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Aryan Salinas

Answered question

2022-03-24

How do I compute the maximum value of the solution of this differential equation?

{\begin{aligned} \frac{d y}{d t} & = f (y) = 1 + k ϵ - k y - e^{- y} \\ y (0) & = ϵ \end{aligned}

Answer & Explanation

Kathleen Sanchez

Beginner2022-03-25Added 7 answers

Step 1
Here is my approach to show that $y_{\infty} = \overset{―}{y}$
For the sake of contradiction assume that $y_{\infty} < \overset{―}{y}$ . Because $y (t)$ is a solution to your IVP we must have for any $T > 0$ that
$\int_{y (0)}^{y (T)} \frac{d x}{1 + k ϵ - k x - e^{- x}} = \int_{0}^{T} d t$
This implies for any $T > 0$
$\int_{ϵ}^{y (T)} \frac{d x}{1 + k ϵ - k x - e^{- x}} = T$
Take $T \mapsto + \infty$ and get
$\int_{ϵ}^{y_{\infty}} \frac{d x}{1 + k ϵ - k x - e^{- x}} = + \infty$
This is a contradiction since $x \mapsto \frac{1}{1 + k ϵ - k x - e^{- x}}$ is continuous on $[ϵ, y_{\infty}]$

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Answer & Explanation

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