# Solving a linear 2. order non-homogeneous differential equation

1. Nov 27, 2012

### 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?

2. Nov 27, 2012

### HallsofIvy

Yes, that is correct.

(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 $e^{-x}$.

3. Nov 28, 2012

### 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!

4. Nov 28, 2012

### 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.

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