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I'm used to the notation $\frac{d}{dx}$ as an opertaor, which operates on some function(say $y$), giving the derivative $y'$ of the function. But I can't figure out what does it mean in the case of differential equations. Say a differential $\frac{dy}{dx} + P(x) = Q$. How can we solve this by stating $ dy+ P(x)\,dx =Q\,dx$ ?

Michael Hardy
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Arnabesh Samadder
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3 Answers3

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Here is a related answer on what "$dx$" means: What is $dx$ in integration?

In some differential equations it is useful to work with differentials explicitly because it makes separation of variables possible. For example: $$ \frac{dy}{dx} = y(1-y). $$ Here one can write $$ \frac{dy}{y(1-y)} = dx $$ thus separating $x$ and $y$ from each other. Then $$ \int \frac{dy}{y(1-y)} = \int dx $$ and so by evaluating both integrals one gets a solution.

Michael Hardy
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Things do not change. Differential equations are a kind of functional equation.

With $\frac{dy}{dx}+P(x)=Q$ the problem is: find a function "$y(x)$" whose derivative is $Q-P(x)$.

trying
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You can view "d" as a differential. In case of function of one variable y(x) at a point $x_0$: $dy(x_0)=y'(x_0)x$, where $x$ is a short writing for the function, that to x associates x. However if you compute differential of this function x, it will simply be still identity, since identity already is linear.

Therefore you can write $dy(x_0)=y'(x_0)dx(x_0)$ and since we are dealing with linear forms, we can simply divide, to get $\frac{dy(x_0)}{dx(x_0)}=y'(x)$ and the final step is to deal with functions, so you forget the variable altogether and end up with a fraction of two differentials $\frac{dy}{dx}$.

If we finally comeback to your original equation: $\frac{dy}{dx}+P=Q$, you can view dy and dx as differentials, hence it makes sense to rewrite it as: $dy+Pdx=Qdx$, which is equivalent, since $dx$ is a nowhere zero as a linear form and linear forms behave like real numbers.

Hope that's clear enough.

Keen
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