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Find the Family of Implicit Solutions

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Find a family of solutions to the differential equation, did he just swear at me?

  • Thread starter mr_coffee
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Ello ello!
confusion, part 3. THe directions are the following:
Find a family of solutions to the differential equation
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/64/86dc9abe51adb1b16452035777ef8b1.png [Broken]
(To enter the answer in the form below you may have to rearrange the equation so that the constant is by itself on one side of the equation.) Then the solution in implicit form is:
the set of points (x, y) where F(x,y) = ENTER ANSWER HERE = constant

I'm abit distraught, how would i go about doing this?
I re-wrote it as:
(x^2+2xy)+x(dy/dx) = 0;

the Ti-89 said the answer is:
y= -2xy*ln(x)-x^2/2 + c
but when i solve for C and get:
C = y + 2xy*ln(x) + x^2/2;
WEbworks says, Can't take Log(0).

Any tip on what i shall do>?
thanks.

Last edited by a moderator:

Answers and Replies

Ello ello!
confusion, part 3. THe directions are the following:
Find a family of solutions to the differential equation
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/64/86dc9abe51adb1b16452035777ef8b1.png [Broken]
(To enter the answer in the form below you may have to rearrange the equation so that the constant is by itself on one side of the equation.) Then the solution in implicit form is:
the set of points (x, y) whereENTER ANSWER HERE = constant

I'm abit distraught, how would i go about doing this?
I re-wrote it as:
(x^2+2xy)+x(dy/dx) = 0;

the Ti-89 said the answer is:
y= -2xy*ln(x)-x^2/2 + c
but when i solve for C and get:
C = y + 2xy*ln(x) + x^2/2;
WEbworks says, Can't take Log(0).

Any tip on what i shall do>?
thanks.


What does log(0) have to do with it? Your original problem said "Find a family of solutions to the differential equation" in the form F(x,y) = C. Okay y+ 2xyln(x)+ x2/2 = C is of that form.

I am not terribly impressed by your letting a TI-89 do the work for you- you are supposed to be smarter than a machine! If you divide xy'+ x2+ 2xy= 0 by x you get y'+ x+ 2y= 0, a linear equation:
y'+ 2y= x. Multiplying the entire equation by e2x, an integrating factor, give e2xy'+ 2e2x= (e2xy)'= xe2x. The integral of the left side, of course, is e2xy and the right hand side you can do with integration by parts.

Last edited by a moderator:
I know i'm ashamed of myself. Thanks for the help though, it worked! You just forgot a minus sign on ur explanation but everything else was right! But i had a question. How did you know to muliply everything by e^(2x)? For integrating facotr i always thought u had to have like:
y'+y/x = x;
then the integrating factor would be:
I = e^(integral(1/x));
I = x;

But in my case, there wasn't anything but a 2 infront of the y.

THe answer that was right was this one:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/74/3b36e53e805bbec1fd92f9d7f9bf841.png [Broken]

Thanks again!

Last edited by a moderator:

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Find the Family of Implicit Solutions

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