Alternate form of Principle of superposition

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SUMMARY

The discussion centers on the Principle of Superposition in linear differential equations, specifically the expression $$(L[c_1y_1 + c_2y_2] = 0) ↔ (L[y_1] = L[y_2] = 0)$$. The participants confirm that including the condition of "any coefficients ##c_1## and ##c_2##" is essential for the validity of the principle. The rewritten form emphasizes that if the linear operator L applied to a linear combination of solutions equals zero, then each individual solution must also satisfy the operator. This establishes a definitive equivalence in the context of linear systems.

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Homework Statement
I am trying to reword my textbook definition of the principle of superposition in terms of propositional logic
Relevant Equations
$$L[y] = y^{''} + p(t)y^{'} + q(t)y = 0$$
The definition is,
1712891868313.png

I rewrite it as $$(L[y_1] = L[y_2] = 0) \rightarrow (L[c_1y_1 + c_2y_2] = 0)$$.

However, I also wonder, whether it could also be rewritten as,

$$(L[c_1y_1 + c_2y_2] = 0) \rightarrow (L[y_1] = L[y_2] = 0) $$

And thus, combining, the two cases,

Principle of superposition. $$(L[c_1y_1 + c_2y_2] = 0) ↔ (L[y_1] = L[y_2] = 0)$$

Is my reasoning correct please?

Thanks for any help!
 
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As long as you include the statement about ”any coefficients ##c_1## and ##c_2##” it is obvious that if ##L[c_1y_1 + c_2y_2] = 0## then each of the ys is a solution.
 
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Orodruin said:
As long as you include the statement about ”any coefficients ##c_1## and ##c_2##” it is obvious that if ##L[c_1y_1 + c_2y_2] = 0## then each of the ys is a solution.
Thank you for your reply @Orodruin!

Yes that is a good idea to quantify my statement with ∀ to give

$$(L[c_1y_1 + c_2y_2] = 0) ↔ (L[y_1] = L[y_2] = 0)$$ $$∀c_1, c_2 ∈ \mathbb{R}$$


Thanks!
 

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