Ordinary Differential Equations Questions and Answers

Answer: D

The largest interval is \((0, 5)\).

Divide by \(e^{x}\) to write the equation as \(y'' + p(x)y' + q(x)y = g(x)\) with

\[ \begin{aligned} p &= \frac{e^{-x}}{x - 5}, \quad q = e^{-x}\sqrt{x} \\ g &= e^{-x}\ln x \end{aligned} \]

The existence and uniqueness theorem guarantees a unique solution on the largest open interval containing the initial point x = 1 on which p, q and g are all continuous. \(\sqrt{x}\) and \(\ln x\) need \(x \gt 0\), and p blows up at x = 5. The interval containing 1 is therefore (0, 5).

\((-\infty, \infty)\) and \((-5, 5)\) include negative x, where \(\ln x\) is undefined, and \((0, \infty)\) crosses the singularity at x = 5.

Answer: A

The sum of the roots is 0.

In standard form the equation is \(y'' + \frac{\sin x}{x^{2}}y' - \frac{\tan x}{x^{3}}y = 0\). At x = 0,

\[ \begin{aligned} p_{0} &= \lim_{x \to 0} x \cdot \frac{\sin x}{x^{2}} = 1 \\ q_{0} &= \lim_{x \to 0} x^{2} \cdot \left(-\frac{\tan x}{x^{3}}\right) = -1 \end{aligned} \]

so x = 0 is a regular singular point. The indicial equation \(r(r - 1) + p_{0}r + q_{0} = 0\) becomes \(r^{2} - 1 = 0\), with roots 1 and -1, whose sum is 0. (In general the sum is \(1 - p_{0}\).) Taking \(p_{0} = 0\) would give a sum of 1, and neither 2 nor -2 can arise from roots of \(r^{2} = 1\).

Answer: C

The second smallest eigenvalue is 7.

Write \(\mu = \lambda - 3\). The problem \(y'' + \mu y = 0\), \(y(0) = y(\pi) = 0\) has non-zero solutions only when \(\mu = n^{2}\), \(n = 1, 2, 3, \ldots\), with \(y = \sin nx\). (For \(\mu \le 0\) the only solution meeting both conditions is \(y = 0\).)

\[ \lambda = n^{2} + 3 = 4,\ 7,\ 12,\ \ldots \]

The smallest is 4 and the second smallest is 7.

4 is the smallest, not the second. 3 would need \(\mu = 0\), which gives only the zero solution. 9 is \(n^{2}\) for \(n = 3\) without the shift of 3.

Answer: D

The value of the integral is \(\frac{4}{3}\).

Try \(y = ax^{2} + bx + c\) with \(a \neq 0\). Substituting \(y'' = 2a\) and \(y' = 2ax + b\):

\[ \begin{aligned} &(4a - ka)x^{2} + (b - kb)x \\ &\quad + (4a - kc) = 0 \end{aligned} \]

for all \(x\). The \(x^{2}\) term gives \(k = 4\); then \(-3b = 0\) gives \(b = 0\), and \(4a - 4c = 0\) gives \(c = a\). So \(y = a(x^{2} + 1)\), and \(y(1) = 2a = 1\) gives \(a = 1/2\):

\[ \begin{aligned} \int_{0}^{1} 2y\,dx &= \int_{0}^{1} (x^{2} + 1)\,dx \\ &= \frac{1}{3} + 1 = \frac{4}{3} \end{aligned} \]

\(\frac{1}{3}\) keeps only the \(x^{2}\) term, and \(\frac{2}{3}\) and 1 do not follow from \(y = (x^{2} + 1)/2\).

Answer: A

The value is \(W(5) = \frac{1}{8}\).

Write the equation in standard form \(y'' + p(x)y' + q(x)y = 0\) with \(p(x) = \frac{2}{x-1}\). Abel's formula gives the Wronskian:

\[ \begin{aligned} W(x) &= C\exp\left(-\int \frac{2}{x-1}\,dx\right) \\ &= \frac{C}{(x-1)^{2}} \end{aligned} \]

W(2) = C/1 = 2, so C = 2 and \(W(5) = 2/16 = 1/8\).

The term \(xe^{x}y\) plays no part, since Abel's formula uses only the coefficient of \(y'\). Getting 8 comes from using \(\exp(+\int p\,dx)\), a sign error that gives \(W = C(x-1)^{2}\); 1/5 and 5 do not follow from the formula.