If $y = 500{e^{7x}} + 600{e^{ - 7x}},$ then show that $\cfrac{{{d^2}y}}{{d{x^2}}} = 49y.$

Let $y = 500{e^{7x}} + 600{e^{ - 7x}}$ ....(i)
Differentiating (i) w.r.t. x, we get

$\cfrac{{dy}}{{dx}} = 500 \cdot {e^{7x}} \cdot 7 + 600 \cdot {e^{ - 7x}} \cdot ( - 7)$

$= 3500{e^{7x}} - 4200{e^{ - 7x}}$ …(ii)

Differentiating (ii) w.r.t. x, we get

$\cfrac{{{d^2}y}}{{d{x^2}}} = 3500 \cdot 7 \cdot {e^{7x}} - 4200 \cdot ( - 7){e^{ - 7x}} = 24500{e^{7x}} + 29400{e^{ - 7x}}$

$= 49(500{e^{7x}} + 600{e^{ - 7x}}) = 49y$
therefore, $\cfrac{{{d^2}y}}{{d{x^2}}} = 49y$