System of linear differential equation

In summary, a projectile is fired at angle theta with the horizontal with initial velocity of v0. Air resistance is equal to k times the velocity, so the differential equations of motion are: mx'' + kx' = 0my'' + ky' + mg = 0.
  • #1
jaredmt
121
0

Homework Statement


A projectile of mass m is fired at angle theta with the horizontal with initial velocity of v0 (ft/s). neglect all forces except for gravity and air resistance and assume air resistance is equal to k times velocity (ft/s)
a) with x-horizontal and y-vertical, show that the differential equations of the resulting motion are:
mx'' + kx' = 0
my'' + ky' +mg = 0

b) find the solution of the system of differential equations of part (a).


Homework Equations





The Attempt at a Solution



there were no similar examples in the book so i don't know how to do part a... and for part b i tried:

(mD^2 + kD)x = 0
(md^2 + kD)y + mg = 0

im not sure what to do from here or even if that was the right step to begin with
 
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  • #2
The actual set of equations come from drawing a free body diagram of the projectile, and summing forces on the x and y-axis (one equation for each axis).

As far as solving the equations...i wrote a solution to a similar problem some time ago here on the forum. In it i used

Force of air resistance = α * v,
and for some reason i defined another constant B such that:
B = α / mass

And I assumed it started from a point (0,0) with velocity Vo. Aside from that it should get you on the right track..if not give you the answer :) (from https://www.physicsforums.com/showthread.php?t=241101 ). I also divided through by mass since it is a constant, then the differential equation is in terms of position/velocity/acceleration, something easy to interpret.

swraman said:
The force of the air resistance
Code:
F = α * v
α = a constant, with units Kg/s This depends on the properties of the fluid the projectile it traveling through.
v = velocity

So, net acceleration due to air resistance
Code:
a =  α * v / m
a = Β * v
Β is a constant equal to α / m

since acceleration isn't constant, in order to write the position equations for the X and Y axis you have to start from the beginning.
I am going to do X and Y at separate times, as they can get long.

Y AXIS POSITION EQUATION

(acceleration on y axis) = g + B*(velocity on y axis)
Code:
d2y/dt2 = g + Β*dy/dt
d2y/dt2 - Β*dy/dt = g

Homogeneous solution of the Diffeq.
Code:
y = C1 *e^(Βt) + C2*t*e^(Βt)
for constants C1 and C2.

Boundary conditions of typical projectile problem are
Code:
y(0) = 0
y'(0) = Vo*Sin(θ)
Code:
0 = C1 *e^(Β*0) + C2*0*e^(Β*0)
C1 = 0
The solution becomes
Code:
y =  C2*t*e^(Βt)
Code:
y' =  C2*(t*Β*e^(Β*t) + e^(Βt))
C2 = Vo*Sin(θ)

So now we have the homogeneous solution
Code:
Yh = Vo*Cos(θ)*t*e^(Β*t)

Now the particular solution to the Diffeq:
Code:
d2y/dt2 - Β*dy/dt = g
Guess Yp = A*t for some undetermined constant A
Code:
Yp = A*t
Yp' = A
Yp'' = 0

d2y/dt2 - Β*dy/dt = g
0 - Β*A = g
A = -g/Β
Yp = -g*t/B

So now we can write the Y position equation for the projectile with air resistance:
Code:
y = Yp + Yh
[B]y = -g*t/B + Vo*Sin(θ)*t*e^(B*t)[/B]

X AXIS POSITION EQUATION:

Again, we have to start with the acceleration to write the position equation:

(acceleration on x axis) = -B*(velocity on x axis)
Code:
d2x/dt2 = -Β*dx/dt
Since this one is a homogeneus problem, it is much easier to solve:
Code:
d2x/dt2 + Β*dx/dt= 0

x = C1 *e^(-Βt) + C2*t*e^(-Βt)
Again, the innitial conditions for projectile motion:
Code:
x(0) = 0
x'(0) = Vo*Cos(θ)
Plug these in and you get
Code:
C1 = 0
C2 = Vo*Cos(θ)

So, the final equation for the motion on the X axis is
Code:
x = Vo*Cos(θ)*t*e^(-B*t)
Code:
[B][SIZE="4"]y = -g*t/B + Vo*Sin(θ)*t*e^(B*t)
x = Vo*Cos(θ)*t*e^(-B*t)[/SIZE][/B]
 
  • #3
thanks for the help

this one was actually easier than it looked. all i had to do was find summation of forces for x and y direction and those were the 2 equations. then solving it was just a matter of doing both equations separately. i was trying to do it as a system of equations, but i don't think it can be done that way since there is no way to even check it as a system of equations
 
  • #4
Thats actually what I did. The two differential equations for the x and y-axis are both independent of each other, and i solved them spearatley ;)
 

1. What is a system of linear differential equations?

A system of linear differential equations is a set of equations that describe the relationship between multiple variables, where each variable is a function of one or more other variables and their derivatives. These equations are linear because they involve only first derivatives and do not contain any nonlinear terms.

2. How is a system of linear differential equations solved?

A system of linear differential equations can be solved using various methods, such as substitution, elimination, or using matrices. The most commonly used method is the Gaussian elimination method, which involves transforming the system into an equivalent triangular system and then solving for the variables using back substitution.

3. What are the applications of a system of linear differential equations?

A system of linear differential equations has many applications in science and engineering, such as in modeling physical systems, analyzing circuits, and predicting population growth. They are also used in economics, finance, and other fields to study complex relationships between variables.

4. What is the difference between a homogeneous and non-homogeneous system of linear differential equations?

A homogeneous system of linear differential equations is one in which all terms have the same degree and all constant coefficients are equal to zero. A non-homogeneous system has at least one non-zero constant coefficient. Homogeneous systems have a unique solution, while non-homogeneous systems may have infinitely many solutions.

5. Can a system of linear differential equations have a non-constant solution?

Yes, a system of linear differential equations can have non-constant solutions. This means that the solution is not a constant value but rather a function of one or more variables. In this case, the solution is called a general solution, as it represents all possible solutions to the system of equations.

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