Homogeneous Differential Equation

Before we learn the concept of the Homogeneous Differential Equation form of a differential equation, let's first understand what a differential equation is. 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. A differential equation of the form f(x,y)dy = g(x,y)dx is said to be a homogeneous differential equation if the degree of f(x,y) and g(x, y) is the same. These equations are commonly encountered in various fields such as physics, engineering, economics, and mathematics.

In this article, we will cover the concept of homogeneous differential equations. This concept falls under the broader category of differential equations, which is a crucial chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of fifteen questions have been asked on this concept, including one in 2013, one in 2019, two in 2020, two in 2021, five in 2022, and four in 2023.

What is a Homogeneous Function?

A function $f(x, y)$ is said to be a homogeneous function of degree $n$ if it satisfies the property
$
f(\lambda x, \lambda y)=\lambda^n f(x, y)
$

Consider the following examples
1. $f(x, y)=x^3-4 x y^2$
2. $f(x, y)=x-3 y$
3. $f(x, y)=\tan \frac{x}{y}$
In the above examples if we replace $\mathrm{x}$ and $\mathrm{y}$ with $\lambda \mathrm{x}$ and $\lambda \mathrm{y}$, where $\lambda$ is non - zero, we get
1. $\mathrm{f}(\lambda \mathrm{x}, \lambda \mathrm{y})=(\lambda \mathrm{x})^3-4(\lambda \mathrm{x})(\lambda \mathrm{y})^2=\lambda^3\left(\mathrm{x}^3-4 \mathrm{xy}^2\right)=\lambda^3 \mathrm{f}(\mathrm{x}, \mathrm{y})$
2. $\mathrm{f}(\lambda \mathrm{x}, \lambda \mathrm{y})=\lambda \mathrm{x}-3(\lambda \mathrm{y})=\lambda(\mathrm{x}-3 \mathrm{y})=\lambda \mathrm{f}(\mathrm{x}, \mathrm{y})$
3. $\mathrm{f}(\lambda \mathrm{x}, \lambda \mathrm{y})=\tan \frac{\lambda \mathrm{x}}{\lambda \mathrm{y}}=\tan \frac{\mathrm{x}}{\mathrm{y}}=\lambda^0 \mathrm{f}(\mathrm{x}, \mathrm{y})$

Now, if the function is given as
4. $f(x, y)=\sin x+\cos y$, then $f(\lambda x, \lambda y) \neq \lambda^n f(x, y)$

Observe that it is possible to write examples 1,2 and 3 in the form of $f(\lambda x, \lambda y)=\lambda^n f(x, y)$.
But example 4 can't be written in this form.
Here, examples 1,2 and 3 are homogeneous equations of degrees 3,1 and 0 respectively and example 4 is not a homogeneous function.

What is a Homogeneous Differential Equation?

Any differential equation of the form $M(x, y) d x+N(x, y) d y=0$ or $\frac{d y}{d x}=-\frac{M(x, y)}{N(x, y)}$ is called homogeneous if $M(x, y)$ and $N(x, y)$ are homogeneous functions of the same degree.

Since, $M(x, y)$ and $N(x, y)$ are both homogeneous function of degree $n$, then $\mathrm{DE}$ can be reduced to a function of $\mathrm{y} / \mathrm{x}$
$
\frac{d y}{d x}=-\frac{M(x, y)}{N(x, y)}=\phi\left(\frac{y}{x}\right)
$

Solution of Homogeneous Differential Equation

This equation can be solved by the substitution $\mathrm{y}=\mathrm{vx}$.

$\begin{aligned}
& y = v x \\
& \Rightarrow \quad \frac{\mathrm{d}y}{\mathrm{dx}} = v + x \frac{\mathrm{d}v}{\mathrm{dx}}
\end{aligned}$

Thus, $\frac{d y}{d x}=\phi\left(\frac{y}{x}\right)$ transforms to
$
\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\phi(\mathrm{v})
$

$
\Rightarrow \quad \frac{d v}{\phi(v)-v}=\frac{d x}{x}
$

The variables have now been separated and the solution is
$
\int \frac{\mathrm{dv}}{\phi(\mathrm{v})-\mathrm{v}}=\ln \mathrm{x}+\mathrm{c}
$

After the integration $\mathrm{v}$ should be replaced by $\mathrm{y} / \mathrm{x}$ to get the required solution.

