# Solving three differential equations

• Aug 26th 2009, 11:02 PM
guest
Solving three differential equations
Hi.

What would be the best way to solve next three differential so that the result it is not dependant from $\displaystyle s$.
I know what the $\displaystyle x$, $\displaystyle y$, and $\displaystyle z$ are. I need $\displaystyle v_x$, $\displaystyle v_y$ and $\displaystyle v_z$.

$\displaystyle \sqrt{v_x^2+v_y^2+v_z^2} \frac{dv_x}{ds}= -a b \cos(\theta) \frac{dx}{ds}$

$\displaystyle \sqrt{v_x^2+v_y^2+v_z^2} \frac{dv_y}{ds}= -a b \cos(\theta) \frac{dy}{ds}$

$\displaystyle \sqrt{v_x^2+v_y^2+v_z^2} \frac{dv_z}{ds}= -a \frac{dz}{ds} -a b \cos(\theta) \frac{dz}{ds}$

$\displaystyle a,\;b$ are constants.
• Aug 28th 2009, 10:30 AM
Jester
Quote:

Originally Posted by guest
Hi.

What would be the best way to solve next three differential so that the result it is not dependant from $\displaystyle s$.
I know what the $\displaystyle x$, $\displaystyle y$, and $\displaystyle z$ are. I need $\displaystyle v_x$, $\displaystyle v_y$ and $\displaystyle v_z$.

$\displaystyle \sqrt{v_x^2+v_y^2+v_z^2} \frac{dv_x}{ds}= -a b \cos(\theta) \frac{dx}{ds}$

$\displaystyle \sqrt{v_x^2+v_y^2+v_z^2} \frac{dv_y}{ds}= -a b \cos(\theta) \frac{dy}{ds}$

$\displaystyle \sqrt{v_x^2+v_y^2+v_z^2} \frac{dv_z}{ds}= -a \frac{dz}{ds} -a b \cos(\theta) \frac{dz}{ds}$

$\displaystyle a,\;b$ are constants.

What about $\displaystyle \theta$ - is it constant? Also, what is the relation between $\displaystyle v_x$ and $\displaystyle x$, $\displaystyle v_y$ and $\displaystyle y$ and $\displaystyle v_z$ and $\displaystyle z$?
• Aug 28th 2009, 11:31 AM
chisigma
Multiplying both terms by $\displaystyle ds$ the first DE becomes...

$\displaystyle \sqrt{v^{2}_{x} + v^{2}_{y} + v^{2}_{z}} \cdot dv_{x} = -a\cdot b \cdot \cos \theta \cdot dx$

... where the variables are separated. Now we obtain the solution integrating both terms with the aid of the formula...

$\displaystyle \int \sqrt{a^{2} + t^{2}}\cdot dt = \frac{1}{2} \{a^{2} \cdot \ln (t + \sqrt{a^{2} + t^{2}})+ t \cdot \sqrt{a^{2} + t^{2}}\} + c$

Ther other two differential equations are solvable in the same way...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$