\[Ly=h(x)\] f is the linear combination of all the functions that are L(function)=0, so that \[Lf=0\] And g is ONE OF the functions that do this: \[Lg=h(x)\] Why is the general (i.e. total and only (minus degrees of freedom for unknowable coefficients)- is this the correct definition?) solution for y equal to this: \[y=f+g\] I understand that it WORKS, but I've read in many places to stop here, as I've found the 'general' solution (whatever that really means) despite only using 1 out of potentially many g that do this \[L(function)=h(x)\]? Does this rule only apply to a subset of differential equations, or are all other g not independent from out 1st g?

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\[Ly=h(x)\] f is the linear combination of all the functions that are L(function)=0, so that \[Lf=0\] And g is ONE OF the functions that do this: \[Lg=h(x)\] Why is the general (i.e. total and only (minus degrees of freedom for unknowable coefficients)- is this the correct definition?) solution for y equal to this: \[y=f+g\] I understand that it WORKS, but I've read in many places to stop here, as I've found the 'general' solution (whatever that really means) despite only using 1 out of potentially many g that do this \[L(function)=h(x)\]? Does this rule only apply to a subset of differential equations, or are all other g not independent from out 1st g?

Calculus1
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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