Math Doubts

Separation of Variables

A method of solving a differential equation by separating the functions of a variable with respective differential on one side and the functions of another variable with corresponding differential on other side of the equation is called the variables separable method of differential equations.

Introduction

Let $x$ and $y$ represent two variables, then $dx$ and $dy$ are corresponding differentials. When the functions in $x$ and $y$ are denoted by $f(x)$ and $g(y)$ mathematically and form a differential equation in terms of them. The differential equation can be solved by the separation of variables to keep functions in $x$ with respective differential $dx$ on one side and to move the functions in $y$ with corresponding differential $dy$ on other side of the equation as follows.

$g(y)dy \,=\, f(x)dx$

Now, integrate both sides of the differential for solving the differential equation by variable separation.

$\implies$ $\displaystyle \int{g(y) \,}dy \,=\, \int{f(x) \,}dx+c$

Now, let us learn how to use the separation of variables method to solve differential equations of first order and first degree.

Example

Solve $xdy+ydx \,=\, 0$

$\implies$ $xdy \,=\, -ydx$

$\implies$ $\dfrac{dy}{y} \,=\, -\dfrac{dx}{x}$

$\implies$ $\dfrac{1}{y}dy \,=\, -\dfrac{1}{x}dx$

$\implies$ $\displaystyle \int{\dfrac{1}{y}\,}dy \,=\, -\displaystyle \int{\dfrac{1}{x}\,}dx$

$\implies$ $\log_e{y}+c_1 \,=\, -(\log_e{x}+c_2)$

$\implies$ $\log_e{y}+c_1 \,=\, -\log_e{x}-c_2$

$\implies$ $\log_e{y}+c_1+\log_e{x}+c_2 \,=\, 0$

$\implies$ $\log_e{y}+\log_e{x}+c_1+c_2 \,=\, 0$

$\implies$ $\log_e{(y \times x)}+c \,=\, 0$

$\implies$ $\log_e{(x \times y)}+c \,=\, 0$

$\,\,\, \therefore \,\,\,\,\,\,$ $\log_e{(xy)}+c \,=\, 0$

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