desolve y'=(x+y)^2 - Printable Version +- HP Forums ( https://www.hpmuseum.org/forum)+-- Forum: HP Calculators (and very old HP Computers) ( /forum-3.html)+--- Forum: HP Prime ( /forum-5.html)+--- Thread: desolve y'=(x+y)^2 ( /thread-3759.html)Pages: 1 2 |

RE: desolve y'=(x+y)^2 - salvomic - 05-02-2015 04:59 PM
(05-02-2015 04:45 PM)Tugdual Wrote: I have been able to achieve some results with Maxima and contrib_ode. well. ClassPad 400 can solve also Riccati equation? \[ y' + \frac{2x+1}{x}y - \frac{1}{x}y^{2} = x+2 \] thank you. RE: desolve y'=(x+y)^2 - Tugdual - 05-02-2015 05:21 PM
(05-02-2015 04:59 PM)salvomic Wrote:Not 100% the same form as the solution you gave but quite close and definitely correct.(05-02-2015 04:45 PM)Tugdual Wrote: I have been able to achieve some results with Maxima and contrib_ode. Off topic note: too bad this little evil doesn't have MES or unit of measuremernt like the 50g. RE: desolve y'=(x+y)^2 - salvomic - 05-02-2015 05:23 PM
(05-02-2015 05:21 PM)Tugdual Wrote: Not 100% the same form as the solution you gave but quite close and definitely correct. quite well RE: desolve y'=(x+y)^2 - parisse - 05-02-2015 06:15 PM
(05-02-2015 12:49 PM)salvomic Wrote: EDIT:They are not the same Code:
RE: desolve y'=(x+y)^2 - salvomic - 05-02-2015 07:32 PM
(05-02-2015 06:15 PM)parisse Wrote: They are not the same right! In about I get "xcas 1.1.4-19 (c) 2000-14, Bernard Parisse...", I presume it's the latest, however... If so, it's ok. RE: desolve y'=(x+y)^2 - salvomic - 05-02-2015 09:34 PM
(05-02-2015 05:30 AM)parisse Wrote: Xcas can solve this equation. It is a Ricatti equation, you can solve it by giving a particular solution, otherwise the system rewrites it as a 2nd order equation. please, can you explain a practical example to try? thank you RE: desolve y'=(x+y)^2 - parisse - 05-03-2015 06:08 AM
Without solution desolve(y'=(x+y)^2) With particular solution (here a complex one) desolve(y'=(x+y)^2,x,y=-x+i) RE: desolve y'=(x+y)^2 - salvomic - 05-03-2015 07:48 AM
(05-03-2015 06:08 AM)parisse Wrote: Without solution desolve(y'=(x+y)^2) ok, but I must set something? I mean, also so, in XCas I get [] (see image) and no solution, always... RE: desolve y'=(x+y)^2 - parisse - 05-04-2015 07:15 AM
Check that version() returns 1.2.0. If not, you must install the unstable version. RE: desolve y'=(x+y)^2 - salvomic - 05-04-2015 08:34 AM
(05-04-2015 07:15 AM)parisse Wrote: Check that version() returns 1.2.0. If not, you must install the unstable version. as I noted above, I had xcas 1.1.4-19 (c) 2000-14 now I've installed 1.2 unstable and with the general equation I get \[ \frac{\mathrm{c\_1} \sin\left(x\right)-\mathrm{c\_2} \cos\left(x\right)-\mathrm{c\_1}\cdot x \cos\left(x\right)-\mathrm{c\_2}\cdot x \sin\left(x\right)}{\mathrm{c\_1} \cos\left(x\right)+\mathrm{c\_2} \sin\left(x\right)} \] then, with trigtan() -> \[ \frac{-\mathrm{c\_1}\cdot x+\mathrm{c\_1} \tan\left(x\right)-\mathrm{c\_2}\cdot x \tan\left(x\right)-\mathrm{c\_2}}{\mathrm{c\_1}+\mathrm{c\_2} \tan\left(x\right)} \] (trying your advise, \( \mathrm{desolve}\left(y'=(\left(x+y\right)^{2}),x,y=(-x+i)\right) \) the result is \( [-x+i,\frac{4}{4\cdot \mathrm{c\_1} e^{-2*i\cdot x}+2*i}-x+i] \)) *** I tried also the Windows version, but the stable is still 1.1.4.19 also... RE: desolve y'=(x+y)^2 - salvomic - 05-11-2015 09:30 PM
also with resolve(y'=-y^2,t,y) in the Prime I get [], while in the last XCas the correct answer is 1/(t+c_0) I hope they release soon a new firmware almost with the new XCas improvements :-) Salvo |