Tanget of differential equation in a point

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My teacher and I have been completely stumped by a differential equation. We solved it in two different ways, and now we aren't sure which result is correct.

The differential of the function f is the solution to the differential equation (dy(x))/(dx) = 2 y(x) - 1. Solve f so the tangent in the point (2, f(2) has the equation y=2x-5/2.

The general solution for f(x) = 1/2 + e^(2 x) * k. Now we get to the point of contention. My teacher solved the specific solution for f(2)=3/2 while I solved it for f'(2)=3/2. I'm pretty confident in my solution, but not enough to make a convincing argument. Any thoughts?

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$$y'(x)=2y(x)-1$$ has the solution $$y(x)=Ce^{2x}+\frac12.$$

The tangent at $x=2$ is

$$Y-y(2)=y'(2)(X-2),$$ $$Y-\left(Ce^{2\cdot2}+\frac12\right)=2Ce^{2\cdot2}(X-2).$$

We identify the coefficients with the given ones and get the system

$$\begin{cases}2Ce^4=2,\\-4Ce^4+Ce^4+\dfrac12=-\dfrac52\end{cases},$$ which is compatible. For both equations, $C=e^{-4}$.

Hence

$$y(x)=e^{2x-4}+\frac12.$$

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