# Elementary connection between linear algebra and differential equations

While studying linear algebra I encountered the following result regarding the solutions of a non homogeneous system of linear equations: if $$Sol(A,\pmb{b})\neq \emptyset$$

$$Sol(A,\pmb{b})= \pmb{x} + Sol(A,\pmb{0})$$

Where x is a particular solution.

Then, while studying differential equations, I found that the solutions for

$$y'=a(x)y+b(x)$$

are all of the form

$$y=y_p+Ce^{A(x)}$$

Where the last term refers to the solutions of the associate homogeneous equation and $$y_p$$ is a particular solution.
These two concepts seem related to me, but I've not been able to find/think about a satisfying reason, so I'd like to know whether there is actually a relation or not.
I should specify that I've not been able to study properly systems of differential equations of the first order or equations of the n-th order yet, I suspect that the answer is there but I can't be sure. However I'd like to get the complete answer, so if it involves the latter topics it's not a problem.

This can be useful to visualize the correlation between linear algebra and systems of differential equations:

If $$A$$ and $$B$$ are two linear spaces, let $$\Gamma:A\to B$$ be a linear operator with $$b\in B$$, then the level set of $$b$$ of $$\Gamma$$ can be obtained translating $$Ker(\Gamma)$$ of a factor $$\Gamma^{-1}(b)$$, hence $$\forall \gamma\in\Gamma^{-1}(b):\Gamma^{-1}(b)=\gamma+Ker(\Gamma).$$ Let's now consider $$y\in\mathcal C^{(n)}(I)$$ and define $$\Gamma_{a_0,\dots,a_{n-1}}(y):=y^{(n)}+\overset{n-1}{\underset{i=0}{\sum}}a_i(x)y^{(i)}.$$ The derivative defines a linear operation, so $$\Gamma$$ is a linear operator such that $$\Gamma:\mathcal C^{(n)}(I)\to\mathcal C^{(0)}(I)$$.
Now solving the system $$(*)\begin{cases}y(x_0)=y_0\\\vdots\\y^{(n-1)}(x_0)=y_0^{(n-1)} \end{cases}$$ is equivalent to $$\Gamma(y)=b$$, and solving the homogeneous equation associated to $$(*)$$ is equivalent to $$\Gamma(y)=0$$.

We have showed that the the general integral of $$(*)$$ is a linear variety and for this reason it can be expressed by the sum of a particular solution of $$(*)$$ with a solution of the associated homogeneous equation.

• Thank you for your answer. I did not understand everything, but I'll try to – JackV Feb 26 at 16:02