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October 2nd, 2013, 12:15 PM   #1
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Is this allowed?? (First order ODE) Exact Diff Eqn

(ycosx + 2xe^y) + (sinx + x^2e^y - 1)y' = 0

(ycosx + 2xe^y) + (sinx + x^2e^y - 1) dy/dx = 0

multiply both sides by dx

(ycosx + 2xe^y)dx + (sinx + x^2e^y - 1)dy = 0

integral (ycosx + 2xe^y)dx + integral(sinx + x^2e^y - 1)dy = 0

ysinx + x^2 * e^y + ysinx +x^2 * e^y - y = c

2ysinx + 2x^2 * e^y - y = c
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October 2nd, 2013, 05:01 PM   #2
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Re: Is this allowed?? (First order ODE) Exact Diff Eqn

Quote:
Originally Posted by hanilk2006
(ycosx + 2xe^y) + (sinx + x^2e^y - 1)y' = 0

(ycosx + 2xe^y) + (sinx + x^2e^y - 1) dy/dx = 0

multiply both sides by dx

(ycosx + 2xe^y)dx + (sinx + x^2e^y - 1)dy = 0
to this point everything is valid.

Quote:
integral (ycosx + 2xe^y)dx + integral(sinx + x^2e^y - 1)dy = 0
This is not. You cannot integrate y cos(x) 2xe^y with respect to x while treating y as a constant. It isn't constant. Similarly, you cannot integrate sinx+ x^2e^y- 1 with respect to y, treating x as if it were constant.

Quote:
ysinx + x^2 * e^y + ysinx +x^2 * e^y - y = c

2ysinx + 2x^2 * e^y - y = c
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