Find the general solution of the differential equations below \( \frac{d^{2} y}{d x^{2}}-25 y=0 \)

To solve the differential equation \( \frac{d^{2} y}{d x^{2}} - 25 y = 0 \), we start by rewriting it as a characteristic equation. The associated characteristic equation is \( r^2 - 25 = 0 \). Solving this gives us \( r = 5 \) and \( r = -5 \). Thus, the general solution takes the form \( y(x) = C_1 e^{5x} + C_2 e^{-5x} \), where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial or boundary conditions. This type of differential equation appears frequently in scenarios such as modeling oscillations or electrical circuits. For example, in mechanical systems, the solutions can describe the behavior of a mass-spring-damper system at natural frequencies. The exponential functions reflect the growth and decay of responses to initial conditions, which is fundamental in fields like engineering and physics. In terms of common mistakes, one might forget to include both exponential parts of the solution or miscalculate the roots of the characteristic equation. Always double-check calculations of the characteristic equation \(r^2 - 25 = 0\) for accuracy. Remember, each root contributes an independent solution, so no blending or skipping is allowed!