Show $v(x) = 0$ is the only solution for $y'' + cy=0, v(0) = 0, v'(0)=0, c\gt 0$
I have this question as the only practice problem I can't do for my exam in thirty minutes.
I can see that it is true, since with these equations, we can normally take:
$y''+ay'+by=2x^2$
$=y'''+ay''+by'=4x$
$=y^{IV} + ay''' + by'' = 4$
and know $by'' = 4$, since $y'''$ and $y^{IV}$ now clearly equal zero. Since we have $c \gt 0$ the $0$ must come from the $y$ term.
But logic isn't a particularly good proof. It is apparently a very trivial problem, but I am clearly missing something. Any help would be appreciated!
Note: I am interested in the solution whether it is before my exam or not! Thank you very much.
$\endgroup$3 Answers
$\begingroup$Let $y$ be a solution to problem: $$ y''+cy=0,~y'(0)=y(0)=0$$ Define $G(x)=(y'(x))^2+c(y(x))^2$. Clearly $G'(x)=2y'(x)(y''(x)+cy(x))=0$, so $G$ must be constant. But $G(0)=0$. Hence $G(x)\equiv0$. But for $c>0$ this can happen if and only if $y(x)\equiv0$.
$\endgroup$ $\begingroup$$y''=y'\frac{d(y')}{dy}$
You can easily prove it. It will transform the equation into first order. Finally, you would have solutions $$y=0$$
$$OR$$ $$y=A\sin(\omega x+\phi)$$
Where $\omega=\sqrt c$ and other 2 are constants.
This is an equation of SHM.
Hope you can take from here. Hope there are better methods out there.
$\endgroup$ $\begingroup$If $f$ is any function such that
$f'' + c f = 0 \tag{1}$
with $c > 0$, we can multiply (1) through by $f'$ to obtain
$f'f'' + cff' = 0, \tag{2}$
and we note that
$(\dfrac{1}{2}(f')^2 + \dfrac{c}{2}f^2)' = f'f'' + cff' = 0, \tag{3}$
from which we conclude that
$\dfrac{1}{2}(f')^2 + \dfrac{c}{2}f^2 = k, \; \text{a constant.} \tag{4}$
If now $v(x)$ satisfies (1) with $v(0) = v'(0) = 0$, then the above shows that
$\dfrac{1}{2}(v'(x))^2 + \dfrac{c}{2}v^2(x) = 0 \tag{5}$
for all $x$; with $c > 0$, each term on the left of (5) must vanish for it to hold; thus, $v(x) = 0$ everywhere.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
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