Unable to Find My Error for First Order Differential Problem

Question

The Question: Given $f'(x) = 5f(x)$ and $f(3) = 2$ find the equation for $f(x)$

My Problem: My attempt to solve it produced the wrong answer. I got $f(x) = \frac{28}{5x - 1}$ but the correct answer is $f(x) = 2e^{5(x-3)}$ (you can follow this link to see how it is solved).

Note: I am not asking how to get the correct answer, as I have already found and understood such an explanation. I am instead asking why my process is incorrect.

My Method:

1. $\frac{df(x)}{dx} = 5f(x)$
2. $df(x) = 5f(x)dx$
3. $\int{df(x)} = \int{5f(x)dx}$
4. $f(x) = 5f(x)\cdot x + C$
5. $f(3) = 5f(3)\cdot 3 + C$
6. $2 = 30 + C$
7. $C = -28$
8. $f(x) = 5f(x)\cdot x - 28$
9. $28 = f(x)(5x - 1)$
10. $f(x) = \frac{28}{5x - 1}$

Explanation: I know my answer is wrong, as it does not satisfy the condition $f'(x) = 5f(x)$, yet I am intrigued by this problem as I cannot seem to figure out what I did wrong. My answer does satisfy the condition $f(3) = 2$ and its derivative - $f'(x) = -\frac{140}{(5x - 1)^2}$ - is suggestive of some similarity to the original differential (eg numerator is multiplied by 5). Perhaps someone can enlighten me regarding where I went astray. Thank you.

Answer 1

Your mistake is in going from (3) to (4).

Since $f(x)$ is a function of $x$, you cannot treat it as a constant in the integral. Basically, you should not say

$$\int 5 f(x) dx = 5 f(x) \cdot x + C$$

Answer 2

as @Michael has pointed out, Line 3 is a bit complicated. Because you don't know how to integrate $\int f(x)dx$

$$f'=5f$$ $$\implies \frac {f'}{f}=5$$ $$(\ln f)'=5$$ Integrate $$\ln f=5\int dx=5x+K$$ Therefore $$f(x)=Ke^{5x}$$ $$f(3)=2 \implies K=\frac 2 {e^{15}}$$ $$f(x)=\frac {2e^{5x}} {e^{15}}=2e^{5(x-3)}$$