Homogeneous Differential Equation

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}
$