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    i've shown the general solution of y" + 3y' + 2y = 6e^x to be...

    y = Ae^(-2x) + B(e^x) + (e^x)

    and the general solution of y" + 3y' + 2y = 2x - 5 to be...

    y = Ae^(-2x) + B(e^x) + x - 4

    and then i am told that the solution of y" + 3y' + 2y = 6(e^x) + 2x - 5 to be...

    y = Ae^(-2x) + B(e^x) + (e^x) + x - 4

    thats all well and good and useful etc... but here is the perennial favourite question 'but why?'

    Without assuming its true i'm not sure how to work it out and therefore show the answers match. How would I prove that the additive property works?

    edit: i think i am looking for something more complicated than it is... bear with me a minute....

    edit again: don't i feel like a fool...

    y" + 3y' + 2y = 6e^x + 2x + 5

    first bit... y = Ae^(-2x) + B(e^-x) easy enough...

    second bit...

    y = @(e^x) + £x + $
    y' = @(e^x) + £
    y" = @(e^x)

    subbing back in gives...

    [email protected](e^x) + 3£ + 2£x + 2$ = 6(e^x) + 2x - 5
    @ =1
    £ = 1
    $ = -4

    y = Ae^(-2x) + B(e^x) + (e^x) + x - 4
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    (Original post by El Stevo)
    thats all well and good and useful etc... but here is the perennial favourite question 'but why?'
    Solutions to differential equations form a Vector Space.

    If u is a solution, and v is another solution, then u + v is indeed another solution.

    Also if u is a solution, and t is a number over the field F (real, complex...), then tu is another solution.

    Galois
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    (Original post by Galois)
    Solutions to differential equations form a Vector Space.

    If u is a solution, and v is another solution, then u + v is indeed another solution.

    Also if u is a solution, and t is a number over the field F (real, complex...), then tu is another solution.

    Galois
    Sorry to be pedantic, but solutions to HOMOGENEOUS LINEAR DEs form a vector space, neither of which these are. Their solution spaces are actually called AFFINE SPACES.

    The solution space to

    y'' + 3y' + 2y = 0

    is a vector space though, and this is the crucial point. The reason El Stevo's method works is that if u_1 is a particular solution to the inhomogeneous equation

    y'' + y' + 2y = f_1

    then every solution to this equation can be written as Y + u_1 where Y is a solution to the original homogeneous equation. This is fairly easily seen to be the case by substituting y = z + u_1 into the inhomogeneous equation and seeing the z must satisfy the homogeneous one.

    So if u_1 and u_2 are solns to the inhomogeneous equation with f_1 and f_2 on the RHS, then u_1+u_2 can easily be seen to be a particular solution of the equation

    y'' + y' + 2y = f_1 + f_2

    by linearity, and so

    Y + u_1 + u_2

    is the general solution, where as before Y is the general solution of the homogeneous equation.
 
 
 
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