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

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:

$9f(-\log 3) = 61$