# Isocline methods

• Nov 4th 2009, 06:02 AM
ViperRobK
Isocline methods
I have the equation $\displaystyle \frac{dx}{dt} = x^2 - t^2 - 1$

I have started to draw the isoclines and found them to be Hyperbolas i have then drawn the direction plot and cant really see what to do from there i have worked out that the line x = -t fits but cannot see any other solutions.

Thank you for you help

ViperRobK
• Sep 9th 2013, 11:41 PM
mathlover10
Re: Isocline methods
do you graph the isoclines with a calculator? often they involve powers of e so it seems tricky to do it in your head
an interesting thing to do is to predict the behavior of the solution as t goes to infinity. the isoclines are slope lines where dy/dt=m=-a(t)y+b(t) for instance. another example of isoclines would be circles for: y'=-y/(x^2+y^2) where the circles become flattened out into the null cline m=y=0
• Sep 10th 2013, 02:13 AM
JJacquelin
Re: Isocline methods
The solutions of the ODE are :
x = -t +exp(-tē)/(c-sqrt(pi/4)*erf(t))
If t tends to infinity, exp(-tē) tends to 0. So, the asymptotic line is x = -t.
If abs(c) < sqrt(pi/4) there is a particular value of t=a so that (c-sqrt(pi/4)*erf(a))=0 and then x tends to infinity. In these cases there is a second asymptotic line t=a.
As a consequence, in some cases (depending on c), the isoclines look like, but are not exactly hyperbolas.
In any case, never they are hyperbolas.
• Sep 11th 2013, 12:47 AM
mathlover10
Re: Isocline methods
how do you do the denominator erf function? so this is a linear Riccati equation that can always be reduced to a 2nd order linear ODE? x'=qo(t)+qo(t)x+qo(t)x^2
• Sep 11th 2013, 04:25 AM
JJacquelin
Re: Isocline methods
The linear Riccati equation can be reduced to a 2nd order linear ODE :
x'=P(t)+Q(t)x+R(t)x^2
Let : x = -y' / (R*y) Where y(t) is the new unknown function.
• Sep 12th 2013, 01:29 AM
mathlover10
Re: Isocline methods
if q2 is nonzero then you can make a substitution v=yq2, S=q2q0, R=q1+(q2'/q2) which satisfies a Riccati, v'=v^2+R(x)v+S(x)=(yq2)'=q0q2+(q1+q2'/q2)v+v^2
with double substitution v=-u'/u, u now satisfies a linear 2nd ODE:
u''-R(x)u'+S(x)u=0, v'=-(u'/u)'=-(u''/u)+v^2, u''/u=v^2-v'=-S+Ru'/u, therefore we have reduced the Ricatti to an equation
u''+Su-Ru'=0, solving this equation will lead to a solution y=-u'/(q2u) You can find this proof in texts, but what was the original motivation behind the invention of the substitutions?
• Sep 22nd 2013, 03:09 PM
mathlover10
Re: Isocline methods
do you plot these isoclines by hand somehow? it can be done for circles, but I'm not sure how to do it for more complex functions