Search 74,048 tutors

# Differential Equations problems

For a problem where you have a vertical damped spring with an applied force, you have a variety of forces. The basic force equation from Newton's second law is, of course, F=ma or, if we want to put everything in terms of position u, Fnet=mu'', where u'' is the second derivative of position with respect to time (some books use x or y for position instead). This net force is equivalent to a sum of many forces. In this case there is the force of gravity, since the spring is vertical, which is just Fg=mg. There is also the spring force which is given by Hooke's law, Fspring=-kx. In our case, x is the length of the spring, which is the distance the weight brings it down to the starting point, L, plus the position where it may be due to a combination of the other forces, u, so that Fspring=-k(L+u). Also, we have a damping force which counteracts any motion of the spring, and is therefore proportional to velocity, so that its force is Fd=-γu'. The last force we have is an arbitrary applied force completely independent of position, velocity or acceleration, Fa. Adding all of these forces up, we have:
Fnet=Fg+Fspring+Fd+Fa
This equation, substituting in terms of u with all of the equations above, turns into:
mu''=mg-k(L+u)-γu'+Fa
Rearranging things, we now have:
mu''+γu'+ku=mg+Fa-kL
Setting the terms on the right side equal to Fb (as one of the terms is arbitrary anyway) as none of them depend on u, u' or u'', we now get an equation:
u''+(γ/m)u'+(k/m)u=Fb/m which is a standard second-order differential equation.

In order to solve something of this kind, we need to go back and solve for γ using Fd=-γu', solve for k using Fspring=-k(L+u), and plug in for m and g as needed. Then we will have the general equation for a damped spring. For example:

A problem for this concept gives us various information. Often, a good thing to do while reading through a problem is to extract all the useful mathematical information from it. In this case, the problem starts by saying "A mass of 5kg stretches a spring 10 cm." We'll just note down m=5kg and L=10cm. The rest of the problem gives:

Fa=10 sin(t/2) N(newtons)
Fd(u'=4cm/sec)= 2 N
Initially: u'=3cm/sec, u=0

Formulate the initial value problem describing the motion of the mass. (i.e. give the general equation for the motion in these circumstances)

Here, we can solve for k using Fspring=-k(L+u), using the initual u=0, Fspring=-k(L) or k=-Fspring/L. For just the elongation before the spring is set in motion, Fspring will be compensated by Fg, so that k=mg/L. Plugging in, this becomes k=(5kg*9.8m/s^2)/0.1m=490kg/s^2 (or N/m, the standard unit for k). This is a somewhat large number for a spring constant, but that makes sense with the given numbers.

We can find γ using Fd=-γu', so that γ=Fd/u', or γ=(2N/0.04m/sec)=50Nm/s

Therefore, the answer becomes:
u''+((50Nm/s)/5kg)u'+(490kg/s^2)/5kg)u=(10sin(t/2)N)/5kg or
u''+10 kgNm/s u'+98s^(-2) u=2sin(t/2)N

