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

From drhuang57@gmail.com@21:1/5 to All on Wed Jan 27 15:05:26 2021
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

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• From Nasser M. Abbasi@21:1/5 to drhu...@gmail.com on Wed Jan 27 20:01:24 2021
On 1/27/2021 5:05 PM, drhu...@gmail.com wrote:
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

I do not try Wolfram alpha, but Mathematica solves this. I assume

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

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• From drhuang57@gmail.com@21:1/5 to Nasser M. Abbasi on Sun Jan 31 04:29:07 2021
On Thursday, 28 January 2021 at 13:01:26 UTC+11, Nasser M. Abbasi wrote:
On 1/27/2021 5:05 PM, drhu...@gmail.com wrote:
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

I do not try Wolfram alpha, but Mathematica solves this. I assume

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

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

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• From Nasser M. Abbasi@21:1/5 to drhu...@gmail.com on Sun Jan 31 09:47:46 2021
On 1/31/2021 6:29 AM, drhu...@gmail.com wrote:
On Thursday, 28 January 2021 at 13:01:26 UTC+11, Nasser M. Abbasi wrote:
On 1/27/2021 5:05 PM, drhu...@gmail.com wrote:
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

I do not try Wolfram alpha, but Mathematica solves this. I assume

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

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

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

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• From Nasser M. Abbasi@21:1/5 to drhu...@gmail.com on Sun Jan 31 11:25:47 2021
On 1/31/2021 6:29 AM, drhu...@gmail.com wrote:
On Thursday, 28 January 2021 at 13:01:26 UTC+11, Nasser M. Abbasi wrote:
On 1/27/2021 5:05 PM, drhu...@gmail.com wrote:
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

I do not try Wolfram alpha, but Mathematica solves this. I assume

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

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

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

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• From Nasser M. Abbasi@21:1/5 to Nasser M. Abbasi on Sun Jan 31 11:27:02 2021
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

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• From drhuang57@gmail.com@21:1/5 to Nasser M. Abbasi on Fri May 7 15:40:07 2021
On Monday, 1 February 2021 at 02:47:51 UTC+11, Nasser M. Abbasi wrote:
On 1/31/2021 6:29 AM, drhu...@gmail.com wrote:
On Thursday, 28 January 2021 at 13:01:26 UTC+11, Nasser M. Abbasi wrote:
On 1/27/2021 5:05 PM, drhu...@gmail.com wrote:
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

I do not try Wolfram alpha, but Mathematica solves this. I assume

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

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

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

mathHand.com

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