# Reduction of Order on Second Order Linear Homogenous ODE Explanation

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?

• 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 $v(t)$. Feb 26, 2018 at 1:11
• Could you elaborate on why we can only solve for $v(t)$ in the linear case? Feb 26, 2018 at 1:18

$$y''+p(x)y'+q(x)y=0$$
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.
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 $y_1(t) \ne 0$).