I have the following differential equation
$x^2u'' + xu' + (2+\lambda)u = 0$
and the solution provided says to assume $u=x^p$ which leads to
$p^2=-(2+\lambda)$ and then jumps to
$u = A\cos(\sqrt{2 +\lambda}\log(x)) + B\sin(\sqrt{2 +\lambda}\log(x))$
I can see that it clearly is a solution but I have no idea how the $\log(x)$ got in there.
$$x^2u'' + xu' + (2+\lambda)u = 0$$
$$p^2=-(2+\lambda)\implies$$
$$ p=\pm i\sqrt {2+\lambda}$$
$$ u= x^p= e^{p \ln x} $$
$$ u_1=e^{ i\sqrt {2+\lambda}\ln x}=\cos (\sqrt {2+\lambda}\ln x)+ i\sin (\sqrt {2+\lambda}\ln x) $$
Where we have used the Euler's
$$ e^{i\theta}=\cos (\theta)+i\sin (\theta)$$
and
$$u_2=e^{ i\sqrt {2+\lambda}\ln x}=\cos (\sqrt {2+\lambda}\ln x)- i\sin (\sqrt {2+\lambda}\ln x) $$
And the rest is clear.