# Help on solving differential equations...

• Oct 18th 2008, 11:53 AM
ben218
Help on solving differential equations...
I am doing a project on projectiles in sport and have set up the following differential equations when investigating the projectile of a golf ball with air resistance:

mx'' = -kx' and my'' = -mg -ky'

with initial conditions:
x(0)=0
x'(0)=u0=ucos(alpha)
y(0)=0
y'(0)=vo=usin(alpha)

Could someone help me solve these differential equations and show me the method in which it is done. I am really not sure how to do them.

Thanks
• Oct 18th 2008, 09:37 PM
Tony2710
Hi,

You need to use matrices to solve this system of differential equations.

First you need to divide both by "m" so you get x'' and y'' on the left.

Then you express these together as a matrix equation:

[x'' y'']^T=[A][x' y']

The matrix A should be formed so that when this is expanded you get the same result as your initial equations.

You then need to find the eigenvalues and eigenvectors of the matrix A. This will then lead you to your general solution which will include a lot of exponents! Then you can substitute in your initial conditions.

This should hopefully push you in the right direction!
• Oct 18th 2008, 10:01 PM
mr fantastic
Quote:

Originally Posted by ben218
I am doing a project on projectiles in sport and have set up the following differential equations when investigating the projectile of a golf ball with air resistance:

mx'' = -kx' and my'' = -mg -ky'

with initial conditions:
x(0)=0
x'(0)=u0=ucos(alpha)
y(0)=0
y'(0)=vo=usin(alpha)

Could someone help me solve these differential equations and show me the method in which it is done. I am really not sure how to do them.

Thanks

$m \frac{d^2 x}{dt^2} = - k \frac{dx}{dt}$.

Substitute $w = \frac{dx}{dt}$:

$m \frac{d w}{dt} = - k w \Rightarrow \frac{dw}{w}$ $= -\frac{k}{m} \, dt \Rightarrow \int \frac{dw}{w} = - \int \frac{k}{m} \, dt \Rightarrow \ln w = -\frac{k}{m} t + A \Rightarrow w = B e^{-kt/m}$.

When $t = 0$ you know that $w = \frac{dx}{dt} = u \cos \alpha$.

Therefore $B = u \cos \alpha$.

Therefore $w = u \cos \alpha e^{-kt/m}$.

Therefore $\frac{dx}{dt} = u \cos \alpha e^{-kt/m} \Rightarrow x = -\frac{u m}{k} \cos \alpha e^{-kt/m} + C$.

When $t = 0$ you know that $x = 0$.

Therefore $C = \frac{u m}{k} \cos \alpha$.

Therefore $x = \frac{u m}{k} \cos \alpha \left( 1 - e^{-kt/m}\right)$.

A similar approach is used to solve $m \frac{d^2 y}{dt^2} = -mg - k \frac{dy}{dt}$.

Then you probably want to treat x = x(t) and y = y(t) as parametric equations and eliminate t to get a Cartesian equation y = y(x) for the path.
• Oct 19th 2008, 11:41 AM
ben218
thanks.

could you go through the other equation also please. Not sure if i need to set say z to be dy/dt then use the same method. I tried doing this and i'm not sure if after rearranging it i have come up with the correct equations to integrate....
• Oct 19th 2008, 03:14 PM
mr fantastic
Quote:

Originally Posted by ben218
thanks.

could you go through the other equation also please. Not sure if i need to set say z to be dy/dt then use the same method. I tried doing this and i'm not sure if after rearranging it i have come up with the correct equations to integrate....

Your approach is correct. If you post your working I'll review it.