Differential Equations Questions and Answers

Answer: B

The function satisfies \(y'' - y = 0\).

\[ \begin{aligned} y &= be^{x} + ae^{-x} \\ y' &= be^{x} - ae^{-x} \\ y'' &= be^{x} + ae^{-x} = y \end{aligned} \]

So \(y'' - y = 0\). \(y'' + y = 0\) is satisfied by \(\sin x\) and \(\cos x\), not by exponentials, and \(y'' - 2y = 0\) would need exponents \(\pm\sqrt{2}x\).

Answer: C

The order and degree are 1 and 4.

Clear the negative power: multiply both sides by \(\left(2y - x\frac{dy}{dx}\right)^{3}\).

\[ \begin{aligned} \left(2\frac{dy}{dx} - 3x\right) \times \\ \left(2y - x\frac{dy}{dx}\right)^{3} = 1 \end{aligned} \]

The highest derivative is \(\frac{dy}{dx}\), so the order is 1. In this polynomial form the highest power of \(\frac{dy}{dx}\) is \(1 + 3 = 4\), so the degree is 4. Reading the degree as 3 from the exponent alone ignores the factor \(2\frac{dy}{dx}\).

Answer: B

The curve is \(y = x(x - 1)\).

Rewrite as a linear equation: \(\frac{dy}{dx} - \frac{y}{x} = x\). The integrating factor is \(e^{-\int dx/x} = \frac{1}{x}\).

\[ \begin{aligned} \frac{d}{dx}\left(\frac{y}{x}\right) &= 1 \Rightarrow \frac{y}{x} = x + c \\ y &= x^{2} + cx \\ (1, 0): 0 &= 1 + c \Rightarrow c = -1 \end{aligned} \]

So \(y = x^{2} - x = x(x - 1)\). \(y = x(x + 1)\) does not pass through (1, 0), and the cubic options do not satisfy the equation.