Question

If $y=e^{mx}\sin(nx)$, then the value of \[ \frac{d^2y}{dx^2}-2m\frac{dy}{dx}+(m^2+n^2)y \] is:

• A. $0$
• B. $y$
• C. $2y$
• D. $-y$

Hint:

Functions involving $e^{mx}\sin(nx)$ and $e^{mx}\cos(nx)$ frequently satisfy linear differential equations. Always look for cancellation patterns after repeated differentiation.

Answer 1

The Correct Option is A

Concept: This problem involves repeated differentiation of exponential-trigonometric functions. Functions of the form \[ e^{mx}\sin(nx) \] and \[ e^{mx}\cos(nx) \] often satisfy standard second-order differential equations. We carefully differentiate the function twice and simplify the expression.

Step 1: Differentiate the given function once.
Given, \[ y=e^{mx}\sin(nx) \] Using product rule: \[ \frac{dy}{dx} = e^{mx}(n\cos nx)+\sin(nx)\left(me^{mx}\right) \] \[ \frac{dy}{dx} = e^{mx}\left(n\cos nx+m\sin nx\right) \]

Step 2: Differentiate again to find $\dfrac{d^2y}{dx^2}$.
Differentiate: \[ \frac{dy}{dx} = e^{mx}(n\cos nx+m\sin nx) \] Again applying product rule: \[ \frac{d^2y}{dx^2} = me^{mx}(n\cos nx+m\sin nx) + e^{mx}(-n^2\sin nx+mn\cos nx) \] Combining terms: \[ \frac{d^2y}{dx^2} = e^{mx}\left(2mn\cos nx+(m^2-n^2)\sin nx\right) \]

Step 3: Substitute into the required expression.
We need: \[ \frac{d^2y}{dx^2} -2m\frac{dy}{dx} +(m^2+n^2)y \] Substituting values: \[ = e^{mx}\left(2mn\cos nx+(m^2-n^2)\sin nx\right) \] \[ -2m\left[e^{mx}(n\cos nx+m\sin nx)\right] \] \[ +(m^2+n^2)e^{mx}\sin nx \] Factor out $e^{mx}$: \[ =e^{mx}\Big[ 2mn\cos nx+(m^2-n^2)\sin nx \] \[ -2mn\cos nx-2m^2\sin nx +(m^2+n^2)\sin nx \Big] \] Now simplify: \[ =e^{mx}\Big[ (2mn-2mn)\cos nx \] \[ +(m^2-n^2-2m^2+m^2+n^2)\sin nx \Big] \] \[ =e^{mx}(0) \] Hence, \[ \boxed{0} \]