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I'm doing the following problem: The differential equation $$ \dot{y} = X(t,y), X(t,y) = \frac{1}{3}y^{1/4} +t^{1/3}$$ defined on $D_X = (0,\infty)\times(0,\infty)$.

I already solved it with: $y(t) = t^{4/3}$.

But here is what i don't understand. The problem says:

For $\eta >0$ let $(I_\eta,y_\eta)$ denote the maximal solution with $y_\eta(1) = \eta$

for:

a) For $0 < \eta < 1 $ Then $y_\eta(t) < t^{4/3}$, for $t \in I_\eta$

b) For $\eta > 1 $ Then $y_\eta(t) > t^{4/3}$, for $t \in I_\eta$

I am very confused about the $y_\eta(1) = \eta$ notation, so i can't understand what the goal with the task is. Can you help?

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  • $\begingroup$ $y_\eta$ denotes the maximal solution of the equation satisfying the initial condition $y(1)=\eta$. $I_\eta$ stands for the domain of that maximal solution. You have obtained a solution of the equation for $\eta=1$ only. $\endgroup$ – user539887 Apr 5 at 11:20
  • $\begingroup$ I see. Thank you $\endgroup$ – Pernk Dernets Apr 8 at 6:12

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