Logarithms Questions and Answers

Answer: D

Read the two features separately.

The square bracket at \( a \) means the endpoint is included (\( x \geq a \)), so the interval is closed at that end. It extends to \( +\infty \) with no upper endpoint, so it is unbounded.

Hence it is closed and unbounded. Infinity is never itself included, which is why that side is always written with a parenthesis.

Answer: C

An ordinary differential equation has a single independent variable, so it describes a one-dimensional system - quantities evolving along one axis, usually time.

Two or more independent variables call for a partial differential equation instead.

Answer: B

Linearity is judged in the dependent variable and its derivatives: they may appear only to the first power and never multiplied together.

\[ (y-x)\,\partial x + 4xy\,\partial y = 0 \]

Rearranged as

\[ \frac{dy}{dx} = \frac{x-y}{4xy} \]

the term \( 4xy\,\tfrac{dy}{dx} \) has \( y \) multiplying its own derivative, so the equation is non-linear.

A linear first-order equation has the form \( \tfrac{dy}{dx} + P(x)y = Q(x) \), with no product of \( y \) and \( y' \).

Answer: B

The word differential points to differentials and hence to derivatives, and equation to a statement of equality. A differential equation is therefore an equation containing derivatives of an unknown function.

\[ \frac{dy}{dx} + 2y = e^{x} \]

An equation containing integrals instead would be an integral equation.

Answer: B

Start from Boyle's law at constant temperature:

\[ V = \frac{c}{p} \]

Differentiating with respect to \( p \):

\[ \frac{dV}{dp} = -\frac{c}{p^{2}} \]

Since \( c = pV \), this becomes

\[ \frac{dV}{dp} = -\frac{pV}{p^{2}} = -\frac{V}{p} \]

The negative sign reflects that volume falls as pressure rises.

Answer: D

Expand the argument first:

\[ \sin(\omega t - \theta) = \sin \omega t \cos\theta - \cos \omega t \sin\theta \]

The transform is linear, and \( \cos\theta \) and \( \sin\theta \) are constants, so each term transforms separately:

\[ \mathcal{L}\{\sin(\omega t - \theta)\} = \cos\theta\left(\frac{\omega}{s^{2}+\omega^{2}}\right) - \sin\theta\left(\frac{s}{s^{2}+\omega^{2}}\right) \]

The minus sign carries over from the expansion; a phase of \( +\theta \) would make it a plus.

Answer: B

An initial value problem is a differential equation together with the value of the unknown function (and enough of its derivatives) at one point:

\[ y' = f(t,y), \qquad y(t_{0}) = y_{0} \]

The equation is what makes it a differentiation problem; the initial condition merely selects one solution from the general family.

Answer: C

The Laplace transform carries the initial conditions into the algebra from the outset:

\[ \mathcal{L}\{y'\} = sY - y(0) \]

so \( y(0) \) appears in the subsidiary equation itself. The particular solution falls out directly, with no general solution and no arbitrary constants to determine afterwards.

Answer: C

The transfer function is the ratio of the transformed output to the transformed input, with zero initial conditions:

\[ H(s) = \frac{Y(s)}{X(s)} = \frac{\text{output}}{\text{input}} \]

It characterises the system independently of any particular input, which is what makes it so useful in control and signal analysis.

Answer: D

\[ \log(ab) = \log a + \log b \]

The logarithm of a product is the sum of the logarithms. The companion rules are

\[ \log\left(\frac{a}{b}\right) = \log a - \log b, \qquad \log(a^{n}) = n\log a \]

Turning multiplication into addition is exactly what made logarithms a calculating tool.