Determine whether this is an exact equation [differential]

[jwxie]

Homework Statement

Determine whether this is an exact equation. If it is, find the solution.

$\[\frac{\mathrm{d}y }{\mathrm{d} x} = -\frac{ax+by}{bx+cy}\]$

Homework Equations

The Attempt at a Solution

If I moved the right side to the left (because an exact equation has the form ZERO on the right side)...
I get $\[\frac{\partial M}{\partial y}(\frac{ax+by}{bx+cy}) \neq 0\]$, where as the partial of the second term $\[1\frac{\partial N}{\partial y} = 0\]$.

But the book said this is an exact equation.

I attempted the problem first by trying solving it as a homogeneous equation. That is, the equation can be written in the form of the ratio $\[\frac{y}{x}\]$

I tried, and I got the followings
$\[\frac{\mathrm{d}y }{\mathrm{d} x} = -\frac{a+b(y/x)}{b+c(y/x)}\]$
and let v = y/x
$\[v+ x\frac{\mathrm{d}v }{\mathrm{d} x} = -\frac{a+b(v)}{b+c(v)}\]$

I tried to solve this by partial fraction integration but unfortunately i can't because i can't separate the terms since they are written in symbolic forms. I tried to use quadratic formulas but still no help.

So how do I solve this problem?

Thanks.

 

[lurflurf]

An exact differential looks like
dU=U_x dx + U_y dy
Where U_x and U_y are partial derivative with respect to x and y respectively
Notice that under mild conditions on U_x and U_y we have
U_xy=U_yx
by equality of mixed partial derivatives
Write the equation in the form
F dx +G dy = 0
If this is an exact differential
F_y=G_x
so check if it is
An exact differential equation such as
dU=0
is simple to solve
U=some constant.

 

[jwxie]

OH. the partial of the two gives b
the form M(x,y) = N(x,y) y' = 0 can be written as F dx +G dy = 0 if we multiply both side by dx, right?

the dU is always 0 if it has to be an exact, am i correct?

 

[lurflurf]

That is right. Now try to find U by integration.