Let f:RR satisfy f(x+y)=f(x)f(y) for all x,yR, f'(0)=4a, and f satisfies f''(x)-3af'(x)-f(…

Question

Let $f:\mathbb{R}	o\mathbb{R}$ satisfy $f(x+y)=f(x)f(y)$ for all $x,y\in\mathbb{R}$, $f'(0)=4a$, and $f$ satisfies $f''(x)-3af'(x)-f(x)=0$, $a>0$. If $F(x)=f(2ax)$, the area of the region $R=\{(x,y): 0\le y\le F(ax),\ 0\le x\le2\}$ is:

Accepted Answer

$f(x)=e^{\lambda x}$; from $f'(0)=4a$, $\lambda=4a$. Substituting into $\lambda^2-3a\lambda-1=0$: $16a^2-12a^2-1=0 \Rightarrow a=\dfrac12$.

So $f(x)=e^{2x}$ and $F(x)=e^{2x}$ (i.e., $F(ax)=e^{x}$ after substitution with $a=1/2$).

$$\text{Area} = \int_0^2 e^x\,dx = e^2-1.$$

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