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I understand that if $y_1$ and $y_2$ are solutions, then $Ay_1+By_2$ is also a solution, and I understand by $A$ and $B$ can take any number. But why are there only 2 things (here $y_1$ and $y_2$, I encountered the word 'degrees of freedom' alluding to the same thing in Feynman's lectures), and we can be sure that there is not a $y_3$, that is not a multiple of $y_2$ of $y_2$, hiding away?

I assume that this problem generalises to having n independent solutions for an nth order linear homogeneous equation.

If it helps in the writing of your answers, am very much a beginner in differential equations.

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What is important here is that $y_1$ and $y_2$ are independent of each other. By independence it is meant that one cannot be equal to a constant multiple of the other. It can be proved that this condition (together with $y_1, y_2$ being solutions) is equivalent to showing that the Wronskian $y_1y_2'-y_2y_1'$ is non zero identically in the domain under consideration.

Once this is guaranteed it is easy to prove that any other solution $y$ can be formed by combining $y_1$ and $y_2$. To prove this, let $x_0$ be a point in the domain of $y$ and consider the system of equations in the variables $A$ and $B$:

$Ay_1(x_0) + By_2(x_0)=y(x_0)$

$Ay_1'(x_0) + By_2'(x_0)=y'(x_0)$

This system has a nontrivial solution provided $y_1y_2'-y_2y_1'$ is non zero which it is and so we can find a $A$ and B satisfying the given equations. Further since $y$ and $Ay_1+By_2$ are both solutions behaving identically at $x_0$ (and likewise their derivatives) and by a uniqueness theorem there can be only one, so $y=Ay_1+By_2$. In this way all other solutions are combinations of $y_1$ and $y_2$.

(For the proofs on the result on the Wronskian and on uniqueness I suggest you consult your text book.)

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