Differential equations with variables separable

What is a Differential Equation?

A differential equation is an equation involving one or more terms and the derivatives of one dependent variable with respect to the other independent variable.

Differential equation: dy/dx = f(x)

Where “x” is an independent variable and “y” is a dependent variable

Example of differential equation: $x \frac{d y}{d x}+2 y=0$
The above-written equation involves variables as well as the derivative of the dependent variable $\mathrm{y}$ with respect to the independent variable $\mathrm{x}$. Therefore, it is a differential equation.
The following relations are some of the examples of differential equations:
(i) $\frac{d y}{d x}=\sin 2 x+\cos x$
(ii) $\mathrm{k} \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2\right]^{3 / 2}$

Separable Differential Equation

Differential equations where the variables can be separated from each other are called separable differential equations. The general form of a separable differential equation is dy/dx = f(x)g(y).

These equations can be easily solved by separating the variables and integrating them individually.

Variable Separable Differential Equation Definition

The differential of the form $\frac{d y}{d x}=f(x) g(y)$ where $f(x)$ is a function of $x$ and $g(y)$ is a function of $y$, are said to be variable separable form.

Steps to Solve differential equations using variable separable methods-

1. Check for any values of y that make g(y)=0. These correspond to constant solutions.
2. Rewrite the differential equation in the form $\frac{d y}{g(y)}=f(x) d x$
3. Integrate both sides of the equation.
4. Solve the resulting equation for y if possible.
5. If an initial condition exists, substitute the appropriate values for x and y into the equation and solve for the constant.

Rewrite the equation as
$
\frac{d y}{g(y)}=f(x) d x \quad[\text { where } g(y) \neq 0]
$

This process is separating the variables. Now, integrating both sides, we get
$
\int \frac{d y}{g(y)}=\int f(x) d x+c
$

By this, we get the solution of the differential equation

Let's see some illustration for a better understanding
(i) Solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}=\left(\mathrm{e}^{\mathrm{x}}+1\right)\left(\mathrm{y}^2+1\right)$

Rewrite the differential equation as
$
\frac{\mathrm{dy}}{1+\mathrm{y}^2}=\left(\mathrm{e}^{\mathrm{x}}+1\right) \mathrm{dx}
$

Integrating both sides, we get
$
\begin{aligned}
& \int \frac{\mathrm{dy}}{1+\mathrm{y}^2}=\int\left(\mathrm{e}^{\mathrm{x}}+1\right) \mathrm{dx} \\
& \Rightarrow \tan ^{-1} y=e^x+x+c \\
& \Rightarrow y=\tan \left(e^x+x+c\right)
\end{aligned}
$

(ii) A differential equation of the form $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}(\mathrm{ax}+\mathrm{by}+\mathrm{c})$ where $\mathrm{a}, \mathrm{b}$, and $\mathrm{c}$ are constants, can be converted into an equation with variables separable by the substitution $\mathrm{v}=\mathrm{ax}+\mathrm{by}+\mathrm{c}$.
$
\begin{aligned}
& \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}(\mathrm{ax}+\mathrm{by}+\mathrm{c}) \\
& \mathrm{v}=\mathrm{ax}+\mathrm{by}+\mathrm{c} \\
& \therefore \frac{d v}{d x}=a+b \frac{d y}{d x} \text { or, } \frac{d y}{d x}=\frac{\frac{d v}{d x}-a}{b} \\
& \Rightarrow \frac{\frac{\mathrm{d} v}{\mathrm{dx}}-\mathrm{a}}{\mathrm{b}}=\mathrm{f}(\mathrm{v}) \Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{bf}(\mathrm{v})+\mathrm{a} \\
& \Rightarrow \frac{\mathrm{dv}}{\mathrm{bf}(\mathrm{v})+\mathrm{a}}=\mathrm{dx}
\end{aligned}
$

In the differential equation (ii), the variables $\mathrm{x}$ and $\mathrm{v}$ are separated.
Integrating (ii), we get
$
\begin{aligned}
& \Rightarrow \quad \int \frac{d v}{b f(v)+a}=\int d x+C \\
& \Rightarrow \quad \int \frac{d v}{b f(v)+a}=x+C \text {, where } v=a x+b y+c
\end{aligned}
$

This represents the general solution of the differential equation (i).

Recommended Video Based on Variable Separable Differential Equations

Solved Examples Based On Variable Separable Differential Equations

Example 1: Which of the following can be solved using the variable separation method?

(1) $\frac{d y}{d x}=1+x y$
(2) $\frac{d y}{d x}=x^2 y$
(3) $\frac{d y}{d x}=\sin (x+y)$
(4) $\frac{d y}{d x}=x^y$

Solution:
In options (1),(3), and (4), $\mathrm{x}$ and $\mathrm{y}$ cannot be completely separated to make quantity involving $\mathrm{x}$ with $\mathrm{dx}$ and quantity involving $y$ with dy. But it can be done in option (2) as:
$
\frac{d y}{d x}=x^2 y \Rightarrow \frac{d y}{y}=x^2 d x
$

Hence, the answer is the option (2).

