Why does seperation of variables work to solve differential equations?

Question

When given this simple differential equation and a condition, $$\frac{dy}{dx}=e^{2x-y} \hspace{1em} , \hspace{1em} C_1 : y(0)=0$$

Using separation of variables,

$$\frac{dy}{dx}=e^{2x}\cdot e^{-y} \Rightarrow \ \frac{dy}{e^{-y}}=e^{2x} \ dx$$

$$\int e^{y} \ dy= \int e^{2x} \ dx \Rightarrow \ e^{y} = \frac{e^{2x}}{2} +c $$

Solving for c ,

$$e^{0} = \frac{e^{2(0)}}{2} + c \ \Rightarrow 1 = \frac{1}{2} +c $$

$$ c = \frac{1}{2}$$

Solving for y ,

$$e^y = \frac{e^{2x}}{2} + \frac{1}{2} \ \Rightarrow e^y = \frac{1}{2}(e^{2x} + 1) $$

$$y = \ln\left(\frac{1}{2}\right) + \ln\left(e^{2x} + 1\right) $$

Finally,

$$\boxed{y=\ln\left(e^{2x} + 1\right) -\ln(2)}$$

Question: In the above example we can see that the method of separation of variables indeed works in solving our differential equation, but why? What actually is going on when the method of separation of variables is used?

Answer 1

Separation of variables really begins with the following:

$$y'(x)=f(x) g(y),y(x_0)=y_0 \\ \int_{x_0}^x \frac{y'(t)}{g(y(t))} dt = \int_{x_0}^x f(t) dt.$$

Now on the left side you use integration by substitution (which as you may recall is ultimately a combination of the chain rule and the fundamental theorem of calculus) to get

$$\int_{y_0}^y \frac{1}{g(u)} du = \int_{x_0}^x f(t) dt.$$

Thus you get $H(y)-H(y_0)=F(x)-F(x_0)$ where $H$ is an antiderivative of $1/g$ and $F$ is an antiderivative of $f$. Assuming $g$ always has one sign between $y_0$ and $y$, you can write $y=H^{-1}(H(y_0)+F(x)-F(x_0))$.

Separating differentials is a notation trick that ultimately accomplishes the same thing as this calculation.