I am learning how to do calculus and was presented with an example I am struggling a bit to understand. Why does $\frac{df}{dx}=3f$ have the general solution of $f(x)=Ce^{3x}$?
General solution to $\frac{df}{dx}=3f$
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calculus
ordinary-differential-equations
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5Have you tried differentiating $Ce^{3x}$? – 2011-11-24
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0I see. $f=e^{3x}$ so $\frac{d}{dx}(e^{3x})=3e^{3x}$ which is $3f$. – 2011-11-24
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1I suppose the deeper interpretation of the question is why this ODE _only_ has solution $f(x) = Ce^{3x}$, i.e. why this is the most general possible solution. – 2011-11-24
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4If we take 7530's "deeper interpretation", note that you can rearrange things as $\dfrac{f^\prime}{f}=3$. Recall that one can logarithmically differentiate: $\dfrac{\mathrm d}{\mathrm dx}\log(f(x))=\dfrac{f^\prime(x)}{f(x)}$... then integrate both sides. – 2011-11-24
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1@J.M. What happens at the points where $f$ could be 0? For calculus, I preffer the following approach: Let $g(x)= \frac{f(x)}{e^{3x}}$. Then $g'(x)=....$... – 2011-11-24
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0I guess I'll worry about that when the exponential function intersects the horizontal axis, @N.S. :) – 2011-11-24
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2@J.M. The point I am trying to make is that the problem only gives $f'=3f$. I don't see any obvious reason (at the level of a first year calculus student) to conclude that $f$ is nowhere vanishing....Yes we know the solution is the exponential function, which doesn't vanish, but how can we use that before solving the problem? – 2011-11-24
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0@N.S. $f(x)=0x^{3x}$ has this property. – 2011-11-24
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0@N.S. I always was a "solve first, check later" guy. Barring the boring f=0 case, you can check that the solution doesn't vanish. If the solution I got did intersect the horizontal axis, then I'll backtrack, since it shows my assumption(s) were faulty. But it works out fine here. I guess we'll have to agree to disagree on these views. – 2011-11-24
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0@N.S. I suppose I could have taken the Frobenius or characteristic equation routes if you really wanted me to avoid the $f=0$ business, but this seems like "nuking mosquitoes" territory for me in this simple business. – 2011-11-24
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0@J.M. Well I don't dissagree with you, and I know that once we solve the equation we can argue that the extra assumption we make doesn't change the outcome of the solution... Anyhow, my oppinion is that in a calculus class I would rather consider $g(x)=\frac{f(x)}{e^{3x}}$ and prove that $g'=0$ instead of worrying about these things ;) – 2011-11-24
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1@N.S.: I think the problem with that approach is that it seems to emphasize the "I already know the answer, I'm just verifying it" approach (or makes it seem like it should be a "guess-and-check"). – 2011-11-24
2 Answers
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If we are worried about the hazards of division by $0$, let $$g(x)=\frac{f(x)}{e^{3x}}.$$ Differentiate, using the Quotient Rule. We get $$g'(x)=\frac{e^{3x}f'(x)-f(x)(3e^{3x})}{(e^{3x})^2}.\qquad\qquad(\ast)$$ Using $f'(x)=3f(x)$ we can see that the numerator in $(\ast)$ is $0$. So $g'(x)$ is identically $0$, and therefore $g(x)$ is a constant function $C$. Thus $$C=\frac{f(x)}{e^{3x}},$$ and $f(x)=Ce^{3x}$.
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You have $ \frac{dy}{dx} = 3y$. Multiply by $x$ and divide by $y$ you'll get $\frac{1}{y}dy=3dx$. By doing integration you'll have: $\int \frac{1}{y}dy=\int3dx$, and after solving the integral you'll have $\log{y} = 3x + c$. By looking at the exponent, you'll get $y = e^{3x}e^{c}$, where $c^{,} = e^{c}$ is a positive constant.