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.

Answer: B

At t = 1, x = 2/e.

The characteristic equation is \(m^{2} + 2m + 1 = 0\), with the repeated root \(m = -1\) (critical damping). So the general solution and the conditions give:

\[ \begin{aligned} x &= (A + Bt)e^{-t} \\ x(0) &= A = 1 \\ \dot{x}(0) &= B - A = 0 \Rightarrow B = 1 \end{aligned} \]

Hence \(x = (1 + t)e^{-t}\), and \(x(1) = 2e^{-1} = 2/e\). The option 1/e is what you get by dropping the \(Bt\) term that a repeated root requires.

Answer: D

The solution is cosh t - sinh t.

The equation \(y'' - y = 0\) has the general solution \(y = Ae^{t} + Be^{-t}\). The condition \(y(\infty) = 0\) forces \(A = 0\), and \(y(0) = 1\) gives \(B = 1\), so \(y = e^{-t}\).

Since \(\cosh t - \sinh t = e^{-t}\), the answer is cosh t - sinh t. The option cosh t + sinh t equals \(e^{t}\), which grows without bound, and the trigonometric options solve \(y'' + y = 0\), a different equation.

Answer: D

The solution is \(y = \frac{x^{4}}{5} + \frac{4}{5x}\).

The left side is an exact derivative, \(x\frac{dy}{dx} + y = \frac{d}{dx}(xy)\), so

\[ \begin{aligned} \frac{d}{dx}(xy) &= x^{4} \\ xy &= \frac{x^{5}}{5} + C \\ y(1) = 1 &\Rightarrow C = \frac{4}{5} \\ y &= \frac{x^{4}}{5} + \frac{4}{5x} \end{aligned} \]

Checking the others: \(y = 5x^{4} - 4\) gives \(x y' + y = 25x^{4} - 4\), not \(x^{4}\); \(\frac{x^{4}}{5} + \frac{4x}{5}\) gives \(x^{4} + \frac{8x}{5}\); and \(\frac{4x^{4}}{5} + \frac{1}{5x}\) gives \(4x^{4}\), although it too satisfies \(y(1) = 1\).

Answer: A

Order counts derivatives; degree counts powers. So \( (y'')^{3} + y = 0 \) is of order two and degree three, while \( y''' + (y')^{5} = 0 \) is of order three and degree one, the degree being read from the *highest* derivative present. Confusing the two is the commonest slip in classifying an equation

Answer: D

The highest derivative present is the second, so the order is two. That derivative appears cubed, so the degree is three. The first derivative's own power is irrelevant to the degree, which is read only from the highest-order term after the equation has been cleared of radicals and fractions in the derivatives