Example 2: Which of the following doesn't represent variable separated form?
(1) $\sin x d x+\cos y d y=0$
(2) $\sin (x+y) d x+\cos y d y=0$
$\left(1+x^2\right) d x+\frac{1}{y} d y=0$
(4) $\sec ^2 x d x+\tan y d y=0$

Solution:
The differential equations in options (1), (3) and (4) are of variable separable form.
But in option (2), we have:
$
\sin (x+y) d x+\cos y d y=0
$

In this, variables cannot be separated as $\sin (\mathrm{x}+\mathrm{y})$ is a composite function.
Hence, the answer is the option (2).

Example 3: Given that the slope of the tangent to a curve $y=y(x)$ at any point $(x, y)$ is $\frac{2 y}{x^2}$. If the curve passes through the centre of the circle $x^2+y^2-2 x-2 y=0$, then its equation is:

Solution:
Given the slope of the tangent $=\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2 y}{x^2}$.
$
\frac{\mathrm{d} y}{\mathrm{y}}=\frac{2 d x}{\mathrm{x}^2}
$

Integrate both sides.
$
\ln (y)=-\frac{2}{x}+C
$

It passes through the center of the circle $x^2+y^2-2 x-2 y=0$ i.e $(1,1)$
$
0=-2+C \Rightarrow C=2
$

Eq. of the curve is
$
\begin{aligned}
& x \cdot \ln |y|=-2+2 x \\
& x \cdot \ln |y|=2(x-1)
\end{aligned}
$

Hence, the answer is $x \log _e|y|=2(x-1)$

Example 4: The solution of the differential equation $x\left(y^2+y\right) d x+(x+1) d y=0$ is:
Solution:
$
x y(y+1) d x+(x+1) d y=0
$

Dividing throughout by $y(y+1)(x+1)$, we get:
$
\begin{aligned}
& \frac{x d x}{(x+1)}+\frac{d y}{y(y+1)}=0 \\
& \Rightarrow \int \frac{x d x}{(x+1)}+\int \frac{d y}{y(y+1)}=c \\
& \Rightarrow \int \frac{(x+1)-1}{(x+1)} d x+\int \frac{(1+y)-y}{y(y+1)} d y=c \\
& \Rightarrow \int\left(1-\frac{1}{(x+1)}\right) d x+\int\left(\frac{1}{y}-\frac{1}{y+1}\right) d y=c \\
& \Rightarrow x-\ln |x+1|+\ln |y|-\ln |y+1|=c \\
& \Rightarrow \ln \left|\frac{y}{(x+1)(y+1)}\right|=c-x \\
& \Rightarrow\left|\frac{y}{(x+1)(y+1)}\right|=e^{c-x} \\
& \Rightarrow\left|\frac{y}{(x+1)(y+1)}\right|=c \cdot e^{-x} \\
&
\end{aligned}
$

Hence, the answer is $\left|\frac{y}{(x+1)(y+1)}\right|=c \cdot e^{-x}$

Example 5: Let $y=f(x)$ be the solution of the differential equation $y(x+1) d x-x^2 d y=0, y(1)=e$. Then $x \rightarrow 0^{+} f(x)$ is equal to

Solution:
$
\begin{aligned}
& y(x+1) d x-x^2 d y=0, \quad y(1)=e \\
& \Rightarrow \frac{d y}{d x}=\frac{y(x+1)}{x^2} \\
& \Rightarrow \int \frac{d y}{y}=\int \frac{(x+1) d x}{x^2} \\
& \ell \text { ny }=\ell \mathrm{nx}-\frac{1}{x}+c \\
& \because y(1)=e \\
& \therefore 1=0-1+C \Rightarrow C=2
\end{aligned}
$

Now, $\ell \mathrm{ny}=\ell \mathrm{nx}-\frac{1}{\mathrm{x}}+2$
$
\begin{aligned}
& \Rightarrow \ell \ln \left(\frac{y}{x}\right)=2-\frac{1}{x} \\
& \Rightarrow \frac{y}{x}=e^{2-\frac{1}{x}} \\
& \Rightarrow y=x, e^{2-\frac{1}{x}}
\end{aligned}
$

So, $\lim _{x \rightarrow 0^{+}} y=\lim _{x \rightarrow 0^{+}} x e^{2-\frac{1}{x}}=0$
Hence, the answer is 0.

Summary

Variable Separable differential equations are a useful and accessible class of differential equations that can be equated in many real-life phenomena, such as population growth and radioactive decay. By separating the variables and integrating both sides independently, we can find solutions that describe how these systems evolve. This method is a fundamental tool in the analysis of dynamic systems in various fields, including mathematics, biology, chemistry, and physics.