Question

(2xy+3)dx+(x^2-1)dy=0
exact equations

I get that they are exact and get to f(x,y)=Integral(2xy+3)dx not sure if it should equal x^2+3x+g(y) or x^2+g(y)

Answer 1

im gonna go with \[x^2+3x+g(y)\]

Answer 2

So you're looking for an intermediate step? Or do you think that these are your final solutions?

Answer 3

no i know that these aren't our final solutions. you continue to

\[x^2+3x+g'(y)=X^2-1\]

\[g'(y)=-3x-1\]

\[g(y)=(-3/2) x^2-1x\]
then i stopped cause i don't think this is right:/

Answer 4

\[\Large f_x+f_y\frac{dy}{dx}=0 \]
where f(x,y) is multivariable. You checked the exactness for this equation already. so you want to solve either:
\[\Large f_x=2xy+3 \] or
\[\Large f_y=x^2-1 \]

Lets choose the first one:
\[ \Large f_x=\frac{\partial f(x,y)}{\partial x}=2xy+3\]
and integrate with respect to x to get one representation for f(x,y)

\[\Large f(x,y)=\int\partial f(x,y)=x^2y+3x+g(y)+k \]

Answer 5

ok so i missed the y in the x^2y. so now we
\[x^2y+3x+g'(y)=x^2-1\] ??

Answer 6

differentiate your temporary expression f(x,y) with respect to partial y.
\[\Large f_y=\frac{\partial f(x,y)}{\partial y}=x^2+g'(y)=x^2-1 \]
Simplify an integrate:
\[\Large g'(y)=-1 \\ \Large \therefore g(y)=-y\]

Answer 7

Awesome! thank you so much!

Answer 8

you're very welcome :-)