I am trying to figure out if this equation is solvable. I was given
$$ y^{(3)} + 3y^{''} - 10y^{'} = 0 $$
with initial conditions
$$ y(0)=7, y'(0)=0, y''(0)=70. $$
I solved the characteristic equation and ended up with m = -5, m = 0, m = 2. Then I found that the general solution was
$$ y = c_1e^{-5t}+c_2+c_3e^{2t}. $$
But when I find the Wronskian to determine linear independence, I end up with
$$ W = \det\begin{pmatrix} e^{-5t} & 1 & e^{2t} \\ -5e^{-5t} & 0 & 2e^{2t} \\ 25e^{-5t} & 0 & 4e^{2t} \\ \end{pmatrix} = 0. $$ Does this mean that $y$ is not a viable solution?
EDIT: Removed $y = y^{(3)}...$
Hint: By using the substitution $v=y'$ the equation becomes a second order ODE $$v''+3v'-10v=0$$ which has an auxiliary equation of $$m^2+3m-10=0$$ $$(m+5)(m-2)=0$$ $$m=-5, 2$$ $$\therefore v=y'=c_1e^{-5t}+c_2e^{2t}$$ $$y=\int c_1e^{-5t}+c_2e^{2t} dt = c_3e^{-5t}+c_4e^{2t}+c_5$$ $$y(0)=c_3+c_4+c_5=7$$ $$y'(0)=-5c_3+2c_4+c_5=0$$ $$y''(0)=25c_3+4c_4+c_5=70$$ $$\therefore \begin{pmatrix} 1 & 1 & 1 \\ -5 & 2 & 1 \\ 25 & 4 & 1\end{pmatrix}\begin{pmatrix} c_3\\ c_4 \\ c_5\end{pmatrix}=\begin{pmatrix} 7\\ 0\\ 70\end{pmatrix}$$ $$\begin{pmatrix} c_3\\ c_4 \\ c_5\end{pmatrix}=\begin{pmatrix} 1 & 1 & 1 \\ -5 & 2 & 1 \\ 25 & 4 & 1\end{pmatrix}^{-1}\begin{pmatrix} 7\\ 0\\ 70\end{pmatrix}=\begin{pmatrix} 2 \\ 5 \\ 0\end{pmatrix}$$