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[SOLVED] 2.2.11 de of y'+y=5sin{2t}

karush

Well-known member
Jan 31, 2012
2,623
Wahiawa, Hawaii
Find the general solution of the given differential equation, and use it to determine how solutions behave as $ t\to\infty$ $y'+y=5\sin{2t}$
ok I did this first
$u(t)y'+u(t)y=u(i)5\sin{2t}$
then
$\frac{1}{5}u(t)y'+\frac{1}{5}u(t)y=u(i)\sin{2t}$
so far .... couldn't find an esample to follow
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book answer
$y=ce^{−t}+\sin 2t−2\cos 2t$
$\textit{y is asymptotic to sin2t−2cos2t as $t\to\infty$}$
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
617
North Texas
$y' + a(t) \cdot y = b(t)$

$y' + y = 5\sin(2t)$

integrating factor is $e^{\int a(t) dt} = e^{\int dt} = e^t$

$e^t y' + e^t y = 5e^t \sin(2t)$

$\displaystyle (e^t y)' = 5e^t \sin(2t) \, dt$

$\displaystyle e^t y = 5 \int e^t \sin(2t) \, dt$

integrate the right side by parts (2 iterations which I'm not typing out)

$e^t y = e^t[\sin(2t)-2\cos(2t)] + C$

$y = \sin(2t)-2\cos(2t) + Ce^{-t}$

$t \to \infty \implies Ce^{-t} \to 0$
 

karush

Well-known member
Jan 31, 2012
2,623
Wahiawa, Hawaii
Appreciate much😎

It's been a year since I started these
So plan on DE next 3 months
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
I have never been a fan of using the "integrating factor" for something as simple as linear differential equations with constant coefficients. The "associated homogeneous equation" is y'+ y= 0 which has characteristic equation r+ 1= 0 so characteristic root r= -1. The general solution to the associated homogeneous equation is $y(t)= Ce^{-t}$.

Now, recognizing "sine" and "cosine" together as one of the types of solutions we expect for such equations, we seek a solution to the entire equation of the form $y= Acos(2t)+ Bsin(2t)$. We need to determine A and B. Then $y'= -2Asin(2t)+ 2Bcos(25)$ so the equation becomes $y'+ y= -2Asin(2t)+ 2Bcos(2t)+ Acos(2t)+ Bsin(2t)= (A+ 2B) cos(2t)+ (B- 2A)sin(2t)= 5 sin(2t)$. Since this is to be true for all t we must have A+ 2B= 0 and B- 2A= 5. 2 times the first equation gives 2A+ 4B= 0 and adding the second equation eliminates A: 5B= 5 so B= 1. Then A+ 2B= A+ 2= 0 so A= -2.

$-2cos(2t)+ sin(2t)$ satisfies the entire equation so the general solution to the entire equation is $y(t)= Ce^{-t}- 2cos(2t)+ sin(2t)$.

As t goes to infinity, $Ce^{-t}$ goes to 0 for all C so this solution "behaves like" $sin(2t)- cos(2t)$.