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New user needs help with a differential equation

Posted 11 years ago
Hello ppl,
I'm newly using Mathematica for my master thesis at my university and I'm simply trying to guess how to solve this simple differential equation and Plot it! Please if you can help me I will appreciate it!  It is very difficult to study all my myself .   I'm not a genius after all :-/
(3*x^2*y+8*x*y^2)dx+(x^3+8*x^2*y+12*y^2)dy=0
POSTED BY: Dani Slavova
8 Replies
Actually Seokin, I believe that the equation that Dani is writing out is the following (in Mathematica)
(3*x^2*y[x] + 8*x*y[x]^2) + (x^3 + 8*x^2*y[x] + 12*y[x]^2) D[y[x], x] == 0
 
And it appears that DSolve can handle this 
 In[1]:= DSolve[(3*x^2*y[x] +
      8*x*y[x]^2) + (x^3 + 8*x^2*y[x] + 12*y[x]^2) D[y[x], x] == 0,
  y[x], x]
 
 
 Out[1]= {{y[x] -> -(x^2/3) - (12 x^3 - 16 x^4)/(
     24 (9 x^5 - 8 x^6 + 27 C[1] +
        3 Sqrt[3] Sqrt[
         x^9 - x^10 + 18 x^5 C[1] - 16 x^6 C[1] + 27 C[1]^2])^(1/3)) +
    1/6 (9 x^5 - 8 x^6 + 27 C[1] +
       3 Sqrt[3] Sqrt[
        x^9 - x^10 + 18 x^5 C[1] - 16 x^6 C[1] + 27 C[1]^2])^(
     1/3)}, {y[x] -> -(x^2/3) + ((1 + I Sqrt[3]) (12 x^3 - 16 x^4))/(
    48 (9 x^5 - 8 x^6 + 27 C[1] +
       3 Sqrt[3] Sqrt[
        x^9 - x^10 + 18 x^5 C[1] - 16 x^6 C[1] + 27 C[1]^2])^(1/3)) -
    1/12 (1 - I Sqrt[3]) (9 x^5 - 8 x^6 + 27 C[1] +
       3 Sqrt[3] Sqrt[
        x^9 - x^10 + 18 x^5 C[1] - 16 x^6 C[1] + 27 C[1]^2])^(
     1/3)}, {y[x] -> -(x^2/3) + ((1 - I Sqrt[3]) (12 x^3 - 16 x^4))/(
    48 (9 x^5 - 8 x^6 + 27 C[1] +
       3 Sqrt[3] Sqrt[
        x^9 - x^10 + 18 x^5 C[1] - 16 x^6 C[1] + 27 C[1]^2])^(1/3)) -
    1/12 (1 + I Sqrt[3]) (9 x^5 - 8 x^6 + 27 C[1] +
       3 Sqrt[3] Sqrt[
        x^9 - x^10 + 18 x^5 C[1] - 16 x^6 C[1] + 27 C[1]^2])^(1/3)}}

Note that the C[1] are the unknown arbitrary constants that would be determined by the initial condition (which were not supplied). Alson note that there are 3 solutions but that two of them are non-real generally. 

If, say, the initial condition was that y[0]==1 then the process process would read 
 In[13]:= DSolve[{(3*x^2*y[x] + 8*x*y[x]^2) + (x^3 + 8*x^2*y[x] + 12*y[x]^2) D[y[x], x] == 0, y[0] == 1}, y[x], x]
 
 
 Out[13]= {{y[x] -> (-3 x^3 + 4 x^4 -
       2 x^2 (108 + 9 x^5 - 8 x^6 +
          3 Sqrt[3] Sqrt[432 + 72 x^5 - 64 x^6 + x^9 - x^10])^(
        1/3) + (108 + 9 x^5 - 8 x^6 +
         3 Sqrt[3] Sqrt[432 + 72 x^5 - 64 x^6 + x^9 - x^10])^(
       2/3))/(6 (108 + 9 x^5 - 8 x^6 +
        3 Sqrt[3] Sqrt[432 + 72 x^5 - 64 x^6 + x^9 - x^10])^(1/3))}}

Which yields a single (generally real) solution.

If you wanted to plot this then you might do something like 
Plot[(-3 x^3 + 4 x^4 -
    2 x^2 (108 + 9 x^5 - 8 x^6 +
       3 Sqrt[3] Sqrt[432 + 72 x^5 - 64 x^6 + x^9 - x^10])^(
     1/3) + (108 + 9 x^5 - 8 x^6 +
      3 Sqrt[3] Sqrt[432 + 72 x^5 - 64 x^6 + x^9 - x^10])^(
    2/3))/(6 (108 + 9 x^5 - 8 x^6 +
      3 Sqrt[3] Sqrt[432 + 72 x^5 - 64 x^6 + x^9 - x^10])^(1/3)), {x,
  1, 10}, PlotRange -> All]
POSTED BY: David Reiss
Hello Dani

I am also a beginner at Mathematica, but I believe I can give you some aid. As a high school student in Calculus class, I have utilized Mathematica for such purposes.

I would use D[] and ContourPlot to work your problem.

This is how I worked your problem:

In[17]:= D[3*x^2*y + 8*x*y^2, x] + D[x^3 + 8*x^2*y + 12*y^2, y] == 0


Out[17]= 8 x^2 + 24 y + 6 x y + 8 y^2 == 0
This should solve the derivatives.

ContourPlot[8 x^2 + 24 y + 6 x y + 8 y^2 == 0, {x, -2, 3}, {y, -4, 0}]
This should give the plot of the equation which is an ellipse.

Of course, I am not a college mathematician, and my knowledge of mathematics is limited. I hope that this helped your problem

Also, as a tip, whenever you want to know how a function works in Mathematica, you can type in ? and the function.

For example,
?ContourPlot
Should tell you how and what the ContourPlot does.

If my explanation is unclear, do feel free to criticize me.
POSTED BY: Seokin Yeh
Ah, I guess I had been mistaken.

I hope my answer had not been misleading. I should've guessed the complexity when he said it was for his mater's degree.

Thank you.
POSTED BY: Seokin Yeh
No problem Seokin!
That is the advantage of a forum with members that advise and teach each other.
POSTED BY: David Reiss
Posted 11 years ago
The equation given can be easily solved with pencil and paper in far less time than required to put it into Mathematica.
(Hint: the left hand side is an Exact Differential.)
Best regards,
David
POSTED BY: David Keith
Posted 11 years ago
Thank you so much. (BTW I'm a lady) emoticon I know that is much easier to solve it manualy but I need to use Mathematica for my master thesis .... and it looks more representative ;)
POSTED BY: Dani Slavova
Posted 11 years ago
Hi Dani,
Then there are perhaps 2 ways to do this.
1) Solve it manually for F=C , where this is the integral of the exact differential. (The fact that the exact differential = 0 implies it's integral is a constant.) Then plot a family of curves depending on C, perhps by solving for y.
2) Solve the original equation for dy/dx=y' = f[x,y] and solve that using Mathematica.
Best,
David
POSTED BY: David Keith
Posted 11 years ago
In thinking on this a little further, I would solve this on paper for the solution F[x,y] ==C, and then use CountourPlot to plot the family of curves associated with C, unless of course initial conditions provide a single value for C. This will allow mathematica to represent the solution set even though neither y nor x is one-to-one.
POSTED BY: David Keith
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