If $y'=y^2+p(x)$ and $z'\le z^2+p(x)$, with $y(0)=z(0)=0$, then $z(x)\le y(x)$, for $x\in[0,1)$. Is this true?
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From $y'=y^2+p(x)$ and $z'\le z^2+p(x)$, we can get that $z'-y'\leq z^2-y^2$. By calculation, $$[e^{-\int(z+y)\mathrm{dx}}(z-y)]'=e^{-\int(z+y)\mathrm{dx}}[(z-y)'-(z+y)(z-y)]\leq0$$ AS $y(0)=z(0)$, so for $x\in[0,1)$, by integral, we can get $e^{-\int(z+y)\mathrm{dx}}(z-y)\leq0$, or $z(x)\leq y(x)$. |
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