dx/dt = x(x^3 - 2y^3) dy/dt = y(2x^3 - y^3) Please help!
Last edited by hazeleyes; Mar 5th 2011 at 07:49 AM.
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. Notice that both the numerator and denominator are of degree 4,ie homogeneous. Do you know how to solve a homogeneous DE? Hint: take
Originally Posted by Sambit . Notice that both the numerator and denominator are of degree 4,ie homogeneous. Do you know how to solve a homogeneous DE? Hint: take hi, thanks for your hint, after i sub y=tx i got dy/dx={t(2-t^3)}/{1-2t^3} but then what should be doing next? thanks alot
Last edited by hazeleyes; Mar 5th 2011 at 05:35 AM.
If y = xv then your ODE becomes which separate to or (this easily integrates!)
what is the 'v' means and what is the reason to sub 'y=vx'? it was dy/dx if sub y=vx dy= v dx ? thanks
Last edited by hazeleyes; Mar 5th 2011 at 06:29 AM.
If you have an ODE of the form (called homogeneous) then a substitution will give another ODE that separates. Note
oh I see!! Thank you so much for your clear explanation.
Last edited by hazeleyes; Mar 5th 2011 at 07:05 AM.
y=y(x) y=vx v=y/x 1/v=x/y dy/dx = x(dv/dx) x(dv/dx)={ v(2-v^3)}/ (1-2v^3) is this form correct? thx
Product rule .
Thank you so much for your remind! after integration, I got v/(v^3 +1) = x + e^c v/(v^3 +1) = x + A ,where A is constant
Last edited by hazeleyes; Mar 5th 2011 at 07:54 AM.