In the previous section, we completed a discussion on how to solve first order first degree differential equations with variables separable. In this section, we will see homogeneous differential equations.
Some basics can be discussed with the help of some examples.
Example 1
This can be written in 3 steps:
1. Consider a function in x and y:
$\small{F_1(x,y)~=~y^2 + 2xy}$
2. Let us replace $\small{x}$ by $\small{\lambda x}$ and $\small{y}$ by $\small{\lambda y}$.
♦ Where $\small{\lambda}$ is a nonzero constant.
• We get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{F_1(x,y)} & {~=~} &{y^2 + 2xy} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{F_1(\lambda x,\lambda y)} & {~=~} &{(\lambda y)^2 + 2(\lambda x)(\lambda y)} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{\lambda^2 y^2 + 2\lambda^2 x y} \\
{~\color{magenta} 4 } &{} &{} & {~=~} &{\lambda^2 \left(y^2 + 2 x y \right)} \\
{~\color{magenta} 5 } &{} &{} & {~=~} &{\lambda^2 \,F_1(x,y)} \\
\end{array}}$
3. We see that:
For the function $\small{F_1}$, it is possible to write:
$\small{F(\lambda x,\lambda y)~=~\lambda^n F(x,y)}$
♦ Where $\small{n}$ is a natural number.
For this example, $\small{n~=~2}$
Example 2
This can be written in 3 steps:
1. Consider a function in x and y:
$\small{F_2(x,y)~=~2x - 3y}$
2. Let us replace $\small{x}$ by $\small{\lambda x}$ and $\small{y}$ by $\small{\lambda y}$.
♦ Where $\small{\lambda}$ is a nonzero constant.
• We get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{F_2(x,y)} & {~=~} &{2x - 3y} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{F_2(\lambda x,\lambda y)} & {~=~} &{2(\lambda x) - 3(\lambda y)} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{\lambda \left(2x - 3y \right)} \\
{~\color{magenta} 4 } &{} &{} & {~=~} &{\lambda \,F_2(x,y)} \\
\end{array}}$
3. We see that:
For the function $\small{F_2}$, it is possible to write:
$\small{F(\lambda x,\lambda y)~=~\lambda^n F(x,y)}$
♦ Where $\small{n}$ is a natural number.
For this example, $\small{n~=~1}$
Example 3
This can be written in 3 steps:
1. Consider a function in x and y:
$\small{F_3(x,y)~=~\cos\left(\frac{y}{x} \right)}$
2. Let us replace $\small{x}$ by $\small{\lambda x}$ and $\small{y}$ by $\small{\lambda y}$.
♦ Where $\small{\lambda}$ is a nonzero constant.
• We get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{F_3(x,y)} & {~=~} &{\cos\left(\frac{ y}{x} \right)} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{F_3(\lambda x,\lambda y)} & {~=~} &{\cos\left(\frac{\lambda y}{\lambda x} \right)} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{\cos\left(\frac{y}{x} \right)} \\
{~\color{magenta} 4 } &{} &{} & {~=~} &{\lambda^0 \,F_3(x,y)} \\
\end{array}}$
3. We see that:
For the function $\small{F_3}$, it is possible to write:
$\small{F(\lambda x,\lambda y)~=~\lambda^n F(x,y)}$
♦ Where $\small{n}$ is a natural number.
For this example, $\small{n~=~0}$
Example 4
This can be written in 5 steps:
1. Consider a function in x and y:
$\small{F_4(x,y)~=~\sin x~+~\cos y}$
2. Let us replace $\small{x}$ by $\small{\lambda x}$ and $\small{y}$ by $\small{\lambda y}$.
♦ Where $\small{\lambda}$ is a nonzero constant.
