For a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$, suppose $f^{\prime}(x)=3 f(x)+\alpha$, where $\alpha \in \mathbb{R}, f(0)=1$ and $\lim _\limits{x \rightarrow-\infty} f(x)=7$. Then $9 f\left(-\log _e 3\right)$ is equal to _________.
Answer (integer)
61
Solution
<p>$$\begin{aligned}
& f^{\prime}(x)=3 f(x)+\alpha \\
& \Rightarrow \frac{d y}{3 y+\alpha}=d x \\
& \Rightarrow \frac{1}{3} \ln (3 y+\alpha)=x+C \\
& y(0)=1 \Rightarrow C=\frac{1}{3} \ln (3+\alpha) \\
& \frac{1}{3} \ln \left(\frac{3 y+\alpha}{3+\alpha}\right)=x \\
& \Rightarrow y=\frac{1}{3}\left((3+\alpha) e^{3 x}-\alpha\right)=f(x) \\
& \lim _{x \rightarrow-\infty} f(x)=7 \Rightarrow \alpha=-21 \\
& \Rightarrow f(x)=7-6 e^{3 x} \\
& 9 f(-\ln 3)=61
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree
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