Question

Consider the initial-value problem: dy/dt= 2te^(y-3), y(0) = 1. (a) Show Equation (1) possesses a unique local solution y(t). (b) Compute the solution y(t) explicitly. (c) Does the solution exist globally (for all times t)? Why? (d) Determine lim_t rightarrow -infinity y(t) and lim_t rightarrow infinity y(t) (if possible).

Answer 1

Answer with explanation:

`(dy)/(dt)=2 t e^(y-3)\\\\ (dy)/(e^(y-3))=2t dt\\\\\text{Integrating both sides}\\\\\int e^(3-y)\, dy=2\int {t}dt\\\\-e^(3-y)=2* (t^2)/(2) +k\\\\ -e^(3-y)=t^2+k`

When , t=0 , then y=1.

`-e^(3-1)=0^2+k\\\\k=-e^(-2)\\\\-e^(3-y)=t^2-e^(-2)\\\\ \text{Taking log on both sides}\\\\3-y=\log(e^(-2)-t^2)\\\\y(t)=3-\log(e^(-2)-t^2)`

`\lim_(t\to-\infty)3-\log(e^(-2)-t^2)\\\\=3-\log(e^(-2)-\infty)\\\\=3- m\\\\=m\\\\\lim_(t \to -\infty)  3-\log(e^(-2)-t^2)=m\\\\\lim_(t\to\infty) 3-\log(e^(-2)-t^2)\\\\=3-\log(e^(-2)-\infty)\\\\=3- m\\\\ \lim_(t \to \infty) 3-\log(e^(-2)-t^2) =m`

where, m=not defined

As, log(-\infinity)=not defined

→log of negative is not defined.So, y(t)=Not defined.