Q. 9 - Calculus

Question

Why do we use the terminology "separable" to describe a differential equation that can be written in the form

Step-by-Step Solution

Answer

Therefore, using the variable separable approach, differential equations of the type $\frac{dy}{dx}=p\left(x\right)q\left(y\right)$ can be solved.

Step 1. Given information

A differential equation that can be written in the form $\frac{dy}{dx}=p\left(x\right)q\left(y\right)$

Step 2. Calculation

The $\frac{dy}{dx}=g(x,y)$, where $g(x, y)$ is a function of the variables x and y, denotes a general first order differential equation. The x and y components of the differential equation are said to be separable in a specific situation where $g(x, y)$ can be represented as a product of functions of the variables x and y in the form $g(x, y)=p\left(x\right) q\left(y\right)$. In other words, just the variable x will have a function on one side of the equation, and only the variable y will have a function on the other. The variables appear independently in this scenario. It follows that the differential equation of the form $g(x, y)=p\left(x\right), q\left(y\right)$ is of the variable separable type.

If each of these operations is feasible, an equation of the form $g(x, y)=p\left(x\right) q\left(y\right)$ can be resolved by dividing both sides by $q\left(y\right)$, integrating both sides, and then solving for y. Thus,

$$\begin{gathered} \frac{dy}{dx}=p\left(x\right)q\left(y\right) \\ \int \frac{1}{q\left(y\right)}\frac{dy}{dx}dx=\int p\left(x\right)dx \\ \int \frac{1}{q\left(y\right)}dy=\int p\left(x\right)dx \end{gathered}$$

Think about the illustration $\frac{dy}{dx}=2xy$. The function $g(x, y)=2 x y$ has the properties $p\left(x\right)=2 x, q\left(y\right)=y$. The variable separable approach can be used to solve this example. Separate the variables, then combine both sides to arrive at the answer as

$$\begin{gathered} \frac{dy}{dx}=2xy \\ \frac{1}{y}dy=2xdx \end{gathered}$$

Integrate both sides

$$\begin{gathered} \ln \left|y\right|=x^2+C \\ y=e^{x^2+C} \\ =A{e}^{x^2} \left(e^C=A\right) \end{gathered}$$

Therefore, using the variable separable approach, differential equations of the type $\frac{dy}{dx}=p\left(x\right)q\left(y\right)$ can be solved.