Question

If the differential equation \(\begin{array}{cc}f(x) & f^'(x) \\ f^'(x) & f^{''}(x)\end{array} = 0\) for all \(x\) and \(f(0) = 1,f^'(0) = 2\), then...

• A. \(f^'(x) = -f(x)\)
• B. \(f^'(x) = f(x)\)
• C. \(f^'(x) = 2f(x)\)
• D. \(f^'(x) = 0\)

Hint:

Show the determinant condition says d/dx of f prime over f is zero, then use the initial values.

Answer 1

The Correct Option is C

Step 1: Expand the determinant:
The condition is \(f f'' - (f')^2 = 0\) for all x, so \(f f'' = (f')^2\).

Step 2: Recognise a derivative:
Divide by \(f^2\) (f is not zero near 0 because \(f(0)=1\)): \(\frac{f f'' - (f')^2}{f^2} = 0\). The left side is exactly \(\frac{d}{dx}\left(\frac{f'}{f}\right)\).

Step 3: Integrate:
So \(\frac{f'}{f}\) is a constant \(c\). Using \(f(0) = 1\) and \(f'(0) = 2\): \(c = 2/1 = 2\).

Step 4: Conclude:
\[ f'(x) = 2f(x) \]
The options \(f' = -f\), \(f' = f\) and \(f' = 0\) would give \(c = -1, 1, 0\), which do not match \(c = 2\).

Final Answer:
\(f'(x) = 2f(x)\), option (C). \[ \boxed{f'(x) = 2f(x)} \]