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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?

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

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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 $v(t)$.

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

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