Solving a linear 2. order non-homogeneous differential equation

[Chem.Stud.]

This is a case where an object is coupled to a spring, laying on a table. The object is moving, friction less, horizontally on the table. We assume the object is moving in an outer forice field which acts in the same direction as the object's motion. The motion is modeled by

y''(t) + y(t) = e$^{-t}$ (1)

where the right hand side is supposed to reflect the force field. (physics is not my field, so I'm not able to explain this in detail, I'm just a chemistry student with some extra maths).

This is a fairly simple problem, and I understand that the general solution can be written as a sum of the complimentary solution (of the homogeneous case) and one particular solution.

I've done the calculations, and found

y(t) = y$_{C}$(t) + y$_{P}$(t)
...= Acos(t) + Bsin(t) + 0,5e$^{-t}$

However, I'm not sure if this is what I'm supposed to do.

I'm asked to

a) Find a particular solution y$_{P}$ on the form

y$_{P}$(t) = e$^{-t}$(Acos(t) + A sin(t)).

and

b) Find the general solution of the differential equation (1).

and finally

c) Assume the object from the start (y(0) = 0) is not moving (laying still). Find the objects position as a function of time in this case. What happens when t$\rightarrow$∞?

What exactly am I supposed to do?

 

[HallsofIvy]

This is a case where an object is coupled to a spring, laying on a table. The object is moving, friction less, horizontally on the table. We assume the object is moving in an outer forice field which acts in the same direction as the object's motion. The motion is modeled by

y''(t) + y(t) = e$^{-t}$ (1)

where the right hand side is supposed to reflect the force field. (physics is not my field, so I'm not able to explain this in detail, I'm just a chemistry student with some extra maths).

This is a fairly simple problem, and I understand that the general solution can be written as a sum of the complimentary solution (of the homogeneous case) and one particular solution.

I've done the calculations, and found

y(t) = y$_{C}$(t) + y$_{P}$(t)
...= Acos(t) + Bsin(t) + 0,5e$^{-t}$

Yes, that is correct.

However, I'm not sure if this is what I'm supposed to do.

I'm asked to

a) Find a particular solution y$_{P}$ on the form

y$_{P}$(t) = e$^{-t}$(Acos(t) + A sin(t)).

(Are the coefficients of sin and cos both "A"?)

I agree with you. I suspect that either you or the text has confused two different problems.

If y were equal to that, then you would have
$$dy_p/dt= -e^{-t}(A cos(t)+ A sin(t))+ e^{-t}(-Asin(t)+ A cos(t)$$
$$dy_p/dt= -2Ae^{-t}sin(t)$$
If both coefficients are "A"!

$$d^2y_p/dt^2= 2Ae^{-t}sin(t)- 2Ae^{-t}cos(t)= e^{-t}(2Asin(t)- 2Acos(t))$$
Now, putting that into the equation,
$$d^2y_p/dt^2+ y= e^{-t}(3Asin(t)- Ae^{-t}cos(t)$$
and NO choice of A will make that equal to $e^{-t}$.

If you meant $y= e^{-t}(A cos(t)+ B sin(t))$, the calculations are a little harder:
$$dy/dt= -e^{-t}(A cos(t)+ B sin(t))+ e^{-t}(-A sin(t)+ B cos(t))= e^{-t}((B-A)cos(t)- (A+B) sin(t))$$
and so
$$d^2y/dt^2= -e^{-t}((B-A)cos(t)- (A+ B) sin(t))+ e^{-t}((A- B)sin(t)-(A+ B)cos(t))$$

So the equation becomes
$$d^2y/dt^2+ dy/dt= e^{-t}(Asin(t)- B cos(t))$$
and still, NO choice of A and B will make that equal to $e^{-x}$.

and

b) Find the general solution of the differential equation (1).

and finally

c) Assume the object from the start (y(0) = 0) is not moving (laying still). Find the objects position as a function of time in this case. What happens when t$\rightarrow$∞?

What exactly am I supposed to do?

 

[Chem.Stud.]

I got together with some friends, and it seems that I've misread the problem (shoot me now!).

The right hand side of (1) is supposed to be e$^{-t}$*sin(t)

This makes sense again! I'm taking a basic course in linear algebra and linear differential equations with initial value problems, and so I suppose our lecturer has been "kind" and provided us with the "correctly guessed" y$_{P}$. All I had to do was to insert it into (1) and find A and B, which respectively is 2/5 and 1/5.

I also understand the rest of the problem. The general solution I already have, as the complimentary solution isn't affected by my misreading the problem's right hand side. I then impose the initial values, and find the unknown constants c1 and c2.

When t approaches ∞, the whole solution will approach
-5/2cos(∞)+1/5sin(∞).

I'm not quite sure how to interpret that. Does this converge to a certain value?

Thank you for taking your time, though!

 

[HallsofIvy]

All that is saying is that the "non-homogeneous part", $e^{-t}sin(t)$, gives a "transient"- a part of the solution that, because of that "$e^{-t}$", rapidly goes to 0, leaving the "steady state" solution, -(5/2)cos(t)+ (1/5)sin(t) for large t. No, that does not have any limit as t goes to infinity.

 

[blue_raver22]

I would first start by analyzing the problem and understanding the physical context in which it is set. In this case, we have an object coupled to a spring and moving horizontally on a frictionless surface under the influence of an outer force field. The differential equation (1) represents the motion of the object and we are asked to find a particular solution and the general solution, as well as the object's position in the case where it is not moving.

To solve this problem, we can use the method of undetermined coefficients to find the particular solution. This method assumes that the particular solution has the same form as the right-hand side of the equation, in this case e^{-t}(Acos(t) + A sin(t)). We can then substitute this into the equation and solve for the coefficients A and B.

To find the general solution, we need to consider the complimentary solution, which is the solution to the homogeneous equation y''(t) + y(t) = 0. This can be written as y_{C}(t) = Acos(t) + Bsin(t). Combining this with the particular solution, we get the general solution y(t) = y_{C}(t) + y_{P}(t).

In the case where the object is not moving, we know that y(0) = 0, which means that A = 0 in the particular solution. This reduces the equation to y(t) = y_{C}(t), and we can see that as t\rightarrow∞, the object's position will approach the value of Bsin(t). This means that the object will oscillate with an amplitude of B and a frequency of 1, as determined by the coefficient of t in the solution.

In summary, to solve this differential equation we need to find the particular solution using the method of undetermined coefficients, combine it with the complimentary solution to get the general solution, and then use the initial conditions to find the object's position in the case where it is not moving. This will give us a complete understanding of the object's motion in this system.