-2.1.9 Find general solution of 2y'+y=3t

[karush]

$\tiny{2.1.9}$
2000
Find the general solution of the given differential equation, and use it to determine how
solutions behave as $t\to\infty$.

$2y'+y=3t$
divide by 2
$y'+\frac{1}{2}y=\frac{3}{2}t$
find integrating factor,
$\displaystyle\exp\left(\int \frac{1}{2} dt\right)=e^{t/2}+c$
multiply thru
$e^{t/2}y'+e^{t/2}\frac{y}{2}
=\frac{3e^{t/2}}{2}t $ok something went
------------------------------------
book answer
$\color{red}\displaystyle y=ce^{-t/2}+3t-6 \\
\textit{y is asymptotic to } 3t-6 \textit{ as } t\to\infty $

 

[MarkFL]

When you determine your integrating factor, you need only determine up to but not including the constant of integration:

$$\mu(t)=\exp\left(\int\frac{1}{2}\,dt\right)=e^{\frac{t}{2}}$$

And so the ODE becomes:

$$e^{\frac{t}{2}}y'+\frac{1}{2}e^{\frac{t}{2}}y=\frac{3}{2}te^{\frac{t}{2}}$$

Or:

$$\frac{d}{dt}\left(e^{\frac{t}{2}}y\right)=\frac{3}{2}te^{\frac{t}{2}}$$

Integrating, we obtain:

$$e^{\frac{t}{2}}y=3e^{\frac{t}{2}}(t-2)+c_1$$

Now does the answer given by the book make sense? :)

Hence:

$$y(t)=3(t-2)+c_1e^{-\frac{t}{2}}$$

 

[karush]

for some reason I thought $\displaystyle e^{t/2}$ stayed the same whether taking the integral or derivative

$\displaystyle\frac{d}{dt} e^{t/2} = \frac{e^{t/2}}{2}$

multiply thru
$\displaystyle e^{t/2}y'+\frac{1}{2}e^{t/2}y
=\left(e^{t/2}y\right)^\prime
=\frac{3}{2}te^{t/2}$
integrate through
$\displaystyle e^{t/2}y
=\int\frac{3}{2}te^{t/2}\ dt=3 e^{t/2}(t-2)+c$
simplify
$\displaystyle y=ce^{t/2}+3t-6$
=========================================
$\textit{ok I didn't know how this the book says y is asymptotic to $3t-6$ as $t\to\infty$}$

Draw a direction field for the given differential equation.
how do you do this with Desmos

 

[MarkFL]

Here's a link to a Desmos slope field for this ODE:

https://www.desmos.com/calculator/my0rwuxl5x

 

[topsquark]

$\displaystyle e^{t/2}y =3 e^{t/2}(t-2)+c$
simplify
$\displaystyle y=ce^{t/2}+3t-6$

Think about these two lines for a moment...

-Dan

 

[karush]

Think about these two lines for a moment...

-Dan

$\displaystyle y=3(t-2)+\frac{c}{e^{t/2}}=ce^{-t/2}+3t-6$

 

[topsquark]

$\displaystyle y=3(t-2)+\frac{c}{e^{t/2}}=ce^{-t/2}+3t-6$

You just aren't doing well with the typos today, huh? (Sun)

Whatever. You got it.

-Dan

 

[karush]

I just wanted to see if you would catch it...