Final answer:
The general solutions for the given differential equations were found using the characteristic equations, yielding y = c1e2x + c2e3x for distinct real roots, y = c1cos(2x) + c2sin(2x) for complex roots, and y = (c1 + c2x)e3x for repeated roots.
Step-by-step explanation:
The task is to find the most general solution for three second-order linear homogeneous differential equations with constant coefficients. The general approach involves finding the characteristic equation associated with each differential equation, determining its roots, and hence the general solution.
Equation 1: y" – 5y' + 6y = 0
The characteristic equation is r2 – 5r + 6 = 0. Factoring gives (r – 2)(r – 3) = 0, so the roots are r = 2 and r = 3. Since these are real and distinct, the general solution is y = c1e2x + c2e3x, where c1 and c2 are constants determined by initial conditions.
Equation 2: y'' + 4y = 0
The characteristic equation is r2 + 4 = 0. Solving for r, we find complex roots r = ±2i. The general solution is y = c1cos(2x) + c2sin(2x).
Equation 3: y'' – 6y' + 9y = 0
The characteristic equation is r2 – 6r + 9 = 0. This factors to (r – 3)2 = 0, giving a repeated root r = 3. The general solution for repeated roots is y = (c1 + c2x)e3x.
Reasoning and Verification
For each equation, we identified the characteristic equation, and by using the discriminant or factoring, we determined the roots. We then wrote down the general solution depending on the nature of the roots: real and distinct, complex, or repeated. Verification involves ensuring the solutions satisfy the original differential equations.