• We get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{F_4(x,y)} & {~=~} &{\sin x~+~\cos y} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{F_4(\lambda x,\lambda y)} & {~=~} &{\sin(\lambda x)~+~\cos(\lambda y)} \\
\end{array}}$
3. Let $\small{\lambda = 3}$ and $\small{x = \frac{\pi}{6}}$
(a) $\small{\sin(x)~=~\sin \left(\frac{\pi}{6} \right)~=~\frac{1}{2}}$
• $\small{\sin(\lambda x)~=~\sin \left(\frac{3 \pi}{6} \right)~=~\sin \left(\frac{\pi}{2} \right)~=~1}$
(b) $\small{\cos(x)~=~\cos \left(\frac{\pi}{6} \right)~=~\frac{\sqrt 3}{2}}$
• $\small{\cos(\lambda x)~=~\cos \left(\frac{3 \pi}{6} \right)~=~\cos \left(\frac{\pi}{2} \right)~=~0}$
4. We see that,
• $\small{\sin(x)}$ need not be equal to $\small{\sin(\lambda x) }$
• $\small{\cos(x)}$ need not be equal to $\small{\cos(\lambda x) }$
5. So for the function $\small{F_4}$, we must write:
$\small{F(\lambda x,\lambda y)~\ne~\lambda^n F(x,y)}$
♦ Where $\small{n}$ is a natural number.
Based on the above four examples, we can write the definition:
•
A function $\small{F(x,y)}$ is said to be a homogeneous function of degree n if
$\small{F(x,y)~=~\lambda^n F(x,y)}$ for any non zero constant $\small{\lambda}$
In the four examples that we saw above,
♦ $\small{F_1}$ is a homogeneous function of degree 2
♦ $\small{F_2}$ is a homogeneous function of degree 1
♦ $\small{F_3}$ is a homogeneous function of degree 0
♦ $\small{F_4}$ is not a homogeneous function
Consider Example 1 again.
We can rearrange $\small{F_1(x,y)}$ as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{F_1(x,y)} & {~=~} &{y^2 + 2xy} \\
{~\color{magenta} 2 } &{} &{F_1(x,y)} & {~=~} &{x^2\left[\frac{y^2}{x^2} + \frac{2y}{x} \right]} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{x^2\left[\left(\frac{y}{x} \right)^2 + 2 \left(\frac{y}{x} \right) \right]} \\
{~\color{magenta} 4 } &{} &{} & {~=~} &{x^2\left[g_1\left(\frac{y}{x} \right) \right]} \\
\end{array}}$
Here, $\boldsymbol{g_1\left(\frac{y}{x} \right)~=~\left(\frac{y}{x} \right)^2 + 2 \left(\frac{y}{x} \right)}$
• We can rearrange $\small{F_1(x,y)}$ in another way also:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{F_1(x,y)} & {~=~} &{y^2 + 2xy} \\
{~\color{magenta} 2 } &{} &{F_1(x,y)} & {~=~} &{y^2\left[1 + \frac{2x}{y} \right]} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{y^2\left[1 + 2 \left(\frac{x}{y} \right) \right]} \\
{~\color{magenta} 4 } &{} &{} & {~=~} &{y^2\left[h_1\left(\frac{x}{y} \right) \right]} \\
\end{array}}$
Here, $\boldsymbol{h_1\left(\frac{x}{y} \right)~=~1 + 2 \left(\frac{x}{y} \right)}$
Consider Example 2 again.
We can rearrange $\small{F_2(x,y)}$ as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{F_2(x,y)} & {~=~} &{2x - 3y} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{F_2( x, y)} & {~=~} &{x^1 \left[2~-~\frac{3y}{x} \right]} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{x^1 \left[2~-~3 \left(\frac{y}{x} \right) \right]} \\
{~\color{magenta} 4 } &{} &{} & {~=~} &{x^1 \left[h_2\left(\frac{y}{x} \right) \right]} \\
\end{array}}$
Here, $\boldsymbol{h_2\left(\frac{y}{x} \right)~=~2~-~3 \left(\frac{y}{x} \right)}$
• We can rearrange $\small{F_2(x,y)}$ in another way also:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{F_2(x,y)} & {~=~} &{2x - 3y} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{F_2(x,y)} & {~=~} &{y^1 \left[\frac{2x}{y}~-~3 \right]} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{y^1 \left[2 \left(\frac{x}{y} \right)~-~3 \right]} \\
{~\color{magenta} 4 } &{} &{} & {~=~} &{y^1\left[g_2\left(\frac{x}{y} \right) \right]} \\
\end{array}}$
Here, $\boldsymbol{g_2\left(\frac{y}{x} \right)~=~2 \left(\frac{x}{y} \right)~-~3}$
Consider Example 3 again.
