Author Topic: Q1-T5101  (Read 1733 times)

Victor Ivrii

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Q1-T5101
« on: January 25, 2018, 08:19:11 AM »
Find the general solution of the given differential equation, and use it to determine how solutions behave as $t\to \infty$.
$$
ty' - y = t^2e^{-t}, \qquad t > 0.
$$

Junya Zhang

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Re: Q1-T5101
« Reply #1 on: January 25, 2018, 09:41:08 AM »
First notice that the given DE is a first order linear differential equation.
Rewrite it in the standard form we have
$$y'- \frac{1}{t}y = te^{-t}$$

Let $\mu(t)$ denote the integrating factor.
$$\mu(t)=e^{\int -\frac{1}{t} dt} = e^{- log(t)} = e^{log((t)^{-1})} = t^{-1}. $$

Multiply both sides of the equation by $\mu(t)$ we get
$$t^{-1}y'-t^{-2}y=e^{-t}$$
$$\frac{d}{dx}[t^{-1}y]=e^{-t}$$

Integrate both sides with respect to $t$ we get
$$t^{-1}y = -e^{-t} + C$$

So,
 $$y=-te^{-t} + Ct$$

When $t \to \infty$:
Case1: if $C=0$
Then by one use of L'Hopital's rule, we get $y \to 0$.

Case2: if $C>0$
Then $y \to +\infty$.

Case3: if $C<0$
Then $y \to - \infty$.
« Last Edit: January 25, 2018, 09:45:45 AM by Junya Zhang »