Superposition principle

Suppose $L[y] = ay''+by'+cy$. Since differentiation satisfies $(u+v)'=u'+v'$ and $(ku)'=ku'$ for a constant $k$, the operator $L$ is linear: $L[c_1y_1+c_2y_2] = a(c_1y_1+c_2y_2)'' + b(c_1y_1+c_2y_2)' + c(c_1y_1+c_2y_2) = c_1(ay_1''+by_1'+cy_1) + c_2(ay_2''+by_2'+cy_2) = c_1L[y_1]+c_2L[y_2]$.

Because $y_1,y_2$ solve the equation, $L[y_1]=0$ and $L[y_2]=0$, so $L[c_1y_1+c_2y_2]=c_1\cdot 0+c_2\cdot 0=0$: the combination is again a solution. This proves the first half.

For the second half, the theory of linear ODEs guarantees that the solution space of a second-order equation is exactly two-dimensional (an initial condition $y(x_0)=y_0$, $y'(x_0)=y_0'$ picks out a unique solution, giving two free parameters). A nonzero Wronskian $W(y_1,y_2)=y_1y_2'-y_2y_1'$ at some point $x_0$ means the linear system $c_1y_1(x_0)+c_2y_2(x_0)=y_0$, $c_1y_1'(x_0)+c_2y_2'(x_0)=y_0'$ has a unique solution $(c_1,c_2)$ for every choice of initial data. So every solution matches some $c_1y_1+c_2y_2$ for suitable constants, which means $\{y_1,y_2\}$ spans the whole two-dimensional solution space.