Question

Find the general solution to the Cauchy-Euler equation: t²y'' + 2ty' - 6y = 0 in t > 0.

Answer 1

The general solution of the Cauchy-Euler equation t²y'' + 2ty' - 6y = 0 can be found by assuming a trial solution of the form y = t^m, leading to characteristic equation m^2 + m - 6 = 0. Solving this using the quadratic formula yields m = 2 and m = -3, thus the general solution is y(t) = C1t^2 + C2t^-3.

Step-by-step explanation:

Finding the General Solution to a Cauchy-Euler Equation

To find the general solution of the Cauchy-Euler equation t²y'' + 2ty' - 6y = 0, we need to make a substitution to turn the equation into a constant coefficient linear differential equation. Assuming the trial solution y = tm, we obtain derivatives y' = mtm-1 and y'' = m(m-1)tm-2. Plugging these into the differential equation gives:

t²(m(m-1)tm-2) + 2t(mtm-1) - 6tm = 0

Simplifying the above, we get the characteristic equation:

m(m-1) + 2m - 6 = 0

Which simplifies to:

m2 + m - 6 = 0

Applying the quadratic formula, we find:

m = (-1 ± √(1 + 24)) / 2

Thus, m = 2 or m = -3. Thereby, the general solution is a combination of these two solutions:

y(t) = C1t2 + C2t-3

Where C1 and C2 are constants determined by initial conditions or additional information.