at2y" + bty' + cy=0, t > 0
where a, b, c R and a 0. Using the substitution y = t" and proceeding as we did for the constant coefficient case, you
can find a characteristic equation for the differential equation. Distinct solutions r₁ and r2 of this characteristic equation can
be used to find a basis for the solution space of the differential equation. Employ this technique to find the general solution of
4t²y" +8ty' - 3y = 0, t > 0.
Fig: 1