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: Use logarithms:
Since $f(0)=1 > 0$, let $g = \ln f$. Then $g' = f'/f$ and $g'' = \frac{f f'' - (f')^2}{f^2}$.

Step 2: Apply the condition:
The determinant is zero, so $g'' = 0$. Hence $g$ is linear: $g = ax + b$, and $f = Ce^{ax}$.

Step 3: Use initial values:
$f(0) = 1$ gives $C = 1$. $f'(0) = a = 2$. So $f(x) = e^{2x}$.

Step 4: Check options:
$f'(x) = 2e^{2x} = 2f(x)$. The other options would need $f = e^{-x}$, $e^x$ or a constant, none of which has $f'(0) = 2$.

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