# Dif. Eq. General solution is sum of homogeneous and particular solutions

Why is the general solution to ordinary first order differential equations a sum of homogeneous(Setting the inhomogeneous term to 0) and particular(satisfies the differential equation but not necessarily the initial conditions) solutions?

$x=x_{h}+x_{p}$

I have tried finding a proof on the Internet but without a result.

Can someone give me a not too complicated mathematical proof(If possible) to why this relation is valid?

Thank you!

I will denote the linear differential operator by $D$. Let $p$ be a particular solution to the equation $D(x)=b$, such that $D(p)=b$. We can also consider a homogeneous solution $h$, such that $D(h)=0$.

Now consider the set $S=\{p+h|h \text{ satisfies the homogeneous equation}\}$ I will show that this is equal to the solution set. We will show set inclusion both ways.

First we will show that if a general solution exists, it is in this set. Assume that $s\in \Omega$, the full solution set to the differential equation, then consider:

$D(s-p)=D(s)-D(p)=b-b=0$.

This means that $s-p$ is a homogeneous solution, so $s$ can be written as a particular solution and some element from the set of homogeneous solutions. So $\Omega \subset S$.

Finally, consider:

$D(p+h)=D(p)+D(h)=b+0=b$.

So $S\subset \Omega$.

Thus $S=\Omega$ $\square$.

Without initial or boundary conditions, the solution to your average DE is a family of solutions. For example, the solution set to $x' = x$ is $\{Ae^t\}$, a set parametrised by the value $A$.

So, if $x_p$ is a particular solution to a given linear DE, and $x_h$ is a generic element of the solution set of the homogeneous DE, then at the very least you can show that $x = x_h + x_p$ represents a solution set to the original DE, because $x' = x_h' + x_p'$ and so on for all further derivatives. In other words, if the original DE is $a_0 x + a_1 x' + a_2 x'' + \ldots + a_n x^{(n)} = c$, then we have:

$\begin{eqnarray}a_0 x + a_1 x' + a_2 x'' + \ldots + a_n x^{(n)} & = & a_0 (x_h + x_p) + a_1 (x_h + x_p)' + \ldots + a_n (x_h + x_p)^{(n)} \\ & = & a_0 x_h + a_0 x_p + a_1 x_h' + a_1 x_p' + \ldots + a_n x_h^{(n)} + a_n x_p^{(n)} \\ & = & \left( a_0 x_h + a_1 x_h' + \ldots + a_n x_h^{(n)} \right) + \left( a_0 x_p + a_1 x_p' + \ldots + a_n x_p^{(n)} \right) \\ & = & 0 + C\end{eqnarray}$

Because $x_h$ is a solution to the homogeneous DE $a_0 x_h + a_1 x_h' + \ldots + a_n x_h^{(n)} = 0$ and $x_p$ is a solution to the linear DE.

Thus, we at least know that if we add a particular solution and a homogeneous solution together, we get another solution to the DE. So if we add the entire homogeneous solution set to the particular solution we generate a set of functions that are also solutions to the original DE. So as long as we can then pick an element of that set that satisfies our boundary/initial conditions, we can generate the required solution to the DE.

Proof that (a) the set of all $x_h + x_p$ is the full solution set to the DE, and (b) that the method generates unique solutions are slightly trickier exercises.

I found this proof in a very nice Dutch\Flemish book on linear algebra, published by the university of Leuven, it's one of my favourite proofs, since it's so simple:

Let $T$ represent a linear map (this is our linear differential operator), acting between two vector spaces (a space of functions). It sends elements (functions) of an m-dimensional vector space to elements (functions) of an n-dimensional vector space.

Let $x_0 \in \ker(T)$, such that $T(x_0)=0 \in \mathbb{R}^n$.

Notice that this is the property that any homogeneous solution satisfies. Given a homogeneous solution $x_0$, the differential operator spits out $0$.

Now $\forall b \in$ Image(T), $\exists a$ such that $T(a)=b$. We will call this the solution for $b$, and say that $a$ solves the linear differential equation. Since T is a linear map, we may write:

$a=a-x_0+x_0$

$a=c+x_0$

$T(a)=T(c+x_0)$

$T(a)=T(c)+T(x_0)=b$

So $c+x_0$ solves our differential equation, since $x_0$ is in the kernel, this is a homogeneous solution, $c$ is not in the kernel, so it's a particular solution.

Notice that $c$ is still an element of the domain of $T$. We may take arbitrary combinations. Notice also that $T(c)=T(a)-T(x_0)=b$, so $c$ is still a solution to the linear differential equation for $b$. We might call this a particular solution, and we can write any general solution in this separated form as described above, consisting of a homogeneous solution and a particular solution. $\square$

• What is the domain of $T$? (Incidentally, please use MathJax consistently.) The vector space of the solutions of the homogeneous ODE? What is $T$ itself? The evaluation of the action of the differential operator at some point? Or something else? With so many unclear statements, this cannot be considered an answer. – user539887 May 23 '18 at 21:03
• In the first paragraph I list what T is, it's a linear map. If we consider linear differential operators as linear maps, the vector space would be a space of functions, or, the solutions to the differential equation and the map gives you the associated differential equation. Example: $T(f(x))=x^2$ would be, for some specific $T(f(x))=3\frac{d^2 f(x)}{dx^2}+\frac{df(x)}{dx}$. This leads to the equation: $3\frac{d^2 f(x)}{dx^2}+\frac{df(x)}{dx}=x^2$. In the more general setting, we can always write the solution f(x) in terms of an element of the kernel and any particular one. – Wesley Strik May 24 '18 at 18:41
• This however, is a specific example of the fact that in any vector space, under some given linear map, we can write an element in terms of some other element and an element from the kernel. That's basically what $y_{general}=y_{homogeneous}+y_{particular}$ is saying. – Wesley Strik May 24 '18 at 19:06
• Indeed I have not shown any uniqueness with this proof, this is slightly trickier, it was merely an illustration that any full solution can be broken up like this. – Wesley Strik May 24 '18 at 19:21