Question

If $y=3e^{2x}+2e^{3x}.$prove that $\frac{d^{2}y}{d{x}^2}-5\frac{dy}{dx}+6y=0$.

Answer 1

Given : $y=3e^{2x}+2e^{3x}$

$⟹\frac{dy}{dx}=3e^{2x}\times 2+2e^{3x}\times 3=6e^{2x}+6e^{3x}$

$\therefore \frac{dy}{dx}=6e^{2x}+6e^{3x}$

$\frac{d^{2}y}{d{x}^2}=6e^{2x}\times 2+6e^{3x}\times 3$

$=12e^{2x}+18e^{3x}$

Consider, $\frac{d^{2}y}{d{x}^2}-5\frac{dy}{dx}+6y$

$=12e^{2x}+18e^{3x}-5(6e^{2x}+6e^{3x})+6(3e^{2x}+2e^{3x})$

$=12e^{2x}+18e^{3x}-30e^{2x}-30e^{3x}+18e^{2x}+12e^{3x}$

$=30e^{2x}+30e^{3x}-30e^{2x}-30e^{3x}=0$

Hence, $\frac{d^{2}y}{d{x}^2}-5\frac{dy}{dx}+6y=0$