Find all values of k for which the differential euqtion y'' + ky' + ky = 0 has a general solution....

Question

Find all values of k for which the differential euqtion y'' + ky' + ky = 0 has a general solution of the given form.

y = c1e^ax + c2e^bx

Apparently, the answer is: k<0 or k>4 ???? how is this the answer? thanks for the help.

Answer 1

The general solution for any k is of the form:

y = c1 e^-((1/2)(k+√(k(k-4)))x + c1 e^((1/2)(k+√(k(k-4)))x

For real solutions, k(k-4) has to be positive, which means k can be anything except between 0 and 4. Plot out k(k-4) if in doubt. Plug in y into the differential equation if in doubt about the form of the solution.