If the differential equation is of the form
$
\frac{d y}{d x}=\frac{a x+b y+c}{d x+e y+f}
$

It can be reduced to a homogeneous differential equation as follows:
Put $x=X+h, y=Y+k$
where $\mathrm{X}$ and $\mathrm{Y}$ are new variables and $\mathrm{h}$ and $\mathrm{k}$ are constants yet to be chosen
From (2)
$
d x=d X, d y=d Y
$

Equation (1), thus reduces to
$
\frac{d Y}{d X}=\frac{a(X+h)+b(Y+k)+c}{d(X+h)+e(Y+k)+f}=\frac{a X+b Y+(a h+b k+c)}{d X+e Y+(d h+e k+f)}
$

In order to have equation (3) as a homogeneous differential equation, choose $\mathrm{h}$ and $\mathrm{k}$ such that the following equations are satisfied :
$
\left.\begin{array}{rl}
a h+b k+c & =0 \\
d h+e k+f & =0
\end{array}\right\}
$

Now, (3) becomes
$
\frac{d Y}{d X}=\frac{a X+b Y}{d X+e Y}
$
which is a homogeneous differential equation and can be solved by putting $Y=v X$.

Separate the variables and integrate them to get the required solution.

What is a Non-Homogeneous Differential Equation?

Any differential equation which is not Homogenous is called a Non-Homogenous Differential Equation. A non-homogeneous differential equation of second order has the form:

$y^{\prime \prime}+a(t) y^{\prime}+b(t) y=c(t)$

Here, $y^{\prime \prime}$ denotes the second derivative of $y$, and $c(t)$ is a non-zero function of $t$. This equation can be converted to a homogeneous differential equation and the related DE is,

$y^{\prime \prime}+a(t) y^{\prime}+b(t) y=0$

This equation is also called the complementary equation to the given non-homogeneous differential equation.

Recommended Video Based on Homogeneous Differential Equations

Solved Examples Based On Homogeneous Differential Equations

Example 1: Which of the following is not a homogeneous function?

(1) $\delta(x, y)=x^3 y+x y^3$
(2) $\delta(x, y)=\sqrt{\frac{x^3}{y}}+\sqrt{\frac{y^3}{x}}$
(3) $\delta(x, y)=x^2 y-y x^3$
(4) $\delta(x, y)=x^2 y+y x^2$

Solution:

For option (1), we have:
$
\begin{aligned}
f(d x, d y) & =d^3 x^3 d y+d x \cdot d^3 y^3 \\
& =d^4\left(x^3 y+x y^3\right) \\
\Rightarrow f(d x, d y) & =d^4 \delta(x, y)
\end{aligned}
$

For option (2), we have:
$
\begin{aligned}
f(d x, d y) & =\sqrt{\frac{(d x)^3}{d y}}+\sqrt{\frac{(d y)^3}{d x}} \\
& =d \sqrt{\frac{x^3}{y}}+d \sqrt{\frac{y^3}{x}} \\
& =d f(x, y)
\end{aligned}
$

For option (3), we have:

$
\begin{aligned}
f(d x, d y) & =d^2 x^2 d y-d y \cdot d^3 x^3 \\
& =d^3 x^2 y-d^4 y x^3
\end{aligned}
$
$\Rightarrow f(d x, d y)$ can't be expressed as $d^4 f(x, y)$ here.
For option (3), we have: $f(d x, d y)=d^2 x^2 d y+d y d^2 x^2$
$
\begin{aligned}
& =d^3 x^2 y+d^3 y x^2 \\
& =d^3\left(x^2 y+y x^2\right) \\
\Rightarrow f(d x, d y)=d^3 f(x, y) &
\end{aligned}
$
$\therefore(1),(2),(4)$ are homogeneous, but $(3)$ is not.
Hence, the answer is the option (3).

Example 2: $\delta(x, y)=x^{1 / 2} y^{3 / 2}+x y$ is a homogeneous function of the degree:
Solution:
$
\begin{aligned}
& \delta(x, y)=x^{1 / 2} y^{3 / 2}+x y \\
& \begin{aligned}
f(d x, d y) & =(d x)^{1 / 2}(d y)^{3 / 2}+(d x)(d y) \\
& =d^2 x^{1 / 2} y^{3 / 2}+d^2 x y \\
\Rightarrow f(d x, d y) & =d^2\left(x^{1 / 2} y^{3 / 2}+x y\right) \\
& =d^2 f(x, y)
\end{aligned}
\end{aligned}
$

Thus, the Degree $=2$.
Hence, the answer is 2.

