Wednesday

April 16, 2014

April 16, 2014

Posted by **allison** on Wednesday, October 8, 2008 at 3:03pm.

the question is (x-y^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 (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

xy^3+y^2cosx-1/2x^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 12:02amdy/dx=(9+20 x)/(xy^2)

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