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


are all of the form


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.

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

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