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 $f(x)=Ce^{3x}$?
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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. |
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