Applications of Derivatives
Differential Equations
Grade 12
Question:
<p>Suppose, \(f(x) = e^{ax} + e^{bx}\), where \(a \neq b\) and \(f''(x) - 2f'(x) - 15f(x) = 0\) for all \(x \in \mathbb{R}\). Then, find \(ab\).</p>
<p>(a) 15</p>
<p>(b) –15</p>
<p>(c) 10</p>
<p>(d) 16</p>
Step-by-Step Solution
Key Concept: If a function satisfies a differential equation with constant coefficients, the characteristic roots determine the exponential form. Each exponential term must satisfy the equation independently.
<p><strong>Solution:</strong></p><p>Given: $f(x) = e^{ax} + e^{bx}$</p><p>$f'(x) = ae^{ax} + be^{bx}$</p><p>$f''(x) = a^2e^{ax} + b^2e^{bx}$</p><p>From the condition $f''(x) - 2f'(x) - 15f(x) = 0$:</p><p>$a^2e^{ax} + b^2e^{bx} - 2(ae^{ax} + be^{bx}) - 15(e^{ax} + e^{bx}) = 0$</p><p>$(a^2 - 2a - 15)e^{ax} + (b^2 - 2b - 15)e^{bx} = 0$</p><p>For this to hold for all $x$:</p><p>$a^2 - 2a - 15 = 0$ and $b^2 - 2b - 15 = 0$</p><p>$(a - 5)(a + 3) = 0$ and $(b - 5)(b + 3) = 0$</p><p>So $a = 5$ or $-3$, and $b = 5$ or $-3$</p><p>Since $a \neq b$: either $a = 5, b = -3$ or $a = -3, b = 5$</p><p>Therefore, $ab = 5 \times (-3) = -15$</p>
Correct Answer: B