I have am not sure how to solve this problem after I have found the solutions of its characteristic equation.
So what I have is:
$\lambda^2+2x+3=0 \Rightarrow \lambda=-1\pm \sqrt{2}i$
then solution would be: $y =c_1e^{-t}cos(\sqrt{2}t)+c_1e^{-t}sin(\sqrt{2}t)+c_1e^{-t}cos(-\sqrt{2}t)+c_1e^{-t}sin(-\sqrt{2}t) $
What would be $ \psi(t)$ and $ \phi(t)$
You should know that the differential equation of the Wronskian determinant $W=φψ'-φ'ψ$ is homogeneous of first order, thus easy to solve \begin{alignat}{1} W'&=\frac{d}{dt}(φψ'-φ'ψ)&&=φψ''-φ''ψ\\&=φ(-2ψ'-3ψ)-(-2φ'-3φ)ψ&&=-2(φψ'-φ'ψ) \\&=-2W \end{alignat} so that $$ W(t)=e^{-2t}W(0)=e^{-2t}(φ(0)ψ'(0)-φ'(0)ψ(0)) $$ and for the given values that is $W(0.4)=39e^{-0.8}=17.52382960...$