8) \( y^{(4)}-8 y^{\prime}=0 \)

To solve the differential equation \( y^{(4)} - 8y' = 0 \), we can first rewrite it as a characteristic equation. Let \( r \) be the characteristic root; this leads to \( r^4 - 8r = 0 \). Factoring out \( r \) gives us \( r(r^3 - 8) = 0 \). The roots are \( r = 0 \) and \( r^3 = 8 \), which gives \( r = 2 \) as the real root (with a multiplicity of 3, since \( r - 2 \) is a factor three times). Therefore, the general solution to this fourth-order differential equation combines these findings: \[ y(t) = C_1 + C_2 t + C_3 e^{2t} + C_4 te^{2t} \] where \( C_1, C_2, C_3, C_4 \) are constants determined by initial conditions. Fun Fact: This equation is linear with constant coefficients, making it quite popular in the study of vibrations and systems in engineering! Real-World Application: Differential equations like this are often seen in physics and engineering, particularly in analyzing systems subjected to forces and in damping oscillators. The solutions help predict behavior over time, crucial for designing everything from buildings to vehicles!