2nd order linear equations constant coefficients

[@peradies7044]

So any combination of solutions multiplied by a constant is itself a solution? Wouldn't you then only need one of these solutions if its already two solutions multiplied?

[@mhoefnagel92]

Yes - any linear combination of solutions to a homogeneous differential equation (with 0 on the right) is itself a solution. Solving such a differential equation means finding all possible solutions. For this reason, finding one solution does not solve the differential equation. However, if we are given initial values, then we can single out one solution - the unique solution. For example, solving the equation x^2 - 1 = 0 means to find all possible x that satisfy the equation (x = 1, -1). For this reason, x=1 is not the solution; it is 'a' solution.

[@peradies7044]

ahhh thank you I understand, now it makes sense why linear independance matters!

[@mhoefnagel92]

@@peradies7044 I am glad!