Solving three differential equations

**guest** · August 26th 2009, 11:02 PM

Hi.

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

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

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

$ \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} $

$a,\;b$ are constants.

**Danny** · August 28th 2009, 10:30 AM

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 $s$ .
I know what the $x$ , $y$ , and $z$ are. I need $v_x$ , $v_y$ and $v_z$ .

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

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

$ \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} $

$a,\;b$ are constants.

What about $\theta$ - is it constant? Also, what is the relation between $v_x$ and $x$ , $v_y$ and $y$ and $v_z$ and $z$ ?

**chisigma** · August 28th 2009, 11:31 AM

Multiplying both terms by $ds$ the first DE becomes...

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

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

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

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

Kind regards

$\chi$ $\sigma$

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