# -b.2.1.11 Find the general solution y'+y=5sin{2t}

• MHB
• karush
In summary, y' in the given equation represents the derivative of y with respect to t, the general solution is needed to find all possible solutions to the differential equation, it is found by combining the complementary function and the particular integral, the term "sin{2t}" represents a sinusoidal function in the equation, and the general solution can be used to find a particular solution for any initial conditions by solving for the constant of integration.
karush
Gold Member
MHB
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
____________________________________________
$y=ce^{−t}+\sin 2t−2\cos 2t$
$\textit{y is asymptotic to sin2t−2cos2t as$t\to\infty$}$

Last edited:
$y' + a(t) \cdot y = b(t)$

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

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)

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

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

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

Appreciate much

Last edited:
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" $sin(2t)- cos(2t)$.

## 1. What is the meaning of the notation y'+y=5sin{2t}?

The notation y'+y=5sin{2t} represents a differential equation, where y' is the derivative of y with respect to t and y is a function of t. The equation states that the sum of the function y and its derivative is equal to 5 times the sine of 2t.

## 2. What is the general solution to this differential equation?

The general solution to this differential equation is y(t) = Ce^{-t} + 5/4 sin{2t} - 5/2 cos{2t}, where C is a constant. This solution can be obtained by using the method of integrating factors.

## 3. How can I verify if a particular function is a solution to this differential equation?

To verify if a particular function is a solution, you can substitute the function into the original differential equation and see if it satisfies the equation. In this case, you would substitute the function y(t) into y'+y=5sin{2t} and see if the equation holds true.

## 4. Can this differential equation be solved using separation of variables?

Yes, this differential equation can be solved using separation of variables. The steps would involve isolating the y and t terms on opposite sides of the equation, integrating both sides, and then solving for y.

## 5. What is the physical interpretation of this differential equation?

The physical interpretation of this differential equation would depend on the context in which it is being used. Generally, it could represent a system where the rate of change of a variable y is affected by its own value as well as an external factor represented by the sine function. For example, it could represent the motion of a damped harmonic oscillator under the influence of an external force.

Replies
5
Views
2K
Replies
4
Views
3K
Replies
3
Views
1K
Replies
9
Views
5K
Replies
2
Views
1K
Replies
7
Views
3K
Replies
2
Views
2K
Replies
2
Views
1K
Replies
11
Views
1K
Replies
7
Views
3K