# to be exact or not exact? this is the question...

• February 5th 2011, 05:58 AM
slapmaxwell1
to be exact or not exact? this is the question...
ok so i have this problem, and im having a little trouble separating it..here is the function:

x(dy/dx) = 2xe^x - y + 6x^2

i cant get this problem separated so that i can solve it for exactness. because there is only one y, i guess i want the My=0 and Nx=0. any thoughts? thanks in advance...
• February 5th 2011, 06:14 AM
Jester
Remember, the standard form for exact equations is

$M(x,y)dx + N(x,y)dy = 0,$ where you need $M_y = N_x$.

For you,

$x dy = \left(2x e^x + 6x^2 - y\right)dx$

so

$\left(2x e^x + 6x^2 - y\right)dx - x dy = 0$.
• February 5th 2011, 08:56 AM
Miss
Another way:

$\left(2x e^x + 6x^2 - y\right)dx - x dy = 0$
$\implies (2xe^x+6x^2)dx-(ydx+xdy)=0$
$\implies (2xe^x+6x^2)dx-d(xy)=0$

Now, just integrate.