Discussion
Which one of the following curves correctly represents (schematically) the solution for the equation \(\frac{df}{dx} + 2f = 3;\ f(0) = 0\)?
- A.
- B.
- C.
- D.
Answer: B
The correct curve is (B).
Solve the linear equation with the integrating factor \(e^{2x}\):
\[ \begin{aligned} f(x) &= \tfrac{3}{2} + Ce^{-2x}, \quad f(0) = 0 \\ f(x) &= \tfrac{3}{2}\left(1 - e^{-2x}\right) \end{aligned} \]
The curve starts at 0 with slope \(f'(0) = 3\), not zero, and rises steadily (always concave down) towards \(\frac{3}{2}\). That is graph (B).
Graphs (A) and (C) level off at \(\frac{1}{2}\), the wrong limit. Graphs (C) and (D) start flat with an S-shape, which would need \(f'(0) = 0\), but the equation gives \(f'(0) = 3 - 2f(0) = 3\).
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