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 (xy^3+ysinx)dx = (3xy^2+2ycosx)dy
i solve and i got that it is exact
because
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 (xy^3+y^2sinx)dx + g(y)
it got x^2/2 xy^3y^2cosc + g(y)
then i took derivative of it wrt y
3xy^22ycosx + 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^3ycosx+xy^3 = c
but when i check the back of the book
their answer was
xy^3+y^2cosx1/2x^2 =c
are the same or did i do something wrong...
please help me understand my mistake
anyone please
thanks in advance:)

differential equations  James, Friday, September 9, 2011 at 12:02am
dy/dx=(9+20 x)/(xy^2)
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