Build a basis of homogeneous solutions

Try

A second-order equation constrains acceleration as well as position and velocity. To select one trajectory, expect two initial measurements. Consider y″−3y′+2y=0. Exponentials are useful trial functions because differentiating e^(rt) only multiplies it by r. Substitute the trial function before guessing any constants.

The result is (r²−3r+2)e^(rt)=0. Since the exponential never vanishes, r must satisfy the characteristic equation r²−3r+2=0. Factor it as (r−1)(r−2), producing two modes, e^t and e^(2t).

Predict

Both modes grow, but does every nonzero combination stay positive? Predict the sign near zero of the solution with y(0)=0 and y′(0)=1. Then ask whether the same initial position alone would be enough to choose that solution. Distinguish the equation’s possible modes from the coefficients selected by initial data.

Experiment

Write y=C₁e^t+C₂e^(2t). The initial position gives C₁+C₂=0, and the initial velocity gives C₁+2C₂=1. Subtracting gives C₂=1 and C₁=−1. Thus y=e^(2t)−e^t.

Expression	Value at zero	Contribution to equation
y=e^(2t)−e^t	0	+2y
y′=2e^(2t)−e^t	1	−3y′
y″=4e^(2t)−e^t	3	+y″

Collect the e^(2t) coefficients: 4−6+2=0. Collect the e^t coefficients: −1+3−2=0. This checks the differential equation independently of solving the coefficient system. For positive time the solution is positive because e^t(e^t−1) is positive.

Understand

For ay″+by′+cy=0 with constant a nonzero, two distinct real characteristic roots r₁,r₂ give y=C₁e^(r₁t)+C₂e^(r₂t). The word homogeneous means the forcing term is zero. Linearity ensures that a linear combination of homogeneous solutions is another solution. Independence ensures the two modes supply separate degrees of freedom.

Their Wronskian, the determinant formed from the functions and their first derivatives, is (r₂−r₁)e^((r₁+r₂)t), which never vanishes when the roots are distinct. Equivalently, the two initial-condition equations have a nonzero determinant. This explains why two initial data select a unique combination.

For y=C₁e^t+C₂e^(−t), initial values y(0)=3 and y′(0)=1 give C₁+C₂=3 and C₁−C₂=1, so C₁=2,C₂=1. The positive growing mode ultimately dominates even though the other mode decays. A tiny nonzero coefficient on an unstable mode can matter at later times.

Superposition needs care for forced equations. If L[y₁]=f and L[y₂]=f for a linear differential operator L, then L[y₁+y₂]=2f, not f. The difference y₁−y₂ is homogeneous. Therefore the general forced solution is one particular solution plus the homogeneous family, rather than an arbitrary sum of particular solutions.

Apply

Choose two different initial velocities with the same initial position and solve their coefficient systems. Explain how the equation allows both trajectories. Then change a sign in one mode and use substitution to detect whether it remains a solution.

In your practice, report the characteristic polynomial, roots, basis, coefficient equations, and final verification. A root alone is not the requested function, and an unassigned pair of constants does not yet solve an initial-value problem.

Continue

Repeated roots and complex roots require different real bases. The next lesson derives those forms and shows why duplicating the same exponential cannot provide the missing second degree of freedom.