- #1
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
(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.
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
(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.
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