# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-1 => Topic started by: Victor Ivrii on September 28, 2018, 03:30:24 PM

Title: Q1: TUT0301
Post by: Victor Ivrii on September 28, 2018, 03:30:24 PM
Find the general solution of the given differential equation, and use it to determine how solutions behave as $t\to\infty$.
\begin{equation*}
y' - 2y = e^{2t},\qquad y(0)=2.
\end{equation*}
Title: Re: Q1: TUT0301
Post by: Jiabei Bi on September 28, 2018, 06:00:22 PM
$$y' - 2y = e^{2t}\\\mu (x) = e^{\int-2\ dt} = e^{-2t}\\\frac{d}{dt} (e^{-2t}y) = e^{-2t}y' -2e^{-2t} y = e^{-2t}e^{2t}= 1\\ \text{integral on both side gives:}\\ e^{-2t}y = t + C\\y = te^{2t} + C{e^{2t}}\\y(0) = 0 + C = 2 \implies C = 2\\\text{thus} \ \ y =e^{2t}(t+2)\\\text{and} \ \ t \to \infty \implies y \to \infty$$