The function $f(x)=ce^{kx}$ satisfies $f(0)=c$ and $f'(x)=kce^{kx}$.
Now I wonder if the following is true: If $g(0)=c$ and $g'(x)=kg(x)$ then $g(x)=f(x)$.
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1Yes, this is an uniqueness proof for an IVP of ODE, which is linear of order 1 with constant coefficients – Fakemistake Oct 05 '17 at 17:05
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Yes - If you have a first order differential equation, and have 1 specified boundary condition, then any solution you find is unique. – John Doe Oct 05 '17 at 17:31
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HINT: write your equation in the form $$\frac{dg}{dx}=Kg$$ and from here $$\frac{dg}{g}=Kdx$$
Dr. Sonnhard Graubner
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@R.A. No, it would yield $$\ln(g)=kx+C\implies g(x)=e^{kx+C}=e^Ce^{kx}=ce^{kx}$$ where $c=e^C$. – John Doe Oct 05 '17 at 17:29
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yes i know, have you not ever made a typo? thanks for your hint! – Dr. Sonnhard Graubner Oct 05 '17 at 17:31
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You may use the following theorem (proved here):
Theorem: There exists a unique function $f:\mathbb{R} \to\mathbb{R} $ such that $f'(x) =f(x), f(0)=1$ and it is traditionally denoted by symbol $\exp(x) $ or $e^{x} $. And any function $g:\mathbb{R} \to\mathbb {R} $ which satisfies $g'(x) =g(x) $ is given by $g(x) =g(0)\exp(x)$.
To solve the current equation $f'(x) =kf(x) $ we can set $t=kx$ and $g(t) = f(x) $ so that $$g'(t) =\frac{d} {dt}g(t) =\frac{d} {dx} f(x)\cdot\frac{dx} {dt} =\frac{kf(x)} {k} =f(x) =g(t) $$ and then by above theorem $$f(x) =g(t)=g(0)\exp(t)=f(0)\exp(kx)=ce^{kx}$$ This works for $k\neq 0$ and if $k=0$ then $f$ is constant equal to $c$ so that the given general solution holds for $k=0$ also.
Paramanand Singh
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