Higher Order Linear Differential Equations Questions and Answers

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.

Answer: A

The Wronskian is a constant function.

For \(y'' + p(x)y' + q(x)y = 0\), Abel's formula gives \(W(x) = W(x_{0})\exp\left(-\int_{x_{0}}^{x} p(t)\, dt\right)\). Here there is no \(y'\) term, so \(p = 0\) and \(W\) is constant. Directly: \(W' = \phi_{1}\phi_{2}'' - \phi_{1}''\phi_{2}\), which is \(\phi_{1}(-a\phi_{2}) - (-a\phi_{1})\phi_{2} = 0\).

Because \(\phi_{1}, \phi_{2}\) are linearly independent, the constant is non-zero, so D (identically zero) is false, and a non-zero constant is neither increasing nor decreasing.

Answer: C

The correct statement is that \(f\) and \(g\) can be solutions of an ordinary differential equation.

They are linearly independent, since \(g/f = \log x\) is not constant, which rules out 'linearly dependent'. Their Wronskian is

\[ \begin{aligned} W &= x^{2}(2x\log x + x) - 2x \cdot x^{2}\log x \\ &= x^{3} \end{aligned} \]

so it is not \(x^{2}\). Since \(W \neq 0\) for \(x \neq 0\), they form a fundamental pair for a second-order linear equation. Indeed \(x^{2}\) and \(x^{2}\log x\) are the solutions of the Euler equation \(x^{2}y'' - 3xy' + 4y = 0\), whose indicial equation \(m^{2} - 4m + 4 = 0\) has the repeated root 2. So 'cannot be solutions of an ordinary differential equation' is false.

Answer: A

Only I is true.

Every solution is a combination of terms \(e^{rx}\), \(x e^{rx}\) or \(e^{\alpha x}\cos\beta x\), \(e^{\alpha x}\sin\beta x\), where \(r\) or \(\alpha \pm i\beta\) are the characteristic roots. If every root has negative real part, each term carries a decaying exponential and \(\phi \to 0\).

If the real parts are positive, the terms grow: for \(y'' - 3y' + 2y = 0\) the roots are 1 and 2 and \(\phi = e^{x}\) tends to infinity. So II is false (only the zero solution would tend to 0).