Logarithms Questions and Answers

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: A

Transforming \( y'' - y = 0 \) with \( y(0) = 12 \) and \( y'(0) = 0 \):

\[ s^{2}Y - s \cdot 12 - 0 - Y = 0 \]

\[ Y(s^{2} - 1) = 12s \quad\Longrightarrow\quad Y = \frac{12s}{s^{2}-1} \]

Since \( \mathcal{L}\{\cosh t\} = \dfrac{s}{s^{2}-1} \):

\[ y = 12\cosh t \]

The hyperbolic cosine appears because the denominator is \( s^{2} - 1 \); an ordinary cosine would need \( s^{2} + 1 \).

Answer: A

Some elementary functions have antiderivatives that cannot be written as elementary functions at all.

\[ \int e^{-x^{2}}\,dx, \qquad \int \frac{\sin x}{x}\,dx \]

Both integrals exist, but neither can be expressed with finitely many algebraic, exponential, logarithmic and trigonometric operations. New named functions - the error function, the sine integral - are defined for them.

Answer: B

Expressing a physical situation in variables, functions and equations produces a mathematical model - the bridge between the engineering problem and the mathematics used to solve it.

Solving the model then gives an answer that has to be interpreted back in the original setting, which is the last step of the modelling cycle.

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: B

Solving a differential equation by Laplace transform runs in three steps:

1. Transform the equation into an algebraic subsidiary equation in \( s \)
2. Solve that subsidiary equation by algebra alone
3. Take the inverse transform to recover the solution in \( t \)

Step 2 is where the method pays off: no calculus is needed, only ordinary algebraic manipulation.

Answer: A

A deductive model is built downward from an accepted theory: the governing equations are derived from established principles rather than fitted to data.

An inductive model works the other way, generalising from observations, and a floating model rests on neither.

Answer: D

Both \( \pi \) and \( e \) are transcendental: neither is a root of any polynomial equation with integer coefficients.

An algebraic number is one that is such a root - \( \sqrt{2} \) qualifies, since it satisfies \( x^{2} - 2 = 0 \). No such equation exists for \( \pi \) or \( e \); Hermite proved it for \( e \) in 1873 and Lindemann for \( \pi \) in 1882, and that second proof is what finally settled the impossibility of squaring the circle.

Every transcendental number is irrational, but the reverse fails: \( \sqrt{2} \) is irrational and still algebraic.

Answer: A

In the Laplace-transform method, the very first step is to apply the transform to the whole differential equation, turning derivatives in \( t \) into powers of \( s \):

\[ \mathcal{L}\{y'\} = sY - y(0), \qquad \mathcal{L}\{y''\} = s^{2}Y - sy(0) - y'(0) \]

The result is an algebraic (subsidiary) equation in \( Y(s) \). Solving it is step 2 and inverting is step 3.

Answer: C

An interval containing one endpoint but not the other is called half open - equivalently half closed, since it is open at one end and closed at the other.

\[ [a,b) \quad\text{and}\quad (a,b] \]

The two names describe the same thing from opposite viewpoints.