Example 3: The curve amongst the family of curves represented by the differential equation, $\left(x^2-y^2\right) d x+2 x y \quad d y=0$ which passes through $(1,1)$, is:

Solution:
$
\left(x^2-y^2\right) d x+2 x y d y=0
$

The D.E. can be written as:
$
\frac{d y}{d x}=\frac{y^2-x^2}{2 x y}
$

From the concept
$
\begin{aligned}
& \frac{y}{x}=v \\
& \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\
& \Rightarrow \int \frac{2 v}{v^2+1} d v=\int \frac{-d x}{x}
\end{aligned}
$

On Integrating, we get:
$
\begin{aligned}
& \ln \left(v^2+1\right)=-\ln (x)+C \\
& \left(y^2+x^2\right)=C x
\end{aligned}
$

Passes through $(1,1)$
$
\begin{aligned}
& \Rightarrow C=2 \\
& \Rightarrow y^2+x^2=2 x
\end{aligned}
$
which is an equation of the circle with a centre on the $\mathrm{X}$-axis.
Hence, the curve amongst the family of curves represented by the differential equation, $\left(x^2-y^2\right) d x+2 x y \quad d y=0$ which passes through $(1,1)$, is a circle with a centre on the $x$-axis.

Example 4: The solution of a differential equation $x \frac{\mathrm{d} y}{\mathrm{~d} x}=y(\ln y-\ln x+1)$ is
Solution:
The given equation can be written as:
$
\begin{aligned}
& x \frac{d y}{d x}=y \ln \left(\frac{y}{x}\right)+y \\
\Rightarrow & \frac{x d y-y d x}{d x}=y \ln \left(\frac{y}{x}\right) \\
\Rightarrow & x d y-y d x=y \ln \left(\frac{y}{x}\right) \cdot d x
\end{aligned}
$

Divide throughout by $x y$, we get:
$
\begin{aligned}
& \frac{x d y-y d x}{x y}=\frac{\ln \left(\frac{y}{x}\right)}{x} d x \\
\Rightarrow & \frac{d\left(\ln \left(\frac{y}{x}\right)\right)}{\ln \left(\frac{y}{x}\right)}=\frac{d x}{x}
\end{aligned}
$

Integrating it gives:
$
\begin{aligned}
& \Rightarrow \ln \left(\frac{\ln \left(\frac{y}{x}\right)}{x}\right)=c \\
& \Rightarrow \frac{\ln \left(\frac{y}{x}\right)}{x}=e^c \\
& \Rightarrow \ln \left(\frac{y}{x}\right)=c x \\
& \Rightarrow \frac{y}{x}=e^{C x} \\
& \Rightarrow y=x \cdot e^{c x}
\end{aligned}
$

Hence, the required answer is $y=x \cdot e^{c x}$

Example 5: The solution of the differential equation $\frac{d y}{d x}=-\left(\frac{x^2+3 y^2}{3 x^2+y^2}\right), y(1)=0$ is
Solution:
$
\begin{aligned}
& y=v x \\
& \frac{d y}{d x}=v+\frac{x d v}{d x} \\
& v+\frac{x d v}{d x}=-\frac{x^2+3 v^2 x^2}{3 x^2+v^2 x^2} \\
& x \frac{d v}{d x}=-\frac{1+3 v^2}{3+v^2}-v \\
& x \frac{d v}{d x}=-\frac{1+3 v^2+3 v+v^3}{3+v^2} \\
& \int \frac{3+v^2}{1+3 v^2+3 v+v^3} d v=-\int \frac{d x}{x} \\
& \Rightarrow \int \frac{3+v^2}{(1+v)^3} d v=-\ln x+C
\end{aligned}
$

Let $\quad v+1=t$
$
\begin{aligned}
& d v=d t \\
& \int \frac{3+(t-1)^2}{t^3} d t=-\ln x+C \\
& \Rightarrow \int \frac{t^2-2 t+4}{t^3} d t \\
& \Rightarrow \int\left(\frac{1}{t}-\frac{2}{t^2}+\frac{4}{t^3}\right) d t=-\ln x+C
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow \ln t+\frac{2}{t}-\frac{4}{2 t^2}=-\ln x+C \\
& \Rightarrow \ln \left(\frac{y}{x}+1\right)^{-1} \frac{2}{\frac{y}{x}+1}-\frac{4}{2(y / x+1)^2}=-\ln x+C \\
& \Rightarrow \ln \left(\frac{y+x}{x}\right)+\frac{2 x}{y+x}-\frac{2 x^2}{(x+y)^2}=-\ln x+C \\
& \Rightarrow \ln \left(\frac{y+x}{x}\right)+\frac{2 x}{y+x}-\frac{2 x^2}{(x+y)^2}=-\ln x+C \\
& \Rightarrow \ln |x+y|+\frac{2 x}{(x+y)^2}(x+y-x)=C \\
& \Rightarrow \ln |x+y|+\frac{2 x y}{(x+y)^2}=C \\
& \text { Hence, the answer required is } \log _e|x+y|+\frac{2 x y}{(x+y)^2}=0
\end{aligned}
$