Superimposition principle in solving ODEs

StefanBU
Hi all,

first of all, I have to admit I have often used this richness of knowledge that permeates through the posts of this forum to find answers to questions that I have come across in my studies. Thanks for all! Now, I have a question to post, for the first time.

I am trying to teach myself diff. equations using the MIT OpenCourseware materials and have bumped into a statement to which I have not found a rigorous answer.

One of the ways to identify the general solution to an ODE of the form $\dot{x}+p(t)x=q(t)$ is by the sum of a particular solution and the solution to a homogeneous "version" of the differential equation where $q(t)=0$.

While I can see how the solution "works" by comparing it to a solution derived from the evaluation by integrating factors and the resulting integral expression, I have not been able to prove the principle (and it's bothering me). I also understand that the linearity of ODEs allows us to treat the solution to $\dot{x}+p(t)x=q(t)$ as equivalent to the sum of the solutions of $\dot{x}+p(t)x=q(t)$ and $\dot{x}+p(t)x=C*0$, which is generally referred to a as the superimposition principle.

I have a hunch that the two solutions are linearly independent and sufficient to span the solution space, but I cannot come up with some formal reasoning to justify it. In other words, why does such a sum always hold? Is it true for all ODEs? Can someone provide some depth to this matter?

Homework Helper
No, the set of all solutions to a non-homogeneous equation do not form a vector space. The set of all solutions to a homogeous d.e. form a vector space which is why it is best to go back to that first.

A simple analogy is this- the set of points, (x, y), in a coordinate system form a (two dimensional) vector space with "coordinatewise" addition and scalar multiplication. The points on a straight line through the origin form a subspace. Points on a line not through the origin do NOT form a vector space. For example, $(x_1, x_1+1)$ and $(x_2, x_2+1)$ are in the set of all (x, y) such that y= x+ 1. But the sum $(x_1+ x_2, x_1+ x_2+ 2)$ does not satisfy that equation (it satisfies y= x+ 2) nor does the scalar product $a(x, x+1)= (ax, ax+ a)$ (it satisfies y= x+ a).

But we can look at the line through the origin parallel to that one: y= x. Choose any single point on the line y= x+ 1, (0, 1), say, or (1, 2). Any point on y= x+ 1 can be written as a point on y= x plus that single point: (x, x+ 1)= (x, x)+ (0, 1) or (x, x+ 1)= (x-1, x-1)+ (1, 2).

Similarly, If "L" is an nth order homogeneous differential operation, then the set of all solutions to Ly= 0 forms and n-dimensional vector space. Now suppose that $y_1(x)$ and $y_2(x)$ both satisfy Ly= f(x) for some function f. Then if we let $y(x)= y_1(x)- y_2(x)$ we have $L(y_1- y_2)= L(y_1)- L(y_2)= f(x)- f(x)= 0$. That is, the difference between any two solutions satisfies Ly= 0 and so can be written as a linear combination of n independent solutions to the Ly= 0. Let's call that $y_a$. The $y_1(x)- y_2(x)= y_a(x)$ from which $y_1(x)= y_a(x)+ y_2(x)$. Since $y_1$ can be any solution to the entire equation, that shows that any solution to the entire equation can be written as a solution to the homogeneous equation plus this $y_2(x)$, a solution to the homogeneous equation.

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Homework Helper
While I can see how the solution "works" by comparing it to a solution derived from the evaluation by integrating factors and the resulting integral expression, I have not been able to prove the principle (and it's bothering me). I also understand that the linearity of ODEs allows us to treat the solution to $\dot{x}+p(t)x=q(t)$ as equivalent to the sum of the solutions of $\dot{x}+p(t)x=q(t)$ and $\dot{x}+p(t)x=C*0$, which is generally referred to a as the superimposition principle.

Actually, it's the superposition principle: super + position. Superimposition principle sounds like the linear sum of a bunch of inconvenient burdens to form a super-inconvenient burden. ;)

(The term superimposition is also an actual printing press term, referring to "the placement of an image or video on top of an already-existing image or video, usually to add to the overall image effect, but also sometimes to conceal something (such as when a different face is superimposed over the original face in a photograph).")

StefanBU
@ HallsofIvy: Thanks for the clear explanation.

@ Mute: <embarrassed> I should have read more carfully. I got carried away by "superimpose", which is used commonly in my mother tongue, Italian.