what is solution of y''-y' y-2x=0 and how to test it by online software?

[drhu...@gmail.com]

what is solution of y''-y' y-2x=0 and how to test it by online software?
it is seems wolfram solution is wrong?

mathHand.com

[Nasser M. Abbasi]

I do not try Wolfram alpha, but Mathematica solves this. I assume
answer it gives is correct.

Here is how to solve this by hand.

Integrate both sides w.r.t. x

int ( y''-y' y ,x) = int( 2 x,x)

-y^2/2 + y' = x^2 + C

This is first order ode

y' = C + x^2 + y^2/2

Compare to https://en.wikipedia.org/wiki/Riccati_equation shows it is RICCATI

y' = q0(x) + q2(x) y^2

with missing q1(x). Here q0=c+x^2 abd q2(x) = 1/2.

This has standard method of solving as shown in the above wiki page.

--Nasser

[drhu...@gmail.com]

if you put the solution of the Riccati equation back to the y''-y' y-2x=0, you got nonzero, so it is wrong.

mathHand.com gives
http://server.drhuang.com/input/?guess=dsolve%28++ds%28y%2Cx%2C2%29-++ds%28y%29*y-2x%29&inp=++ds%28y%2Cx%2C2%29-++ds%28y%29*y-2x

[Nasser M. Abbasi]

Well, it works for me. I solved it and put the solution here

https://www.12000.org/my_notes/solving_ODE/current_version/insu3827.htm

I used Maple to solve the generated riccati ODE for now.

Maple verified the solution.

=====================
ode:=diff(y(x),x$2)-diff(y(x),x)*y(x)-2*x=0;

mysol:=y(x) = (_C3*(sqrt(2)*I - 6)*WhittakerM(-sqrt(2)*I/8 + 1, 1/4, I/2*sqrt(2)*x^2)
+ 8*WhittakerW(-sqrt(2)*I/8 + 1, 1/4, I/2*sqrt(2)*x^2)*_C4
- 2*(_C4*WhittakerW(-I/8*sqrt(2), 1/4, I/2*sqrt(2)*x^2)
+ _C3*WhittakerM(-I/8*sqrt(2), 1/4, I/2*sqrt(2)*x^2))*
(-1 + (x^2*I + I/2)*sqrt(2)))/(2*x*(_C4*WhittakerW(-I/8*sqrt(2), 1/4, I/2*sqrt(2)*x^2)
+ _C3*WhittakerM(-I/8*sqrt(2), 1/4, I/2*sqrt(2)*x^2)));

odetest(mysol,ode)
0
===========================

Zero means correct solution.

Since you did not show what you, and just said it is wrong, it is
hard for someone to say what the issue is with what you did.

I am sure Maple solved the Riccati equation correctly there.
Mathematica also solves it, but use different special functions. Instead
of WhittakerM, it uses ParabolicCylinderD functions.

--Nasser

[Nasser M. Abbasi]

On 1/31/2021 6:29 AM, drhu...@gmail.com wrote:

I think I know what you might have done.

The solution of the RICCATI is _not_ the solution to the original ODE.
So that is why you will not get zero if you just plugin this solution
as is in the original ODE.

The solution of the RICCATI is u(x).

The solution in y(x) is obtained from this u(x) using the
transformation y= -u'/(2 u).

Constants of integrations are renamed to keep only 2 constants in
the final solution.

--Nasser

[Nasser M. Abbasi]

On 1/31/2021 11:25 AM, Nasser M. Abbasi wrote:
> The solution in y(x) is obtained from this u(x) using the
> transformation y= -u'/(2 u).

typo

The solution in y(x) is obtained from this u(x) using the

transformation y= -u'/(u/2).

--Nasser

[drhu...@gmail.com]

I put y''-y' y-2x=0 into WolframAlpha to get 2/(c1-x)
I am talking WolframAlpha, not mma.

mathHand.com