# How to prove linear combination solves homogeneous equation

• I have a third-order homogeneous linear differential equation: $$A_3(u) f^{\prime\prime\prime} + A_2(u) f^{\prime\prime} + A_1(u) f^\prime + A_0(u) f = 0,$$

with three linearly-independent solutions $$\phi_1, \phi_2, \phi_3$$.

• I have a function $$F$$ which is a linear combination of the $$\phi_i$$ and hence solves the differential equation. It happens that $$F$$ is also a linear combination $$F = c_1 g_1(u) + c_2 g_2(u) + c_3 g_3(u)$$ of an unrelated set of three functions, which are themselves linearly-independent.

• I am trying to establish that each $$g_i$$ must itself solve the differential equation (or under what conditions this holds).

So far, I've tried using matrix notation, writing the wronskians:

$$\Phi \equiv \begin{bmatrix}\phi_1 & \phi_2 & \phi_3\\ \phi_1^\prime & \phi_2^\prime & \phi_3^\prime \\ \phi_1^{\prime\prime} & \phi_2^{\prime\prime} & \phi_3^{\prime\prime}\end{bmatrix}$$

$$G \equiv \begin{bmatrix}g_1 & g_2 & g_3\\ g_1^\prime & g_2^\prime & g_3^\prime \\ g_1^{\prime\prime} & g_2^{\prime\prime} & g_3^{\prime\prime}\end{bmatrix}$$

If the $$\phi_i$$ are linearly-independent then $$\Phi$$ is nonsingular/invertible; similarly for the $$g_i$$ and $$G$$. So we can write

$$G = (\Phi\Phi^{-1})G = \Phi(\Phi^{-1}G)$$

Then the columns of $$(\Phi^{-1}G)$$ should be coefficients which show that each $$g_i$$ is a linear combination of $$\phi_i$$ and hence a solution to the differential equation.

My problem is I don't see why the entries of $$\Phi^{-1}G$$ should be constants, as I'm hoping, rather than functions of $$u$$ in their own right.

If it helps, my third order homogeneous linear differential equation is not arbitrary, but instead comes from the definition of $$F$$ as a linear combination $$F = c_1 g_1 + c_2 g_2 + c_3 g_3$$. This equation shows that $$F$$ and the $$g_i$$ linearly depend on each other. The first three derivatives of this equation show that $$F^\prime$$ linearly depends, with the same coefficients, on the $$g_i^\prime$$, and so on.

Hence this wronskian vanishes because its columns are linearly-dependent: $$\det\begin{bmatrix}F & g_1 & g_2 & g_3 \\ F^{\prime} & g_1^{\prime} & g_2^{\prime} & g_3^{\prime} \\ F^{\prime\prime} & g_1^{\prime\prime} & g_2^{\prime\prime} & g_3^{\prime\prime} \\ F^{\prime\prime\prime} & g_1^{\prime\prime\prime} & g_2^{\prime\prime\prime} & g_3^{\prime\prime\prime}\end{bmatrix} = 0$$

Expanding the determinant along the first column yields a third-order homogeneous linear differential equation satisfied by $$F$$. (The coefficient on $$F^{\prime\prime\prime}$$ is nonzero because the three $$g_i$$ are linearly-independent.) The $$\phi_i$$ are then three linearly-independent solutions to this equation.

I can imagine you might plug in each solution $$\phi_i$$ in for $$F$$ in this determinant equation to obtain three equations each of which proves that $$\phi_i$$ is a linear combination of the $$g_j$$. So you have $$\Phi = G A$$, where $$A$$ is a matrix of only constants. But in this case, I am not sure why $$A$$ is invertible so that we can obtain $$G = \Phi A^{-1}$$.

Is it enough to say that $$\det{\Phi} \neq 0$$, and $$\det{\Phi} = \det{G}\det{A}$$ so $$\det{A} \neq 0$$? Can you find the coefficients?

$$\det\begin{bmatrix}y & g_1 & g_2 & g_3 \\ y^{\prime} & g_1^{\prime} & g_2^{\prime} & g_3^{\prime} \\ y^{\prime\prime} & g_1^{\prime\prime} & g_2^{\prime\prime} & g_3^{\prime\prime} \\ y^{\prime\prime\prime} & g_1^{\prime\prime\prime} & g_2^{\prime\prime\prime} & g_3^{\prime\prime\prime}\end{bmatrix} = 0$$ where $$y$$ represents the unknown.
$$F$$ satisfies this equation by assumption. But if we replace $$y$$ with any $$g_i$$ in the matrix, we get a matrix with two matching columns. Hence the columns are linearly dependent, so the determinant vanishes, which is equivalent to saying that $$g_i$$ solves the equation.
You can find the coefficients by finding three points $$x_i$$ such that $$[g_1(x_i), g_2(x_i), g_3(x_i)]$$ are linearly independent. Then solve the matrix equation $$[g_j(x_i)]_{i,j} \cdot [c_j] = [y(x_i)]_i$$ for the unique solution. The solution exists and is unique because the $$g_i$$ are independent and the constants make the vectors $$g_j(x_i)$$ independent of one another.