When is the given ODE exact? 
Let $$y'=\frac{cx-ay}{ex+by}.$$ Find the values of the constants
  $a,b,c,e\in\mathbb {R}$ such that the DE is exact. Give one particular
  solution and the equation of the tangent that passes thru some point $(x_0,y_0)$. 

What I have tried: 
Let $P(x,y)=\frac{cx-ay}{ex+by}$ and $Q(x,y)=-1.$ Then the DE is equivalent to $P(x,y)+Q(x,y)y'=0$. P and Q are continuously differentiable in some region $D\subset\mathbb{R}^2$ (as they are polynomials). So I don't need to worry too much about that.
Obviously $ex+by\neq0$ must hold 
and
$ \frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$ $\iff$ $\frac{-a\cdot(ex-by)-(cx-ay)\cdot b}{(ex+by)^2}=0 \iff -a\cdot(ex-by)-(cx-ay)\cdot b=0 \iff -x(ae+bc)=0$.
Therefore, $ae\neq-bc$. 
My answer: $a,b,c,e\in\mathbb{R}$ such that $ae\neq-bc$ and $ex+by\neq0$. Is this correct and complete so far? Then I should give one particular solution which makes me a bit confused. Then I probably need to solve the ODE in order to point out one particular solution, right?
 A: The OP has all the necessary background, but had some trouble applying it. In the end, they have gotten the correct solution on their own (in the comments at this point).
I hope the following might clear up some things. First, let us rewrite the equation in a more 'symmetric' form (it is very important, since what you tried at first would not do - the functions were too different):
$$(cx-ay)dx-(ex+by)dy=0$$
Or, introducing some functions:
$$P(x,y)dx+Q(x,y)dy=0$$
For the equation to be exact, the following should hold (as you correctly point out):
$$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$$
Applying this:
$$-a=-e \qquad \Rightarrow \qquad e=a$$
Now we have:
$$P(x,y)=cx-ay \\ Q(x,y)=-ax-by$$
The reason there's a special cathegory of 'exact equations' is that they can be represented as a full differential of some function:
$$dU(x,y)=\frac{\partial U}{\partial x}dx+\frac{\partial U}{\partial y}dy=0$$
This also means the function itself is equal to a constant. This will be our implicit solution.
Since:
$$P=\frac{\partial U}{\partial x}$$
$$U=\int P dx=\frac{1}{2}cx^2-ayx+f(y)=C$$
Where $f(y)$ is any function of $y$. While $C$ is a constant in both $x$ and $y$. 
On the other hand:
$$Q=\frac{\partial U}{\partial y}$$
$$U=\int Q dy=-ayx-\frac{1}{2}by^2+g(x)=C$$
Where $g(x)$ is any function of $x$. 
Obviously both expressions need to be equal:
$$\frac{1}{2}cx^2-ayx+f(y)=-ayx-\frac{1}{2}by^2+g(x)$$
Which makes the implicit solution:

$$U(x,y)=\frac{1}{2}cx^2-ayx-\frac{1}{2}by^2=C$$

Now the question asks for a particular solution (we can simply take $C=0$) and for the equation of the tangent (which I leave to the OP unless they have some trouble at this step).