Another thing that came up was finding complex roots of numbers. You can graph complex numbers on the complex number plane, which gives the real numbers on the x axis and the imaginary numbers on the y axis. If you draw a unit circle with respect to these axes, so that it intersects at 1, i, -1 and -i, then you can get roots of 1 by taking equally spaced angles from the x axis. The first root of 1 is 1, just on the x axis, and then the second roots of 1 are 1 and -1, which are an angle pi away from each other, and then the third roots of 1 are: 1, a number 2pi/3 away from that, and then 4pi/3 away. Then the fourth roots of 1 are 1, i, -1 and -i, and it continues on from there. Taking the nth root of any number always gives n roots (this is the fundamental theorem of arithmetic). I sort of skipped over the third roots by saying that there is a number "2pi/3 away". This number is complex, and has both real and imaginary roots. I can easily express this number in polar coordinates, as I know that it has a magnitude of 1 and is 2pi/3 away. But if I want it in rectangular coordinates, I must convert. The polar expression for this number would be 1*e^(2pi*i/3), as it has a magnitude of 1 and e^ix gives a rotation around the axis. In fact e^(ix) sometimes has a different notation, cis(x). This is because e^(ix)=cos(x)+i*sin(x), since cosine gives the x coordinates for the unit circle and sine gives the y coordinates for the unit circle. That means that in order to get ordered pairs for the unit circle that we've already drawn between the real and imaginary number lines, we can just expand and find the sine and cosine. So, in this case, e^(2pi*i/3)=cos(2pi/3)+i*sin(2pi/3)=-1/2+i*sqrt(3)/2, which gives the regular old rectangular coordinates. Another trick with the multiple roots of 1 is that the first root is 1, and the rest are equally spaced, so there will be symmetry about the x axis. Therefore, once you have found the roots above, you can just take the corresponding roots below, just by letting the imaginary part be negative instead (or, as they like to say, "taking the complex conjugate").

The last thing that we covered was higher order differential equations. These involve simplifying the equation to its characteristic equation, so that you are just left with a polynomial to factor. In these cases, since they are higher order, the polynomial will be more than quadratic, so you must find a factor and divide it out. The example was:
y'''-y''-y'+y=0
The characteristic equation is, of course:
x^3-x^2-x+1=0
Since there are so few coefficients other than 1 and -1, it was a fair guess that the roots are low, probably 1 or -1 (or perhaps i or -i) since they easily produce each other when multiplied. So, arbitrarily, we tried 1 as a root, so the polynomial had to be divided by x-1. The most important thing is to match the order and then the coefficient of the highest order term of the thing you are dividing, then you can just hope it will work out well enough in the lower orders.

x^2 -1
_____________
x-1)x^3-x^2-x+1
x^3-x^2
_______
0-x+1

In this case, the division works out quite well. Thus the factorization is
x^3-x^2-x+1=(x-1)(x-1)(x+1).

Since there is a repeated root, one of the solutions for the differential equation is of one higher order, (perhaps analogous to how you have a higher order denominator for partial fraction decomposition with repeated roots?) so that the general solution includes the linear combination of all three roots:
y=ae^-t+be^t+cte^t where a,b,c are constants that may be found using initial conditions.

An aside about long division of polynomials:

Long division can be used also for imaginary polynomials. If instead of dividing using 1 as a possible root, we divided using the third root of 1 as a possible root, which we saw above was -1/2+i*sqrt(3)/2, we should be able to find an answer (though perhaps not a pretty one). It is helpful to set up the division including all orders and all imaginary and real terms, even if their coefficients are zero, as they may come up later in the division, and to carry subtraction through to all terms you bring down.

x^2+1/2x+i*(sqrt(3)/2)x-2
___________________________
x+1/2-isqrt(3)/2)x^3-x^2 + 0ix^2                                                - x+0ix+1+0i
-(x^3-(1/2)x^2-i*(sqrt(3)/2)x^2)
___________________________
1/2 x^2 + i*(sqrt(3)/2)x^2
-(1/2 x^2 + (1/4)x - i*(sqrt(3)/4)x)
______________________________
i*(sqrt(3)/2)x^2 - (1/4)x + i*(sqrt(3)/4)x
-(i*(sqrt(3)/2)x^2 + (3/4)x + i*(sqrt(3)/4)x) -x+1
_________________________________________
-2x+1
-(-2x+1-isqrt(3)/2)
____________________
isqrt(3)/2
So that:
x^3-x^2-x+1                                                                isqrt(3)/2
_________________ = x^2+(1/2)x+i*(sqrt(3)/2)x-2+____________________
x+1/2-isqrt(3)/2)                                                        x+1/2-i*(sqrt(3)/2)

The messiness of the answer supports the method's generality.  You could divide it again by x+1/2-i(sqrt(3)/2) and get something much more elegant.