Posted by allison on Wednesday, October 8, 2008 at 3:03pm.
the directions are to determine whether or not the equation is exact if so then solve it.
the question is (x-y^3+ysinx)dx = (3xy^2+2ycosx)dy
i solve and i got that it is exact
My is -3y^2+2ysinx and Nx is -3y^2+2ysinx
since m and n are equal then they are exact
then i started with M
integral of (x-y^3+y^2sinx)dx + g(y)
it got x^2/2 -xy^3-y^2cosc + g(y)
then i took derivative of it wrt y
-3xy^2-2ycosx + g'(y) = N
then -3xy^2 + integral g'(y)= integral of -3xy^2 dy
then g(y) = xy^3 -xy^3 +c
then i got
f(x,y)= 1/2x^2 - xy^3-ycosx+xy^3 = c
but when i check the back of the book
their answer was
are the same or did i do something wrong...
please help me understand my mistake
thanks in advance:)
- differential equations - James, Friday, September 9, 2011 at 12:02am
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