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I'm struggling to understand why the reduction of order works. I'm reading the guide found here, and it starts with the assumption that, given a solution $y_1(t)$ we may be able to find another solution by multiplying with another function: $y_2(t) = v(t)y_1(t)$.

Previously, such derivations of other solutions, like the principle of superposition, were a result of the linearity of linear ODEs, but in this case, we are multiplying by a non-constant and no rules of linearity apply.

I still have a gut feeling that something relating to linearity is at play, but if this is not the case, then what property of second-order linear homogeneous ODEs allows us to make such an assumption?

Edit: In a linked section, the author justifies this assumption with

To find the second solution we will use the fact that a constant times a solution to a linear homogeneous differential equation is also a solution: $y_2(t) = v(t)y_1(t)$

And yet, the "constant" is another function? How can this be?

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  • $\begingroup$ For an arbitrary equation, you can always make the assumption $y_2(t)=v(t)y_1(t)$, but only in the linear case, we can solve $v(t)$. $\endgroup$
    – W. mu
    Commented Feb 26, 2018 at 1:11
  • $\begingroup$ Could you elaborate on why we can only solve for $v(t)$ in the linear case? $\endgroup$
    – jeanluc
    Commented Feb 26, 2018 at 1:18

2 Answers 2

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$$y''+p(x)y'+q(x)y=0$$

We know that $$y_1''+p(x)y_1'+q(x)y_1=0$$

We plug in $$ y_2 = v y_1$$ and use the above equality to simplify and find an equation for $v$ to solve.

The method works due to the linearity of the equations involved and the fact that $y_1$ is a solution to the homogeneous equation.

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Any solution $y(t)$ to your equation can be written as $y(t) = v(t) y_1(t)$ for some $v(t)$, namely $v(t) = y(t)/y_1(t)$ (at least on an interval where $y_1(t) \ne 0$).

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