Lecture Notes in Differential Equations - Bruce E. Shapiro

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Example 29.6. Solve t 2 y ′′ + ty ′ + y = 0. The indicial equation is
0 = r(r − 1) + r + 1 = r 2 + 1 =⇒ r = ±i (29.52)
Thus the roots are a complex conjugate pair with real part λ = 0 and
imaginary part µ = 1. Hence the solution is
y = At 0 cos(1 · ln t) + Bt 0 sin(1 · ln t) = A cos ln t + B sin ln t (29.53)
Summary of Cauchy-Euler Equation
To solve t 2 y ′′ + αty ′ + βy = 0 find the roots of
r(r − 1) + αr + β = 0 (29.54)
There are three possible case.
1. If the roots are real and distinct, say r 1 ≠ r 2 , then the solution is
y = C 1 t r1 + C 2 t r2 (29.55)
2. If the roots are real and equal, say r = r 1 = r 2 , then the solution is
y = C 1 t r + C 2 t r ln |t| (29.56)
3. If the roots form a complex conjugate pair r = λ±iµ, where λ, µ ∈ R,
then the solution is
y = C 1 t λ cos(µ ln t) + C 2 t λ sin(µ ln t) (29.57)
where, in each case, C 1 and C 2 are arbitrary constants (that may be
determined by initial conditions).