# \frac{dy}{dx}=-\frac{ax+by}{bx+cy}

$\frac{dy}{dx}=-\frac{ax+by}{bx+cy}$
You can still ask an expert for help

## Want to know more about Laplace transform?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Hector Roberts
Given exact differential equation is
$\frac{dy}{dx}=-\frac{ax+by}{bx+cy}$
The standard exact differential equation is represented as,
$Mdx+Ndy=0$
It should satisfy the condition of exactness
$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
Lets

Robert Pina
$\frac{dy}{dx}={y}^{\prime }⇒{y}^{\prime }=-\frac{ax-by}{bx-cy}$, multiply equation with $bx-cy$
$\left(ax-by\right)+\left(bx-cy\right){y}^{\prime }=0$
$M\left(x,y\right)=ax-by⇒{N}_{y}=-b$
$N\left(x,y\right)=bx-cy⇒{N}_{x}=b$
Since ${M}_{y}\ne {N}_{x}$ given equation is not exact.

Vasquez

$dy/dx=-\left(ax+by\right)/\left(bx+cy\right)+{y}^{\prime }=0$
Then, $\left(\frac{ax+by}{bx+cy}{\right)}_{x}^{{}^{\prime }}\ne 0$
in general, while $\left(1{\right)}_{y}^{{}^{\prime }}=0.$