If y = (A + Bx)e⁻²ˣ, prove that: (d²y)/(dx²) + 4(dy)/(dx) + 4y = 0 (Application)

If $y = (A + Bx)e^{-2x}$, prove that: $\frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 4y = 0$ (Application)

Answer

Official answer key

Differentiate twice and eliminate $B e^{-2x}$ between the two results.

Given $y = (A+Bx)e^{-2x}$
Differentiate (product rule): $\frac{dy}{dx} = (A+Bx)(-2e^{-2x}) + Be^{-2x} = -2y + Be^{-2x}$
So $\frac{dy}{dx} + 2y = Be^{-2x}$ ... (i)
Differentiate (i) again: $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = -2Be^{-2x}$ ... (ii)
From (i) and (ii): $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = -2\left(\frac{dy}{dx} + 2y\right)$
Hence $\frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 4y = 0$