Question

If y=eˣ/³ is a solution of 6y′′+y'−y=0, use reduction of order to find a second solution.

Answer 1

Explanation:

Let y = ue^(x/3). Then y' = (u/3)e^(x/3) + u'e^(x/3) and y" = (u/9)e^(x/3) + (2u'/3)e^(x/3) + u"e^(x/3).

We have 6y" + y' - y = 6[(u/9)e^(x/3) + (2u'/3)e^(x/3) + u"e^(x/3)] + [(u/3)e^(x/3) + u'e^(x/3)] - ue^(x/3) = (5u' + 6u")e^(x/3) = 0.

Clearly, we must have 5u' + 6u" = 0, which implies that either u = 0 or u = -(6/5)e^(-5/6)x.

Hence, a second non-trivial solution for y is y = -(6/5)e^(-x/2).