Toronto Math Forum
MAT2442018S => MAT244Tests => Quiz1 => Topic started by: Victor Ivrii 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 $t\to \infty$.
$$
ty'  y = t^2e^{t}, \qquad t > 0.
$$

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 $y \to  \infty$.