Question

For a differentiable function \( f : \mathbb{R} \to \mathbb{R} \), suppose \[ f'(x) = 3f(x) + \alpha, \] where \( \alpha \in \mathbb{R} \), \( f(0) = 1 \), and \[ \lim_{x \to -\infty} f(x) = 7. \] Then \( 9f(-\log_2 3) \) is equal to __________ .

Answer 1

Correct Answer: 61

Given the differential equation:

\[\frac{dy}{dx} - 3y = \alpha\]

Let the integrating factor be:

\[I = e^{\int -3dx} = e^{-3x}\]

Multiplying through by the integrating factor:

\[y \cdot e^{-3x} = \int e^{-3x} \cdot \alpha dx\]

Solving for \(y\):

\[y \cdot e^{-3x} = \frac{\alpha e^{-3x}}{-3} + C\]

Multiplying through by \(e^{3x}\):

\[y = \frac{\alpha}{-3} + C \cdot e^{3x}\]

Using the initial condition \(x = 0, y = 1\):

\[1 = \frac{\alpha}{-3} + C \cdot e^0\]

\[C = 1 + \frac{\alpha}{3}\]

As \(x \to -\infty, y \to 7\):

\[y = 7 - 6e^{3x}\]

Finally, evaluating:

\[9f(-\log 3) = 61\]