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In resolving ODE, I often have the following problem :

Let $$xy'-y=0.$$ Then, if $x\neq 0$ and $y\neq 0$, $$\frac{y'}{y}=\frac{1}{x}\iff \ln|y|=\ln|x|+K, \quad K\in \mathbb R$$ and thus $|y|=D|x|$ for $D>0$ and thus $$y=\pm D |x|\iff y=M|x|,\quad M\in \mathbb R$$ because we also have that $0$ is solution. But I think that we can say more : $$y=\pm D |x|\iff y=Mx,\quad M\in \mathbb R,$$ i.e. we can take the absolute value off because $$|y|=|x|\iff y=\pm x$$ and thus if $D>0$ (as previously) we have $$|y|=D|x|\iff y=\pm D x\iff y=Mx,\quad M\in \mathbb R$$ Q1) Am I correct ?

In the solution my teacher always does it. The problem it's that in one equation, he didn't do it and did as following :

$$u'-\frac{8}{3x}u=0.$$ He did as : $$\frac{u'}{u}=\frac{8}{3x}\iff \ln|u|=\frac{8}{3}\ln|x|+C\iff |u|=D|x|^{8/3}\iff u=K|x|^{8/3}$$ with $C\in \mathbb R$, $D>0$ and finally $K\in \mathbb R$.

Q2) Why he let the absolute value ? Can't we write $u=K x^{8/3}$ ? Indeed, as before $$|u|=D|x|^{8/3}\iff u=\pm D x^{8/3}\iff u=Kx^{8/3}$$ with $D>0$ and $K\in \mathbb R$

MSE
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It isn't true that if $|y|=D|x|$ for $D>0$ then $y=M x$ for some $M\in\mathbb{R}$. For example, when $y=|x|$ the equation $|y|=|x|$ is satisfied but we can't find a $M$ such that $y=Mx$ for all $x$. If we try to make it work for $x>0$ we find that $M=1$ whereas for $x<0$ we get $M=-1$.

coconut
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