I'm to show that if $f:\mathbb{R} \to \mathbb{R}$ is differentiable and $f'(x) = \lambda f(x)$ for $\lambda \in \mathbb{R}$, then $f(x) = f(0) e^{\lambda x}$. I'm not really sure where to start with proving this. I was trying to start with the case that $f(x)$ is constant, so $f'(x) = 0 \implies f(x)=c = c \exp(0) = c$. I'm just not sure how to start this proof.
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1What's the derivative of $e^{-\lambda x}f(x)$? – Angina Seng May 03 '17 at 18:23
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Related – Pedro May 03 '17 at 19:33
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Let $g(x)=e^{-\lambda x}f(x)$. Then $g'(x)=e^{-\lambda x}f'(x)-\lambda e^{-\lambda x}f(x)=0$, so $g\equiv c$ is a constant. Thus $f(x)=ce^{\lambda x}$, and putting $x=0$ shows $c=f(0)$.
Jason
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$$ f'(x) = \lambda f(x) $$ $$ \frac{f'(x)}{f(x)} = \lambda $$ $$ (\ln f(x) ) ' = \lambda$$ $$ \ln f(x) = \lambda x + c $$ $$ f(x) = e^{\lambda x + c} = e^{c} \cdot e^{\lambda x} $$ $$f(0) = e^c \cdot 1 \Rightarrow f(x) = f(0) e^{\lambda x} $$
Ofek Gillon
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$$\begin{array}{rcl} f'(x) &=& \lambda f(x) \\ f' &=& \lambda f \\ \dfrac{\mathrm df}{\mathrm dx} &=& \lambda f \\ \dfrac{\mathrm dx}{\mathrm df} &=& \dfrac1{\lambda f} \\ x &=& \displaystyle \int \dfrac{\mathrm df}{\lambda f} \\ x &=& \dfrac1\lambda \ln(\lambda f) + C \\ \lambda x &=& \ln(\lambda f) + \lambda C \\ e^{\lambda x} &=& e^{\lambda C}\lambda f \\ f &=& \dfrac1\lambda e^{\lambda (x-C)} \\ f(x) &=& \dfrac1\lambda e^{\lambda (x-C)} \\ \end{array}$$
Substitute $x=0$ to wrap up.
Kenny Lau
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Let $$g(x)=\frac{f(x)}{e^{\lambda x}}$$
Then $$g'(x)= \frac{f'(x) e^\lambda(x)-\lambda f(x) e^\lambda x}{e^{2 \lambda x}}=0$$
Therefore, $g(x)$ is constant.
N. S.
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