Confused about one Concept Today in my differential equation lecture, the teacher begin to write some examples in which differential equations appear, first begin with the classic example of a rocket, the example of the mass and the resort, and then he request to us give to he a example of a family of lines, and I tell to he
$y=mx+b$ where $m,b$ are parameters and then he put
$\frac{dy}{dx}=m$ and then he remplaze it in the original equation
$y=\frac{dy}{dx}x+b$
and put that this are the "Equivalent differential form of the family of curves.
Then he  put an exercise and exactlly say the following
Consider $y=A \sin x+ B \cos x$ familly of curves with two parameters $A,B$.Find it´s equivalent differential form.
I Never watch these kind of substitution in my life, and since he say that is a differential form I´m confused because the differential form of one equation is an expression like a $dz=f(x,y)dx+g(x,y)dy$
Is correct the teacher or it´s wrong, in the second case, can someone explain me or put the name of the concept.
I think that is wrong with the word differential form because if I don´t make a mistake
the differential of the equation $y=mx+b$ is $dy=mdx$
 A: Given that your teacher referred to $y=\frac{dy}{dx}X + b$ as the "differential form" of $y=mx + b$, and given that you're just encountering some simple differential equations, as a way of introducing differential equations, I think you're hearing a false similarity.  The teacher should have more correctly said the "differential equation form", and either wasn't thinking of differential forms or didn't know about them.  All the teacher is looking to do (and we can tell from the beginning and result) is to show $y=mx+b$ as a differential equation.  We do that by differentiating, in order to find an equivalent expression for $\frac{dy}{dx}$ and somehow work that $\frac{dy}{dx}$ back into the original equation.
Thus, what your teacher would be expecting (if I'm right) is that you express
$$
y=A\sin x + B\cos x
$$
as a differential equation.  Differentiate once or twice to get an expression  in terms of a derivative that you can substitute back into this equation.
A: \begin{align*}
y&=A\sin(x)+B\cos(x)\\
\frac{d}{dx}(y)&=\frac{d}{dx}(A\sin(x)+B\cos(x))\\
\frac{dy}{dx}&=A\frac{d\sin(x)}{dx}+B\frac{d\cos(x)}{dx}\\
\frac{dy}{dx}&=A\cos(x)-B\sin(x)\\
dy&=(A\cos(x)-B\sin(x))dx
\end{align*}
Therefore :
$$
dy-(A\cos(x)-B\sin(x))dx=0
$$
Is the equivalent differential form of $y$
