reason behind finding particular solution for non-homogeneous differential equation I was wondering about the reason behind having to find a particular solution for a non-homogeneous linear differential equation and having to combine it with the general solution to get the whole solution. Thanks in advance.
 A: It's essentially a result from linear algebra, using the fact that the functions in question form a real vector space and the derivative is a linear operator. So I'm going to start with a little bit of linear algebra:
Let $V,W$ be vector spaces over $\mathbb R$ or $\mathbb C$ (or any field, really, but for differential equations these two are of interest), and let $L:V\longrightarrow W$ be a linear map. Also let $b\in W$. If $v\in V$ is a solution of the equation $L(x)=b$, then the solution set of the equation is exactly $v+\ker L:=\{v+x~|~x\in\ker L\}$.
Proof: we first show that every element of $v+\ker L$ is a solution, and then we show that every solution is in $v+\ker L$.
The first part is easy: all elements of that set are of the form $v+x$, where $x\in\ker L$, that is $L(x)=0$. So
$$L(v+x)=L(v)+L(x)=b+0=b,$$
so we do in fact have a solution.
The second part is slightly trickier: if $v'$ is a solution of the equation, then $L(v')=b$. Since $L(v)=b$ as well, this gives us $L(v')=L(v)$, and by linearity we get $L(v'-v)=0$, so $v'-v\in\ker L$. But then $v'=v+(v'-v)$, which is $v$ plus an element of the kernel, so $v'\in v+\ker L$.
Now let's look at linear differential equations. They are of the form
$$\frac{\mathrm d}{\mathrm dt}y-Ay=b,$$
where $A$ is a matrix and $b$ some function (the inhomogenous part). Now the derivative and the matrix are linear, so we could instead define $L:=\frac{\mathrm d}{\mathrm dt}-A$ and write this as
$$L(y)=b.$$
Since $L$ is linear, we know by the theory above that the solution set is some arbitrary solution plus the kernel of $L$. The arbitrary solution is the particular solution. The kernel is the set of all functions satisfying $L(y)=0$, but that's just the homogenous equation, so $\ker L$ is the set of all solutions of the homogenous equations. You could say that it is the general solution to the homogenous equation. So we take a particular solution and add the general solution to the homogenous problem.
