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

**slapmaxwell1** · February 5th 2011, 05:58 AM

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...

**Danny** · February 5th 2011, 06:14 AM

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

$\left(2x e^x + 6x^2 - y\right)dx - x dy = 0$ .

**Miss** · February 5th 2011, 08:56 AM

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.

Thread: to be exact or not exact? this is the question...

LinkBack

Thread Tools

Display

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

Similar Math Help Forum Discussions

transforming from a non-exact to an exact linear differential equation

Tricky question involving exact differentials

Why are exact ODEs called exact?

exact values practice question

exact value question

Search Tags