WHY is the general solution a combination of two solutions?

Hey forum, I've been asking myself this question ever since I first was introduced to differential equations. Suppose we have a nonhomogeneous differential equation of the form



where is a constant. Exactly WHY is the general solution a sum of the particular solution and the solution to the homogeneous solution? Why? Is there a proof?

E.g. the general solution to is the sum of the particular solution which is and the solution to the homogeneous equation, which is . Combining these, we arrive at the general solution



Sure, you could verify that it satisfies the equation but that doesn't answer my question. What justifies combining different solutions to acquire a general one? Why do we add the particular and homogeneous one? Proof? Any intuitive explanation?

I suggest doing an online search for the "superposition principle" to see why this works.

Let p(x) be any solution to L(y)= f(x) where "L" represents some linear combination of derivatives and "f" is a function of x. Now, if Y(x) is any solution to the entire differential equation, then L(Y- p)= L(Y)- L(p)= f(x)- f(x)= 0 so that Y- p satisfies the associated homogeneous equation.

Also you should, when you were first learning about "linear homogeneous differential equations" have learned the "theorem": "The set of all solutions to a nth order linear homogeneous differential equation" form a vector space of dimension n". That means, from the definition of "dimension" for vector spaces, that if we can find n independent solutions (a basis for the vector space) then any solution can be written as a linear combination of those independent solutions.

it has to do with linear algebra more than differential equations.

if L is ANY linear transformation, the general solution to LX = B is any (particular) solution x plus any element of ker(L) (the solutions to LX = 0).

clearly, if Lx = B, and y is in ker(L), then L(x+y) = Lx + Ly = B + 0 = B.

now suppose that z is ANY solution to LX = B, and x is the particular solution we happened to find.

then z = x + (z - x), and: L(z - x) = Lz - Lx = B - B = 0, so z - x is in ker(L). so we can always put z in terms of: particular + homogeneous solution.

the result for linear differential equations then follows from the fact that D (taking the derivative) is a linear operator on the vector space of (suitably many times) differentiable functions.