Find the general solution of the differential equation.

Question

Find the general solution of the differential equation.

y''' - 5y'' - 25y' +125y = 0

Note = Use c₁, c₂, and c₃ for arbitrary constants.

y(t) =

Answer 1

Note that the given homogeneous linear equation with constant coefficients implies the auxiliary equation m³ - 5m² - 25m + 125 = 0. Since m³ - 5m² - 25m + 125 = m²(m - 5) -25(m - 5) = (m - 5)(m² - 25) = (m - 5)(m - 5)(m + 5) = (m - 5)²(m + 5) from the factoring by grouping method and the difference in squares formula a² - b² = (a - b)(a + b), we also know that (m - 5)²(m + 5) = 0. We can finally conclude from the preceding work that y(t) = c₁te^5t + c₂e^5t + c₃e^-5t is the general solution to the given homogeneous linear equation with constant coefficients.