# System of linear differential equation

jaredmt

## 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).

## 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

swraman
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]

jaredmt
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

swraman
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 ;)