# Differential Equations- Reduction of order [duplicate]

Why is the equation in the red rectangle true? Why is it that if I have one solution y1(x), the second solution can be written as y2(x)=v(x)*y1(x)?Is it because they are linearly independent?

Someone wrote there "Any function y2(t) can be written as v(t)y1(t), at least on an interval where y1(t)≠0: you just take v(t)=y2(t)/y1(t). "- But the question is-why is this true?

## marked as duplicate by Moo, Tony S.F., Shailesh, Yanior Weg, CesareoApr 29 at 8:18

Forget about differential equations for a second. Imagine someone gives you a function $$f(t)$$ which is nonzero on the interval $$[a,b]$$. Then $$f(t)/f(t) = 1$$ on the interval $$[a,b]$$ (we restrict to this interval because we are worried about $$f(t)=0$$). If I want to write any arbitrary function $$g(t)$$ as a product $$v(t)f(t)$$ I can do it if $$f(t)$$ is nonzero by choosing $$v(t)=g(t)/f(t)$$. Plug it into the expression and see for yourself,

$$v(t)f(t) = \left(\frac{g(t)}{f(t)}\right)f(t) = g(t) \left(\frac{f(t)}{f(t)}\right) = g(t)\cdot 1=g(t)$$

There's nothing special about the functions you are considering being solutions to differential equations, this is an algebraic maneuver that we can always do (if $$f(t)$$ is nonzero).

Maybe it's simpler with numbers, imagine someone gives you a number $$x$$ and you want to write $$y$$ as something times $$x$$, i.e. $$y=\lambda x$$. If $$x$$ is not zero then you can always just choose $$\lambda = \frac{y}{x}$$. We are doing the same thing but at each $$t$$ we have a possibly different $$y$$ and a possibly different $$x$$ (because now we are looking at functions that depend on $$t$$) so that we must choose $$\lambda$$ for each $$t$$ (this is why $$v$$ is a function).

Back to your questions, no they don't have to be linearly independent. In fact, if they are linearly dependent then we have that $$v(t)$$ is a constant. The point is that you are allowing $$v$$ to vary with $$t$$ and so we can create any function we want. I guess the simplest answer to your question of "why is this true" is that $$f(t)/f(t)=1$$ when $$f(t)\neq 0$$.

Now that we've established that you can write any function $$g(t)$$ as $$v(t)$$ times some known, nonzero function $$f(t)$$ the question you really ought to be asking is "why do we choose the first solution" and the answer is that Bernoulli was a clever guy and this choice simplifies the problem of finding $$v(t)$$.

• Thank you very much for the explanation! – noam Azulay Apr 28 at 20:11