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I'm trying to understand differentiability and to do so i'm trying to answer a question but cannot work it out:

(a). Suppose that $f$ : $\mathbb{R}$ $\rightarrow (0, \infty)$ is differntiable and satisfies $f'(x) = af(x)$ for all $x \in \mathbb{R}$, for some constant $a \in \mathbb{R}$. Prove that $f(x) = Ce^{ax}$, for some constant $C \in \mathbb{R}$.

(b). Now suppose that $h:\mathbb{R} \rightarrow \mathbb{R}$ is twice differentiable and, for all $x \in \mathbb{R}$

$h''(x) + 3h'(x) + 2h(x) = 0$.

Let $F(x) = h'(x) + 2h(x)$ and $G(x) = h'(x) + h(x)$. Show that $F$ and $G$ both satisfy the hypotheses of part (a). Hence prove that:

$h(x) = Ce^{-x} + De^{-2x}$

Any help would be appreciated.

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The (differential-equations) tag might be appropriate –  The Chaz 2.0 Mar 12 '12 at 20:09
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You have to know something in advance. For example: if the derivative of a function is zero everywhere, then the function is constant. So, knowing your solution is supposed to be $f(x)=Ce^{ax}$, why not try showing that $e^{-ax}f(x)$ is constant. –  GEdgar Mar 12 '12 at 20:17
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1 Answer 1

up vote 2 down vote accepted

For (a), let $$g(x)=\frac{f(x)}{e^{ax}}.$$ Differentiate $g(x)$ as usual, using the Quotient Rule. We get $$g'(x)=\frac{e^{ax}f'(x)-ae^{ax}f(x)}{(e^{ax})^2}.$$ Substituting $af(x)$ for $f'(x)$ in the top, we get $$g'(x)=0,$$ since the bottom is never $0$.

Now use the theorem (a consequence of the Mean Value Theorem) that a function whose derivative is identically $0$ in an interval is constant on that interval. It follows that $g(x)$ is a constant $C$ on our interval, and therefore $$f(x)=Ce^{ax}.$$

For (b), verifying that $F(x)$ and $G(x)$ satisfy the hypotheses of (a) is mechanical. For example, for $F(x)$, rewrite our differential equation as $h''(x)+2h'(x)=-\left(h'(x)+2h(x)\right)$, which just says that $F'(x)=aF(x)$ with $a=-1$. Finally, note that $h(x)=F(x)-G(x)$.

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Euden, did you find this answer helpful? If this solves your question, you might consider accepting this answer –  The Chaz 2.0 Mar 16 '12 at 15:52
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