# Solving differential equations by Variables separable

The separation of variables is a method of solving a differential equation in which the functions in one variable with respective differential is separable on one side from the functions in another variable with corresponding differential element. There are two possible cases in the variables separable method. So, let’s learn each case with proofs and their mathematical expressions to understand how to solve a differential equation by separating the variable functions.

The literals $x$ and $y$ are considered as variables and their corresponding differential elements are $dx$ and $dy$ respectively in both cases.

### Simple separation

In this case, the functions in terms of $x$ and $y$ are written as $f(x)$ and $\phi(y)$ respectively.

In this case, the functions in one variable with respective differential are shifted easily on one side and the functions in another variable with corresponding differential are kept on the other side of the equation as follows.

$\phi{(y)}\,dy$ $\,=\,$ $f(x)\,dx$

Now, integrate both sides of the equation. $c_1$ and $c_2$ are constants of the integration.

$\implies$ $\displaystyle \int{\phi{(y)}\,}dy + c_1$ $\,=\,$ $\displaystyle \int{f(x)\,}dx + c_2$

$\implies$ $\displaystyle \int{\phi{(y)}\,}dy$ $\,=\,$ $\displaystyle \int{f(x)\,}dx + c_2-c_1$

The difference of the constants is also a constant. Hence, it is simply denoted by a constant $c$.

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

### Separation by factoring

In this case, the functions in terms of $x$ and $y$ are expressed as $g(x)$ and $h(y)$ respectively.

In some other cases, it is not easy to separate the functions in one variable from the functions in another variable but it is possible to separate them by factorization (or factorisation) as follows.

$\dfrac{dy}{dx} \,=\, g(x).h(y)$

Now, multiply the both sides of the equation by the differential element $dx$.

$\implies$ $\dfrac{dy}{dx} \times dx \,=\, g(x).h(y) \times dx$

$\implies$ $dy \,=\, g(x).h(y)\,dx$

Now, the functions in one variable can be separated from the functions in another variable as follows.

$\implies$ $\dfrac{1}{h(y)}\,dy \,=\, g(x)\,dx$

We can repeat the same procedure for solving the differential equation as follows.

$\implies$ $\displaystyle \int{\dfrac{1}{h(y)}\,}dy+c_1$ $\,=\,$ $\displaystyle \int{g(x)\,}dx+c_2$

$\implies$ $\displaystyle \int{\dfrac{1}{h(y)}\,}dy$ $\,=\,$ $\displaystyle \int{g(x)\,}dx+c_2-c_1$

$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \int{\dfrac{1}{h(y)}\,}dy$ $\,=\,$ $\displaystyle \int{g(x)\,}dx+c$

#### Problems

Let’s learn how to solve the differential equations by the separation of variables from some understandable problems with solutions.

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