Q1-T0101

[Victor Ivrii]

Find the general solution of the given differential equation, and use it to determine how solutions behave as $t\to \infty$.
\begin{equation*}
y'+3y=t+e^{-2t}.
\end{equation*}

[Meng Wu]

$$y'+3y=t+e^{-2t}$$
Since given differential equation has the form $$y'+p(t)y=g(t)$$
Hence $p(t)=3$ and $g(t)=t+e^{-2t}$ $\\$
First, we find the integrating factor $\mu(t)$$\\$
As we know, $\mu(t)=\exp^{\int p(t)dt}$ $\\$
$\mu(t)=\exp^{\int 3dt}=e^{3t}$$\\$
Mulitply $\mu(t)$ to both sides of the equation, we get:$\\$
$e^{3t}y'+3e^{3t}y=te^{3t}+e^{-2t}\cdot e^{3t}=te^{3t}+te^{t}$ $\\$
and $(e^{3t}y)'=te^{3t}+te^{t}$ $\\$
Integrating both sides: $\\$
$$e^{3t}y=\int{te^{3t}+te^{t}}$$ $\\$
which is also:
$$e^{3t}y=\int{te^{3t}}+\int{te^{t}}$$ $\\$
Now, $\int{te^{t}}=e^{t}+c$, where $c$ is arbitrary constant $\\$
For $\int{te^{3t}}$ we use Integration By Parts: $\\$ Let $u=t, dv=e%{3t}$. $\\$Then we get: $du=dt$ and $v={1\over3}e^{3t}$. $\\$
Thus, $$\int{te^{3t}}=uv-\int{vdu}$$
$$\int{te^{3t}}={t\over3}e^{3t}-\int{{1\over3}e^{3t}dt}$$
$$\int{te^{3t}}={t\over3}e^{3t}-{1\over9}e^{3t}$$ $\\$
Thus, $$e^{3t}y={t\over3}e^{3t}-{1\over9}e^{3t}+e^{t}+c$$ where $c$ is arbitrary constant. $\\$
Now divide both side by $e^{3t}$, we get the general solution: $$y={t\over3}-{1\over9}+e^{-2t}+ce^{-3t}$$ $\\$
Note: $y$ is asymptotic to ${t\over3}-{1\over9}$ as $t\rightarrow \infty$.

[David Chan]

This is a first order, linear differential equation.  So, it can be solved by method of integrating factor.  First, $$\mu(t) = \exp\left(\int 3\,dt\right) = e^{3t}$$  So, multiplying by the integrating factor, we get that $$\frac{d}{dt}(e^{3t}y) = te^{3t} + e^{t}$$  Integrating both sides, $$\begin{align*}ye^{3t} &= \int 3e^{3t} dt + \int e^t dt\\ &= \frac{t}{3}e^{3t} - \frac{1}{9}e^{3t} + e^{t} + C\end{align*}$$ Therefore, $$y = \frac{t}{3} - \frac{1}{9} + e^{-2t} + Ce^{-3t}$$ is the general solution.  As $$t \to \infty, \qquad y \to \frac{t}{3} - \frac{1}{9}$$