Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade None
Question:
The function $f(x)$ satisfying the equation $[f'(x)]^2+4f'(x)f(x)+[f(x)]^2 = 0$
$f(x) = Ce^{(2-\sqrt{3})x}$
$f(x) = Ce^{(2+\sqrt{3})x}$
$f(x) = Ce^{(\sqrt{3}-2)x}$
$f(x) = Ce^{-(2+\sqrt{3})x}$
Step-by-Step Solution
Key Concept: The differential equation is quadratic in $f'(x)$, which can be solved using the quadratic formula to obtain two possible first-order linear ODEs.
We have the differential equation $[f'(x)]^2 + 4f'(x)f(x) + [f(x)]^2 = 0$. Treating this as a quadratic in $f'(x)$: $f'(x) = \frac{-4f(x) \pm \sqrt{16[f(x)]^2 - 4[f(x)]^2}}{2} = \frac{-4f(x) \pm 2\sqrt{3}f(x)}{2} = f(x)(-2 \pm \sqrt{3})$. This gives two cases: $f'(x) = (-2+\sqrt{3})f(x)$ or $f'(x) = (-2-\sqrt{3})f(x)$. Solving these separable ODEs: $f(x) = Ce^{(\sqrt{3}-2)x}$ or $f(x) = Ce^{-(2+\sqrt{3})x}$.
Correct Answer: 3,4