Find the solution to the $y''' + 4y'' + 5y' = 0$ using the wronskian.
What I have done:
$$y^3+4y^2+5y=0 \\ \implies y(y^2+4y+5) \\ \implies y^2(y+4)+5y=0 \\ \implies y+4 = -\frac{5}{y} \\ \implies \cdots \\ y = i-2$$
The solution is $$y = e^{-2x}(A\cos(x)+B\sin(x))$$
Using the wronskin for $1, e^{-2x}\cos(x), e^{-2x}\sin(x)$, I get the solution $W = -8e^{-4x}$.
$$\begin{pmatrix}1 & e^{-2x}\cos(x) & e^{-2x}\sin(x) \\ 0 & 2e^{-2x}\sin(x) & -2e^{-2x}\cos(x) \\ 0 & -4e^{-2x}\cos(x) & -4e^{-2x}\sin(x) \end{pmatrix} = 1\left(-8e^{-4x}\sin^2(x) - 8 e^{-4x}\cos^2(x) \right)=-8e^{-4x}$$ The answer in my booklet gives $-5e^{-4x}$, how did they get this?