Toronto Math Forum
MAT2442018F => MAT244Tests => Quiz1 => Topic started 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*}

answer to tut0301

$$
y'  2y = e^{2t}\\\mu (x) = e^{\int2\ dt} = e^{2t}\\\frac{d}{dt} (e^{2t}y) = e^{2t}y' 2e^{2t} y = e^{2t}e^{2t}= 1\\ \text{integral on both side gives:}\\ e^{2t}y = t + C\\y = te^{2t} + C{e^{2t}}\\y(0) = 0 + C = 2 \implies C = 2\\\text{thus} \ \ y =e^{2t}(t+2)\\\text{and} \ \ t \to \infty \implies y \to \infty
$$