Substitute in y=ekx and use the fact that the derivative of ekx is kekx. This gives(k2ekx+kekx)(kekx−ekx)=ke2kxNow ekx is never zero, so we may divide each side by e2kx, giving an cubic equation for k;k(k+ 1) (k−1)=k. This has three solutions; k=0 or k=2 or k=−2. The answer is (d).