Interval of the maximum solution of a nonlinear equation

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Mentor note: Thread moved from the Technical Math section, so there is no template.
@yamata1, , in the future, please post homework problems or exercises in the Homework & Coursework sections, not in the Technical Math sections. I have moved your post.

Hello,

I would like some help with an exercise, specifically the last question:
Let [itex]y_0 \in \mathbb{R} [/itex] and [itex]y'(t)=exp(y(t))(1-exp(y(t)))[/itex] with the initial condition [itex]y(0)=y_0[/itex]
[itex]1-[/itex]Justify that this equation has a unique maximum solution [itex]y[/itex] on an interval [itex] I[/itex] that contains [itex]0[/itex].
[itex]2-[/itex]Show that if [itex]y_0=0[/itex] then [itex]I=\mathbb{R}[/itex] and [itex]u(t)=0[/itex], [itex]\forall t \in
\mathbb{R}[/itex].
[itex]3-[/itex]Show that if [itex]y_0>0[/itex] then [itex]y(t)>0[/itex] with [itex]\forall t \in I[/itex]
[itex]4-[/itex]Let [itex]y_0>0[/itex]
[itex]a)[/itex]Show that [itex]\frac{1}{exp(x)(1-exp(x))}=exp(-x)+\frac{exp(-x)}{exp(x)(exp(-x)-1)}[/itex].
[itex]b)[/itex] [itex]H(x)=-exp(-x)-ln(1-exp(-x))[/itex] for [itex]x>0[/itex] show that [itex]H(y(t))=H(y_0)+t[/itex],[itex]\forall t \in I[/itex].
[itex]c)[/itex] Show that the function H is a bijection of [itex]\mathbb{R}^*_+[/itex] in [itex]\mathbb{R}^*_+[/itex].
[itex]d)[/itex]Deduce that [itex]I=]-H(y_0),+\infty[[/itex].

Ideas for answers:
[itex]1-[/itex]Apply Picard's existence theorem or Cauchy–Lipschitz theorem
[itex]2-[/itex]If [itex]y_0=0[/itex] then [itex]y'(0)=0[/itex] then [itex]y(t)=0[/itex]
we consider the maximal solution Xmax of the equation , defined on an interval [itex]I_{max}[/itex].
We suppose there exists a function[itex] f : \mathbb{R} \rightarrow \mathbb{R}+[/itex], continuous,such that
[itex]||X(t)|| \leqslant f(t) \forall t [/itex] [itex]\in I_{max}[/itex] : then [itex]I_{max}=\mathbb{R}[/itex].
3-The solution would be monotonous from [itex]y'(t)[/itex]>0 ,increasing from 0 but I'm not certain on the way to prove it.
4-a)simplify exp(-x) on both sides , put the R.H.S terms to a common denominator add them and multiply the R.H.S ;both the numerator and denominator; by exp(x).
b)Prove [itex]H(y(t))'=y'(t)*H'(y(t))[/itex]=[itex]\frac{H(y(t))-H(y_0)}{t-0}=1[/itex]
c)The monotony of H(y(t)) is enough I think.
d) I would like help for this , all I can find is [itex] H(y(t))>0 \Longrightarrow H(y_0)+t >0\Longrightarrow t> -H(y_0)[/itex]

I welcome your suggestions or answers ,especially [itex]4-d)[/itex].
 
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Ray Vickson
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Mentor note: Thread moved from the Technical Math section, so there is no template.
@yamata1, , in the future, please post homework problems or exercises in the Homework & Coursework sections, not in the Technical Math sections. I have moved your post.

Hello,

I would like some help with an exercise, specifically the last question:
Let [itex]y_0 \in \mathbb{R} [/itex] and [itex]y'(t)=exp(y(t))(1-exp(y(t)))[/itex] with the initial condition [itex]y(0)=y_0[/itex]
[itex]1-[/itex]Justify that this equation has a unique maximum solution [itex]y[/itex] on an interval [itex] I[/itex] that contains [itex]0[/itex].
[itex]2-[/itex]Show that if [itex]y_0=0[/itex] then [itex]I=\mathbb{R}[/itex] and [itex]u(t)=0[/itex], [itex]\forall t \in
\mathbb{R}[/itex].
[itex]3-[/itex]Show that if [itex]y_0>0[/itex] then [itex]y(t)>0[/itex] with [itex]\forall t \in I[/itex]
[itex]4-[/itex]Let [itex]y_0>0[/itex]
[itex]a)[/itex]Show that [itex]\frac{1}{exp(x)(1-exp(x))}=exp(-x)+\frac{exp(-x)}{exp(x)(exp(-x)-1)}[/itex].
[itex]b)[/itex] [itex]H(x)=-exp(-x)-ln(1-exp(-x))[/itex] for [itex]x>0[/itex] show that [itex]H(y(t))=H(y_0)+t[/itex],[itex]\forall t \in I[/itex].
[itex]c)[/itex] Show that the function H is a bijection of [itex]\mathbb{R}^*_+[/itex] in [itex]\mathbb{R}^*_+[/itex].
[itex]d)[/itex]Deduce that [itex]I=]-H(y_0),+\infty[[/itex].

Ideas for answers:
[itex]1-[/itex]Apply Picard's existence theorem or Cauchy–Lipschitz theorem
[itex]2-[/itex]If [itex]y_0=0[/itex] then [itex]y'(0)=0[/itex] then [itex]y(t)=0[/itex]
we consider the maximal solution Xmax of the equation , defined on an interval [itex]I_{max}[/itex].
We suppose there exists a function[itex] f : \mathbb{R} \rightarrow \mathbb{R}+[/itex], continuous,such that
[itex]||X(t)|| \leqslant f(t) \forall t [/itex] [itex]\in I_{max}[/itex] : then [itex]I_{max}=\mathbb{R}[/itex].
3-The solution would be monotonous from [itex]y'(t)[/itex]>0 ,increasing from 0 but I'm not certain on the way to prove it.
4-a)simplify exp(-x) on both sides , put the R.H.S terms to a common denominator add them and multiply the R.H.S ;both the numerator and denominator; by exp(x).
b)Prove [itex]H(y(t))'=y'(t)*H'(y(t))[/itex]=[itex]\frac{H(y(t))-H(y_0)}{t-0}=1[/itex]
c)The monotony of H(y(t)) is enough I think.
d) I would like help for this , all I can find is [itex] H(y(t))>0 \Longrightarrow H(y_0)+t >0\Longrightarrow t> -H(y_0)[/itex]

I welcome your suggestions or answers ,especially [itex]4-d)[/itex].
I would suggest you look at the DE for ##z(t) = \exp(-y(t))##; it is easily solvable.
 

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