Proof of separation of variables

Let’s say we have a differential equation:
$$f(y) \dod{y}{x} = g(x)$$

Separation of variables says that we can simply “split” the derivative $\dod{y}{x}$ and write
$$f(y) \dif{y} = g(x) \dif{x}$$

Then, we can integrate both sides: $$\begin{gather*}
\int f(y) \dif{y} = \int g(x) \dif{x} \\
F(y) = G(x) + C
\end{gather*}$$ where $F(y)$ and $G(x)$ are antiderivatives of $f(y)$ and $g(x)$ respectively. And voila, we can use algebra to solve the differential equation.

But the “splitting” part always bugged me. Now my professor has explained why this is okay to do. Consider the antiderivative of $f(y)$, and differentiate it with respect to $x$.

We know that
\dod{}{x}F(y) &= \dod{}{y}F(y) \cdot \dod{y}{x} \text{ by the chain rule} \\
&= f(y) \dod{y}{x}
\end{align*}$$because $F(y)$ is the antiderivative of $f(y)$, so the derivative of $F(y)$ must be the function $f(y)$ itself.

Now, substituting into the differential equation for $f(y) \dod{y}{x}$, we have
$$\dod{}{x}F(y) = g(x)$$

Now, if we integrate the function on the left ($\dod{}{x}F(y)$) and the function on the right ($g(x)$) with respect to $x$, we should get the same antiderivative, up to a constant, so
$$ \int \left[\dod{}{x}F(y)\right] \dif{x} = G(x) + C$$

The antiderivative of a derivative of a function is just the function itself (up to a constant), so
$$ F(y) = G(x) + C$$ which is the result we get from doing it the “intuitive” way.

March 15, 2013, 1:45am by Casey
Categories: Math | Tags: , | Permalink | 3 comments

Comments (3)

  1. Good One thx!!!

  2. Very helpful, thanks!

  3. Helped me lot in my studies. Thank u!!

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