Q98 / 150 · MHT CET 2026 (17 April Shift 1) · Mathematics
If $y = e^{-mx}$ is a solution of the differential equation $\dfrac{d^2y}{dx^2} + 4\dfrac{dy}{dx} + 3y = 0$, then the values of $m$ are
- Correct answerA$1, 3$
- B$-1, 3$
- C$-1, -3$
- D$1, -3$
Solution
Given $y = e^{-mx}$
Differentiating with respect to $x$, we get:
$\dfrac{dy}{dx} = -m e^{-mx}$
$\dfrac{d^2y}{dx^2} = m^2 e^{-mx}$
Substituting these into the given differential equation $\dfrac{d^2y}{dx^2} + 4\dfrac{dy}{dx} + 3y = 0$:
$m^2 e^{-mx} + 4(-m e^{-mx}) + 3(e^{-mx}) = 0$
$e^{-mx}(m^2 - 4m + 3) = 0$
Since $e^{-mx} \neq 0$ for any finite $x$, we have:
$m^2 - 4m + 3 = 0$
$(m - 1)(m - 3) = 0$
$m = 1, 3$
Answer: $1, 3$
Differentiating with respect to $x$, we get:
$\dfrac{dy}{dx} = -m e^{-mx}$
$\dfrac{d^2y}{dx^2} = m^2 e^{-mx}$
Substituting these into the given differential equation $\dfrac{d^2y}{dx^2} + 4\dfrac{dy}{dx} + 3y = 0$:
$m^2 e^{-mx} + 4(-m e^{-mx}) + 3(e^{-mx}) = 0$
$e^{-mx}(m^2 - 4m + 3) = 0$
Since $e^{-mx} \neq 0$ for any finite $x$, we have:
$m^2 - 4m + 3 = 0$
$(m - 1)(m - 3) = 0$
$m = 1, 3$
Answer: $1, 3$
Hence, the correct answer is option: 'A'.