Homogeneous Differential Equation - Practice Questions & MCQ

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 x and 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. $\mathrm{f}(\mathrm{x}, \mathrm{y})=\sin \mathrm{x}+\cos \mathrm{y}$, then $\mathrm{f}(\lambda \mathrm{x}, \lambda \mathrm{y}) \neq \lambda^{\mathrm{n}} \mathrm{f}(\mathrm{x}, \mathrm{y})$

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. $^2$.
Here, examples 1, 2, and 3 are homogeneous equations of degree 3,1 and 0 respectively and example 4 is not a homogeneous function.

We can define homogeneous differential equation as follows:
Any DE 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 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)
$

This equation can be solved by the substitution $\mathrm{y}=\mathrm{vx}$.
$
\Rightarrow \quad \begin{aligned}
\mathrm{y} & =\mathrm{vx} \\
\Rightarrow \quad \frac{\mathrm{dy}}{\mathrm{dx}} & =\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}
\end{aligned}
$

Thus, $\frac{d y}{d x}=\phi\left(\frac{y}{x}\right)$ transforms to
$
\begin{aligned}
\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}} & =\phi(\mathrm{v}) \\
\Rightarrow \quad \frac{d v}{\phi(v)-v} & =\frac{d x}{x}
\end{aligned}
$

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 v should be replaced by $y / x$ 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 X and Y are new variables and h and 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 h and 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 $\mathrm{Y}=\mathrm{vX}$.

Note:
If $\mathrm{d} / \mathrm{a}=\mathrm{e} / \mathrm{b}$ (=t say), the above method does not apply.
In such cases, Equation (1) becomes
$
\frac{d y}{d x}=\frac{a x+b y+c}{t(a x+b y)+f}
$

which can be solved by putting $a x+b y=v$

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