the solution would give 1, but that function is not harmonic, so 1 is wrong; \(-1\) and \(-\frac{1}{4}\) have the wrong sign.