# Linear combination of solns of differntial solns, any geometric explanation?

Just learned that if $y_1$ and $y_2$ are solutions to a homogeneous equation, then so a linear combination of $y_1$ and $y_2$. Now, I am sure, but don't know enough if there is some geometric reasoning.

The reason that I am asking is because for the solutions for the constants $c_1$ and $c_2$ uses matrix.

$y(t_0)=y_0, \quad y^{\prime}(t_0)=y^{\prime}_0$

and $c_1$ and $c_2$ satisfy the equations

\begin{aligned} c_1y_1(t_0)+c_2y_2(t_0)&=y_0 \\[0.4em] c_1y^{\prime}_1(t_0)+c_2y^{\prime}_2(t_0)&=y^{\prime}_0 \end{aligned} \qquad \implies \qquad \begin{aligned} c_1=\frac{\begin{vmatrix} y_0 & y_2(t_0)\\ y^{\prime}_0 & y^{\prime}_2(t_0) \end{vmatrix}}{ \begin{vmatrix} y_1(t_0) & y_2(t_0)\\ y^{\prime}_1(t_0) & y^{\prime}_2(t_0)\end{vmatrix} } \\[0.4em] c_2=\frac{\begin{vmatrix} y_1(t_0) & y_0\\ y^{\prime}_1(t_0) & y^{\prime}_0 \end{vmatrix}}{ \begin{vmatrix} y_1(t_0) & y_2(t_0)\\ y^{\prime}_1(t_0) & y^{\prime}_2(t_0) \end{vmatrix}} \end{aligned}

I am thinking that there has to be a relation between vector spaces and these family of solutions, but not sure.

Am I incorrect?

The same holds of course for any system of independent solutions, where you then have to solve a system of linear equations to translate the initial conditions into coefficients in that basis. But using Cramers rule is the most inefficient way to do that if the order of the differential equation becomes larger than $3$.