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:

 $\frac{dy}{dt}=2 t e^{y-3}\\\\ \frac{dy}{e^{y-3}}=2t dt\\\\\text{Integrating both sides}\\\\\int e^{3-y}\, dy=2\int {t}dt\\\\-e^{3-y}=2\times \frac{t^2}{2} +k\\\\ -e^{3-y}=t^2+k$

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.