# How exactly is the Wronskian tied with Cramer's rule in a system of ODEs?

If we have a nonhomogeneous system of linear differential equations: $$\frac{d\vec{y}}{dx}=A\vec{y}+\vec{b}$$ where: $$\frac{d\vec{y}}{dx}=\begin{bmatrix}y_1'\\y_2'\\y_3'\end{bmatrix},\vec{y}=\begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix},A=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}, \vec{b}=\begin{bmatrix}b_1(x)\\b_2(x)\\b_3(x)\end{bmatrix}$$ the particular solution involving these three functions could be written: $$y_{particular}=c_1(x)e^{\lambda_1x}\vec{u_1}+c_2(x)e^{\lambda_2x}\vec{u_2}+c_3(x)e^{\lambda_3x}\vec{u_3}$$ where $$\lambda_i, \vec{u_i}$$ are the eigenvalues and eigenvectors of the constant matrix $$A$$ (I chose it constant to make things simpler.)

When trying to evaluate these $$c_i(x)$$ functions, I was told to evaluate them by: $$c_i(x)=\int\frac{W_i}{W}dx$$ Where $$W$$ is the Wronskian determinant whose columns are the elements of the vectors $$c_i(x)e^{\lambda_3i}\vec{u_i}$$, and $$W_i$$ is the Wronskian determinant whose $$i$$th column was replaced by the contents of the $$\vec{b}$$ vector.

My questions are: How does the integral show up? Why does the process look like Cramer's rule? How did we lead up to that result? Where did this 'Wronskian' determinant come from?

If I have missed something, please point it out and I'll add it to the post.

Write the general solution of the homogeneous linear ODE $$\tag{1} \frac{\mathrm{d}\vec{y}}{\mathrm{d}x} = A(x) \vec{y}$$ in the matrix form as $$\Phi(x; x_0) \vec{c},$$ where $$\vec{c} = \mathrm{col}(c_1, \dots, c_n)$$ is a constant matrix (that is, independent of $$x$$), and $$\Phi$$ is the state-transition matrix such that $$\Phi(x_0, x_0) = I$$ (the identity matrix).
Now we are looking for a general solution of the nonhomogeneous linear ODE $$\tag{2} \frac{\mathrm{d}\vec{y}}{\mathrm{d}x} = A(x) \vec{y} + \vec{b}(x)$$ in the form $$\Phi(x, x_0) \vec{c}(x).$$ Incidentally, here is the source of the term "variation of constants". We look for conditions for the above function of $$x$$ to be a solution of $$(2)$$, that is, to have $$\frac{\mathrm{d}}{\mathrm{d}x}(\Phi(x, x_0) \vec{c}(x)) = A(x) (\Phi(x, x_0) \vec{c}(x)) + \vec{b}(x).$$ But $$\frac{\mathrm{d}}{\mathrm{d}x}(\Phi(x, x_0) \vec{c}(x)) = \frac{\mathrm{d}}{\mathrm{d}x} \Phi(x, x_0) \cdot \vec{c}(x) + \Phi(x, x_0) \cdot \vec{c}'(x)$$ and $$\frac{\mathrm{d}}{\mathrm{d}x} \Phi(x, x_0) = A(x) \Phi(x, x_0),$$ so after cancellation we obtain $$\Phi(x, x_0) \cdot \vec{c}'(x) = \vec{b}(x),$$ which is, for each fixed $$x$$, a Cramer (because $$\det{\Phi(x, x_0)} = W(x, x_0) \ne 0$$) system of $$n$$ linear equations with $$n$$ unknowns, $$c'_1(x), \dots, c'_n(x)$$. The unknowns are given by the Cramer rule. We have thus obtained the derivative of the matrix function $$\vec{c}(x)$$, so we have to integrate that derivative. And that is the source of the integral.
• One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $\vec{c}'=\Phi^{-1}(x,x_0)\vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix. – Gradient Entropy Mar 17 at 23:01
• It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $\Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $\Phi(x,x_0)=\Psi(x)\Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System. – user539887 Mar 18 at 7:18