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

In summary, The general solution of the given differential equation is $\displaystyle y=ce^{-t/2}+3t-6$ and it behaves asymptotically as $3t-6$ as $t\to\infty$. The direction field for the differential equation can be visualized using Desmos.
  • #1
karush
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$\tiny{2.1.9}$
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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 :confused::confused::confused::confused::confused:
------------------------------------
book answer
$\color{red}\displaystyle y=ce^{-t/2}+3t-6 \\
\textit{y is asymptotic to } 3t-6 \textit{ as } t\to\infty $
 
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  • #2
When you determine your integrating factor, you need only determine up to but not including the constant of integration:

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

And so the ODE becomes:

\(\displaystyle e^{\frac{t}{2}}y'+\frac{1}{2}e^{\frac{t}{2}}y=\frac{3}{2}te^{\frac{t}{2}}\)

Or:

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

Integrating, we obtain:

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

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

Hence:

\(\displaystyle y(t)=3(t-2)+c_1e^{-\frac{t}{2}}\)
 
  • #3
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
 
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  • #5
karush said:
$\displaystyle e^{t/2}y =3 e^{t/2}(t-2)+c$
simplify
$\displaystyle y=ce^{t/2}+3t-6$
Think about these two lines for a moment...

-Dan
 
  • #6
topsquark said:
Think about these two lines for a moment...

-Dan

$\displaystyle y=3(t-2)+\frac{c}{e^{t/2}}=ce^{-t/2}+3t-6$
 
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  • #7
karush said:
$\displaystyle y=3(t-2)+\frac{c}{e^{t/2}}=ce^{-t/2}+3t-6$
You just aren't doing well with the typos today, huh? (Sun)

Whatever. You got it.

-Dan
 
  • #8
I just wanted to see if you would catch it...
😎
 

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

1. What is the general solution of the differential equation 2y'+y=3t?

The general solution of the differential equation 2y'+y=3t is y = C*e^(-2t) + (3t-2)/4, where C is an arbitrary constant.

2. How do you find the general solution of a differential equation?

To find the general solution of a differential equation, you need to solve the equation by separating the variables and integrating both sides. This will result in an equation with an arbitrary constant, which represents the family of solutions. To find the specific solution, you need to use initial or boundary conditions.

3. Can you explain the meaning of the arbitrary constant in the general solution?

The arbitrary constant in the general solution represents the family of solutions that satisfy the differential equation. It is necessary because the general solution does not include any specific values, and the constant allows for the inclusion of all possible solutions.

4. What is the role of the initial or boundary conditions in finding the specific solution of a differential equation?

The initial or boundary conditions are specific values given in the problem that help determine the value of the arbitrary constant in the general solution. These conditions act as constraints on the general solution and allow us to find the specific solution that satisfies both the differential equation and the given conditions.

5. Can the general solution of a differential equation have more than one arbitrary constant?

Yes, the general solution of a differential equation can have more than one arbitrary constant. The number of arbitrary constants depends on the order of the differential equation. For a first-order differential equation, there will be one arbitrary constant, while for a second-order differential equation, there will be two arbitrary constants, and so on.

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