MHB -2.1.1 DE Find the general solution

[karush]

[a] Find the general solution of $y^\prime + 3y=t+e^{-2x}\quad \dfrac{dy}{dx}+f(x)y=g(x)$
\$\begin{array}{lll}
\textsf{Similarly} & \dfrac{dy}{dx}+Py=Q\\
\textsf{hence} & \mu(x)=\exp\left(\int f(x)\,dx\right)\\
\textsf{then} & \mu^\prime(x)=\exp\left(\int f(x)\,dx\right)f(x) \\
\textsf{then} & \mu(x)+y'=\mu(x)g(x)\\
\textsf{integrating factor} & \mu(x)=e^{3x}\\
\textsf{multiplying} & e^{3x}y'+3e^{3x}y=xe^{3x}+e^{x}\\
\textsf{rewriting the LHS} & \dfrac{d}{dx}\left(e^{3x}y\right)=xe^{3x}+e^{x}\\
\end{array}$
determine how the solution behave as $t \to \infty$
$ce^{-3x}+\dfrac{x}{3}-\dfrac{1}{9}+e^{-2x}$
y is asymptotic to $\dfrac{t}{3} −\dfrac{1}{9} \textit{ as } t \to \infty$

ok i think this is correct just could be worded better
maybe some typos
suggestions, complaints, or ?

 

[skeeter]

I’m assuming the independent variable is $t$ …

$y’ +3y = t + e^{-2t}$

$\mu(t) = e^{3t}$

$y’ \cdot e^{3t} + y \cdot 3e^{3t} = t \cdot e^{3t} + e^t$

$(y \cdot e^{3t})’ = t\cdot e^{3t} + e^t$

$y \cdot e^{3t} = \dfrac{t}{3} \cdot e^{3t} - \dfrac{1}{9} \cdot e^{3t} + e^t + C$

$y = \dfrac{3t-1}{9} + e^{-2t} + Ce^{-3t}$

$y$ is asymptotic to the line you cited.

 

[karush]

here is where it came from #1

asymptotic ??

 

[skeeter]

here is where it came from #1
View attachment 11299
asymptotic ??

$y=\dfrac{3t-1}{9}$ is a slant asymptote.

as $t \to \infty$, the value of $y$ is very close to the value of the line $y=\dfrac{3t-1}{9}$ and it gets very close rather quickly because the two exponential terms approach zero very quickly.

 

[HOI]

$y'+ 3y= t+ e^{-2t}$

(The original post has $y'+ 3y= t+ e^{-2x}$ but that just doen't make sense! The independent variable has to be either x or t. I have chosen t.)

Another method:
The "associated homogenous equation" is y'+ 3y= 0.
$\frac{dy}{dt}= -3y$
$\frac{dy}{y}= -3dt$
$ln(y)=- 3t+ C$
$y= e^{-3t+ C}= e^Ce^{-3t}= C'e^{-3t}$
where $C'= e^C$

Since that does not involve either t or $e^{-2t}$ we try a solution to the entire equation of the form $y= At+ B+ Ce^{-2t}$.
(The "method of undetermined coefficients")

$y'+ 3y= A- 2Ce^{-2t}+ 3At+ 3B+3Ce^{-2t}=3At+ (A+ 3B)+ Ce^{-2t}= t+ e^{-2t}$

So 3A= 1, A+ 3B= 0, and C= 1. A= 1/3, and B= -1/9.
$y(t)= C'e^{-3t}+ t/3- 1/9+ e^{-2t}$

 

[HOI]

$y'+ 3y= t+ e^{-2t}$

Yet another method (variation of parameters):
Calculating as before that the general solution to the associated homogenous equation is $C'e^{-3t}$ we seek a solution to the entire equation of the form $y= u(x)e^{-3t}$ where u(x) is a function to be determined.
(We are allowing the "parameter", C', to "vary".)

$y'+ y= u'(x)e^{-3t}- 3u(x)e^{-3t}+ 3u(x)e^{-3t}= u'(x)e^{-3t}= t+ e^{-2t}$
$u'(t)= te^{3t}+ e^t$

And now just integrate. The integral of $e^t$ is of course $e^t$. To integrate $te^{3t}$ use "integration by parts", taking u= t and $dv= e^{3t}dt$. Then du= dt and $v= \frac{1}{3}e^{3t}$ so
$\int te^{3t}dt= \frac{1}{3}te^{3t}- \frac{1}{3}\int e^{3t}dt= \frac{1}{3}te^{3t}- \frac{1}{9}e^{3t}$.
$u(t)= \frac{1}{3}te^{3t}- \frac{1}{9}e^{3t}+ e^t$.

Then $u(t)e^{-3t}= \frac{1}{3}t- \frac{1}{9}+ e^{-2t}$
and
$y(t)= C'e^{-3t}+ \frac{1}{3}t- \frac{1}{9}+ e^{-2t}$.