Mathematica Solutions for First Order Differential Equation

[pat666]

Mathematica Disagrees with ME!

Homework Statement

(2xy-5)dx+(x^2+y^2)dy=0 y(3)=1

Homework Equations

The Attempt at a Solution

Ive solved this by hand and now where required to get Mathematica to solve to.
My solution: -17/3=x^2y-5x+y^3/3
I've attached what mathematica has done. Now the question
Am I wrong ?
Is my code wrong?
Is mathematica wrong?

Thanks

P.S I rearranged to this form to sick in mathematica:
dy/dx=(-2xy+5)/(x^2+y^2)

 

[Mark44]

Your thumbnail is too small for me to be able to read. In any case, your work is correct. You can verify it by doing two things:
1) Checking that the initial condition is satisfied; i.e., that when x = 3 and y = 1, then x²y - 5x + y³/3 = -17/3 is a true statement.
2) Differentiating the equation x²y - 5x + y³/3 = -17/3 implicitly to arrive at the differential equation dy/dx = (5 - 2xy)/(x² + y²).

Checking a solution to a differential equation is something you should do as a matter of course. You've already done all the hard work. It's only a little more work to verify that your work is correct.

 

[pat666]

good to know that I did it right by hand, do you know the correct mathematica code so that it will do it.

Thanks

 

[Mark44]

No. I don't have mathematica.

 

[lurflurf]

That mathematica output looks correct. The cubic in y has been factored, what kind of output were you expecting?

 

[pat666]

one the same as mine---- is that not possible here?

 

[I like Serena]

Homework Statement

(2xy-5)dx+(x^2+y^2)dy=0 y(3)=1

Homework Equations

The Attempt at a Solution

Ive solved this by hand and now where required to get Mathematica to solve to.
My solution: -17/3=x^2y-5x+y^3/3
I've attached what mathematica has done. Now the question
Am I wrong ?
Is my code wrong?
Is mathematica wrong?

Thanks

P.S I rearranged to this form to sick in mathematica:
dy/dx=(-2xy+5)/(x^2+y^2)

Mathematica did a little more work than you did.

What you have is a third order equation of y.
Mathematica solved it to be of the form y = f(x).
I haven't checked it, but it looks about right: 1 real solution and 2 imaginary ones.
Presumably you can rewrite the first solution to match your solution.

I'm not a Mathematica specialist, but I suspect it is not equipped to output the kind of solution you gave.

 

[SammyS]

I used WolframAlpha to solve your result for y.

http://www.wolframalpha.com/input/?i=Solve+for+y%3A+-17%2F3%3Dx^2y-5x%2By^3%2F3

Here's what it gives for input Mathematica style -- for solving your result for y.:

Solve[-17/3 == -5 x + x^2 y + y^3/3, y]

 

[SteamKing]

pat666: your solution from sec. 3 of the OP does not satisfy the initial condition y(3)=1

 

[SteamKing]

pat666: Delete previous post. The initial condition is satisfied.

 

[Bill Simpson]

At least with the version of Mathematica I'm using, I believe it returns exactly the same result with or without the initial condition y(3)=1, containing an arbitrary constant denoted by C[1] by Mathematica. That is odd. If I'm not mistaken then usually with a first order differential equation and a single initial condition it is sufficient to eliminate the arbitrary constants. And I've never seen another differential equation behave like this when given to DSolve, but I obviously haven't seen everything DSolve can do. As others have pointed out, you found an implicit solution for y(x) where Mathematica factored the cubic to find three explicit solutions. Two of those when I attempt to solve for C[1] using your initial condition have no solution. The third does does satisfy your differential equation and your initial condition, so it isn't clear why Mathematica didn't return only that solution with arbitrary constant solved for. If I experiment with changing your initial condition I can provoke error messages in some cases and get it to return a single solution with no arbitrary constant in some others. I don't know what it is about your example that makes it do this. And I have no answer for your final question, how to turn an explicit solution into the implicit form that you found by hand.