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
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:)
dy/dx=(9+20 x)/(xy^2)
To determine whether the given equation is exact, we check if the partial derivatives of M(x, y) with respect to y and N(x, y) with respect to x are equal.
You correctly found that My = -3y^2 + 2ysinx and Nx = -3y^2 + 2ysinx. Since My = Nx, the equation is indeed exact.
Now let's proceed with solving it. You started by integrating M(x, y) with respect to x:
∫ (x - y^3 + y^2sinx) dx + g(y) = x^2/2 - xy^3 - y^2cosx + g(y)
Next, you took the derivative of this expression with respect to y to find g'(y):
-3xy^2 - 2ycosx + g'(y) = N(x, y)
You correctly integrated -3xy^2 with respect to y, but there seems to be a mistake in the integrated term, which should be -3xy^3/3 = -xy^3, not -xy^3 + c.
Hence, the correct expression for g(y) should be:
g(y) = -xy^3 + c
Now, substituting this back into the previous expression, we have:
x^2/2 - xy^3 - y^2cosx - xy^3 + c = N(x, y)
Simplifying, we get the solution to be:
f(x, y) = x^2/2 - 2xy^3 - y^2cosx + c
Comparing your answer with the given answer in the book:
xy^3 + y^2cosx - 1/2x^2 = c
There seems to be a sign error in the second term from the given answer. It should be -2xy^3 instead of +2xy^3. So, your answer and the given answer in the book are slightly different, but both are valid solutions to the exact equation.