Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How would you go about solving the following system of ODEs:

\begin{align*} & x''(t) - \frac{2}{y}x'(t) \ y'(t) = 0 \ & y''(t) + \frac{1}{y} \big(x'(t) - y'(t)\big) = 0 \end{align*}

Any help would be very much appreciated!

share|improve this question
    
I have managed to reduce it to the single equation $y''(t) + \frac{1}{y}((y'(t))^2 + 1) = 0.$ –  Alex Kite Apr 7 '12 at 2:42
    
So the first equation gives $\ln x' = 2\ln y+C$, plugging that into the second equation gives $y''(t)+\frac{1}{y}(C'y'(t)^2-y')$. How did you get the $+1$ term? –  Alex R. Apr 7 '12 at 2:48
add comment

1 Answer

up vote 0 down vote accepted

From the first equation we can conclude :

$\ln x'(t)= 2\ln y +C$ , so $x'(t)=C_1\cdot y^2$

plugging this into second equation gives :

$y''(t)+\frac{1}{y}(C_1 \cdot y^2-y'(t))=0$

Now substitute $y'(t)=v$ , where $v$ is a function in terms of variable $y$ ,so:

$y''(t)=v'_{y}\cdot v$

Hence :

$v'_{y}\cdot v+\frac{1}{y}(C_1 \cdot y^2-v)=0$

this equation is equivalent to the first order non-linear ODE :

$v'_y+C_1\cdot y \cdot v^{-1} -\frac{1}{y}=0$

which can be solved using numerical methods .

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.