Difference between two solution of inhomogeneous linear equation Show that the difference between two solutions of an inhomogeneous linear equation $Lu =g$ with the same $g$ is the solution of the homogenous equation $Lu=0$
I know the definition of linearity, but how can I use it to show they have the same solution. Can someone please help me with it. 
 A: I think the problem is more a problem of basic proof-strategy/proof-writing than really a problem about linearity or differential equations. Therefore, I will try to show how mechanical this kind of proof can be. An interesting lecture on that topic is this blogpost by the Field medalist Tim Gowers.

The difference between two solutions of an inhomogeneous linear equation $Lu=g$ with the same $g$ is the solution of the homogenous equation $Lu=0$



*

*The proposition starts by "The difference between two solutions of an inhomogeneous linear equation $Lu=g$.". So, any proof should starts by picking two such solutions: Let $u_1$ and $u_2$ be two solutions of $Lu=g$ and $u=u_2-u_1$ their difference.

*The proposition finishes by "is the solution of the homogenous equation $Lu=0$" so our goal should be to show that $Lu=0$. Therefore, our last line of the proof should be $Lu=0$. Or maybe the next to last line, since I think it is better to write a conclusion in English if possible.

*Now, we try to fill the blank between the first and last lines. It may be useful to imagine moving from an initial state, hypothesis, to a final state, conclusion. And "backward reasoning" may help: ask yourself "From which position can I be sure to reach the desired outcome?" (playing chess or other boardgames helps). 

*How can I obtain $Lu=0$? Surely, since $u=u_1-u_2$, if I know that $L(u_1-u_2)=0$, I win!

*How can I obtain $L(u_1-u_2)=0$? Since $L$ is linear, if I know that $Lu_1-Lu_2=0$, I win!

*How can I show that $Lu_1-Lu_2=0$? If $Lu_1=Lu_2$, that's surely true!

*But I know that $Lu_1=Lu_2=g$, so I am done!


Note that the above list is mostly a draft and it should be written properly as follows:

proof: Let $u_1$ and $u_2$ be two solutions of $Lu=g$ and $u=u_2-u_1$ their difference. Then $Lu_1=Lu_2=g$, so $Lu_1-Lu_2=g-g=0$. Therefore, since $L$ is linear, we have $L(u_1-u_2)=Lu_1-Lu_2=0$. This means that $Lu=0$, so $u$ is a solution of the homogeneous equation. $\square$

Notes:


*

*There are certainly variations on the method, but I am pretty confident that this one will solve most of the proof problems.

*"From which position can I be sure to reach the desired outcome?" is different from "If I assume the desired outcome (conclusion), what can I deduce?". For example, writing "$Lu=0$ so $L(u_1-u_2)=0$, so $Lu_1-Lu_2=0$. Therefore, $Lu_1=Lu_2$. So $u_1$ and $u_2$ are solutions of the same inhomogeneous equation" would not prove what you want to prove. It would actually prove the converse.

*The drawback of this mechanic method is that you don't focus on the intuitive meaning of the theorem (that the space of the solution of an inhomogeneous linear equation is an affine space whose direction is the space of solution of the associated homogeneous equation). The intuitive meaning is of course very important, but you don't go very far if you are not able to write proofs of theorems...


I hope it helps.
A: You can use linearity like this:
If
$Lu_1 = g \tag{1}$
and
$Lu_2 = g, \tag{2}$
then
$Lu_1 - Lu_2$
$= g - g = 0; \tag{3}$
but, by linearity,
$Lu_1 - Lu_2$
$= L(u_1 - u_2), \tag{4}$
so
$L(u_1 - u_2) = 0, \tag{5}$
as per request.
