Final answer:
The general solutions to the given differential equations involve finding the roots of their characteristic equations; these roots may lead to exponential, trigonometric, or polynomial functions that comprise the solution.
Step-by-step explanation:
Finding the General Solution of Differential Equations
To find the general solutions of the given differential equations, one would typically start by looking for characteristic equations associated with each differential equation. These are polynomials whose roots will indicate the form of the solutions. The solutions to these differential equations can be exponential functions, trigonometric functions, or combinations thereof depending on the roots of the characteristic equation.
- For the first equation y'' - y' - 6y = 0, the characteristic equation is r² - r - 6 = 0, which has roots r=3 and r=-2, leading to a general solution of y = C1e³⁴ + C2e⁻²⁵.
- The second equation y'' + 8y' + 16y = 0 has a characteristic equation r² + 8r + 16 = 0, with a repeated root r=-4, giving the general solution y = (C1 + C2x)e⁻⁴x.
- In the third equation y'' + 4y' + 5y = 0, the characteristic equation r² + 4r + 5 = 0 has complex roots, resulting in a solution involving trigonometric functions y = e⁻²x(C1cosx + C2sinx).
- The fourth equation involves a third-order differential, which is more complex to solve. However, the characteristic polynomial for y''' + y'' - 2y = 0 would provide the necessary roots to construct the general solution.
- For the fifth equation, which seems to have a typo, one would need to correctly identify the differential equation before proceeding to solve it.
- The sixth equation, y''' + 12y'' + 36y' = 0, is another higher-order differential equation with repeated roots, resulting in a solution that includes polynomial terms in addition to the exponential functions.
To verify a solution, substitute the first and second derivatives of the function back into the original differential equation to ensure that it satisfies the equation.