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When finding the general solution of second order constant-coefficient ODE, the variables must be independent.

The general solution for a second order constant-coefficient ODE is $y = c_1y_1 + c_2y_2$, where $y_1$ and $y_2$ are solutions.

This means that $y_1$ and $y_2$ cannot be dependent. In other words, they cannot be constant multiples of each other.

Please explain why this requirements exists. Use examples to illustrate your argument. Thank you.

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    $\begingroup$ Did you mean that they can't be dependent instead? $\endgroup$ – Marra Oct 19 '16 at 9:34
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    $\begingroup$ @Marra mistyped it. Thank you! $\endgroup$ – The Pointer Oct 19 '16 at 9:35
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If you have found, by whichever means, two solutions $y_1$ and $y_2$ then any linear combination $c_1y_1+c_2y_2$ with real (or complex) constants is again a solution – no question about that.

But when it comes to get hold of the totality of solutions then two solutions $y_1$, $y_2$ which are constant multiples of each other are no better than just one solution $y_1$, and are just not enough: One solution $y_1$ and all its constant multiples are not yet "all solutions".

The theory of such (i.e. second order linear homogeneous) ODEs guarantees us a two-dimensional vector space $V$ of solutions. Such a $V$ is spanned by any two linearly independent elements of $V$, but you definitely need two of these.

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  • $\begingroup$ Thanks. Can you please clarify what you mean in this segment: 'But when it comes to get hold on the totality of solutions then two solutions $y_1$, $y_2$ which are constant multiples of each other are no better than just one solution $y_1$, and are just not enough.'? $\endgroup$ – The Pointer Oct 19 '16 at 13:23
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    $\begingroup$ @ThePointer If you have two solutions $y_1,y_2$ and $y_1=cy_2$ for a constant $c$, then the set $\{ c_1y_1+c_2y_2 : c_1,c_2 \in \mathbb{R} \}$ is the same as just $\{ c y_1 : c \in \mathbb{R} \}$. In particular it is one dimensional, not two dimensional. The theory guarantees that the solution set is two dimensional. You can find this development in any ODE text (it is a bit long and tedious for a MSE answer). $\endgroup$ – Ian Oct 19 '16 at 13:36

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