Розв'язування диференційних рівнянь із розділеними змінними

This type of differential equation is called "separable" because of the following trick (which will be proven later): $$\begin{align*} \frac{dy}{dx} &= g(x)h(y)\\ \frac{dy}{h(y)} &= g(x) dx\\ \int\frac{dy}{h(y)} &= \int g(x) dx\end{align*}$$ Let's see a concrete example: try to solve the differential equation $y' = 2xy$ using the separating technique above.

$$\begin{align*} \frac{dy}{dx} &= 2xy\\ \frac{dy}{y} &= 2x dx\\ \int\frac{dy}{y} &= \int 2x dx\\ \ln(y) &= x^2+C_1\\ y &= Ce^{x^2} \end{align*}$$

where the $C$ in the final result is $e^{C_1}$.

An application/example of this type of differential equation in circuits:

Consider a series circuit with a battery (voltage $V$), a capacitor (capacitance $C$), and a resistor (resistance $R$). Suppose $Q(t)$ is the charge on the capacitor at time $t$. Then the current is $I(t) = Q'(t)$. Then by Kirchoff's voltage law: $$V = \frac{Q}{C} + Q'R$$ Now, whether or not you understood how we got here, this is a differential equation which you can solve. Give it a try!

It's not immediately obvious that this is separable, but with a little rearranging, it is: $$Q'=\frac{dQ}{dt} = \left(\frac{V}{R} - \frac{Q}{RC}\right)$$ so $$\int\frac{dQ}{\left(\frac{V}{R} - \frac{Q}{RC}\right)} = t + C_1$$ The integral on the left is solved with a simple substitution $u = \left(\frac{V}{R} - \frac{Q}{RC}\right)$, $du = -\frac{1}{RC}dQ$. So $$\ln\left(\frac{V}{R} - \frac{Q}{RC}\right) = -\frac{t}{RC} + C_2.$$ Rearranging, with the shorthand $\tau = RC$: $$\left(\frac{V}{R} - \frac{Q}{RC}\right) = C_3 e^{-t/\tau}$$ Thus $$Q = CV - Ae^{-t/\tau}$$ is our set of solutions. (Here $A$ is the arbitrary constant, since $C$ is capacitance).

With the initial conditions $Q(0) = 0$, we get $A = CV$, so $$Q(t) = CV(1-e^{-t/\tau})$$ which if I recall correctly is the "standard" equation for this particular circuit.

A separable differential equation can be written as a product of a function of $x$ and a function of $y$.

$$\frac{dy}{dx}=g(x)f(y)$$

If $f(y)\neq0$, we can write the above differential equation as

$$\frac{dy}{dx}=\frac{g(x)}{h(y)}$$

where $h(y)=\frac{1}{f(y)}$.

We can then rewrite it in the differential form

$$h(y)dy=g(x)dx$$

and integrate both sides

$$\int h(y)dy=\int g(x)dx$$

We can now examine an application of differential equations in the chemistry of kinetics.

Suppose that we are given the differential rate law of a first-order reaction

$$\frac{d[A]}{dt}=k[A]$$

We can solve this differential equation to get the integrated rate law.

$$\frac{d[A]}{[A]}=kdt$$ $$\int^t_0\frac{d[A]}{[A]}=\int^t_0kdt$$ $$\ln[A]\bigg\vert^t_0=kt\bigg\vert^t_0$$ $$\ln[A]_t-\ln[A]_0=kt$$ $$\boxed{\ln[A]_t=kt+\ln[A]_0}$$

A separable differential equation is an equation of the form $$\frac{dy}{dx} = g(x)h(y),$$ where $g,h$ are given functions of $x,y$, respectively. For example, when $h$ is constant, we recover the anti-derivative equation. In general, you can solve separable equations essentially by "separating the two variables $x$ and $y$": $$\frac{dy}{h(y)} = g(x) dx \implies \int\frac{dy}{h(y)} = \int g(x) dx \implies H(y) = G(x) + C,$$ where $G$ and $H$ are anti-derivatives of $g$ and $1/h$, respectively, and $C$ is some constant depending on the initial conditions. Alternatively, we could have solved the equation using the chain rule: $$g(x) = \frac{1}{h(y)}\frac{d}{dx}y(x) = \frac{d}{dx} H(y(x)) \implies H(y(x)) = G(x) + C.$$ Either way, you can then usually solve for $y(x)$ in terms of $x$ by applying the inverse function $H^{-1}$ to both sides.