We can rearrange $\small{F_3(x,y)}$ as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{F_3(x,y)} & {~=~} &{\cos\left(\frac{ y}{x} \right)} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{F_3(x,y)} & {~=~} &{x^0 \left[\cos\left(\frac{y}{x} \right) \right]} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{x^0\left[h_3\left(\frac{y}{x} \right) \right]} \\
\end{array}}$
Here, $\boldsymbol{h_3\left(\frac{y}{x} \right)~=~\cos\left(\frac{y}{x} \right)}$
Consider Example 4 again.
Let us try to rearrange $\small{F_4(x,y)}$:
$\small{F_4(x,y)~=~\sin x~+~\cos y}$
• Suppose that, n = 3.
$\small{F_4(x,y)~\ne~x^3 \left[\sin \left(\frac{y}{x} \right)~+~\cos \left(\frac{y}{x} \right) \right]}$
$\small{F_4(x,y)~\ne~y^3 \left[\sin \left(\frac{x}{y} \right)~+~\cos \left(\frac{x}{y} \right) \right]}$
• So we can write:
$\small{F_4(x,y)~\ne~x^n \left[g_4 \left(\frac{y}{x} \right) \right]}$
$\small{F_4(x,y)~\ne~y^n \left[h_4 \left(\frac{x}{y} \right) \right]}$
Based on the above examples, we can write:
• A function $\small{F(x,y)}$ is said to be a homogeneous function of degree n if any one of the following two conditions is satisfied:
Condition I:
$\small{F(x,y)~=~x^n g \left(\frac{y}{x} \right)}$
Condition II:
$\small{F(x,y)~=~y^n h \left(\frac{x}{y} \right)}$
So now we have two methods to check whether a given $\small{F(x,y)}$ is a homogeneous function of degree n. Let us see some solved examples:
Solved example 25.43
Show that $\boldsymbol{F(x,y) = \frac{x+2y}{x-y}}$ is a homogeneous function and write the degree.
Solution:
1. We are given:
$\small{F(x,y) = \frac{x+2y}{x-y}}$
2. Let us replace $\small{x}$ by $\small{\lambda x}$ and $\small{y}$ by $\small{\lambda y}$.
♦ Where $\small{\lambda}$ is a nonzero constant.
• We get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{F(x,y)} & {~=~} &{\frac{x+2y}{x-y}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{F(\lambda x,\lambda y)} & {~=~} &{\frac{\lambda x+2(\lambda y)}{\lambda x - \lambda y}} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{\frac{\lambda(x+2y)}{\lambda(x-y)}} \\
{~\color{magenta} 4 } &{} &{} & {~=~} &{\frac{x+2y}{x-y}} \\
{~\color{magenta} 5 } &{} &{} & {~=~} &{\lambda^0 \,F(x,y)} \\
\end{array}}$
3. We see that:
For the function $\small{F(x,y)}$, it is possible to write:
$\small{F(\lambda x,\lambda y)~=~\lambda^n F(x,y)}$
♦ Where $\small{n}$ is a natural number.
•
So it is a homogeneous function.
• For this function, $\small{n~=~0}$. So degree is zero.
Solved example 25.44
Show that $\boldsymbol{F(x,y) = \frac{x^2 + y^2}{2xy}}$ is a homogeneous function and write the degree.
Solution:
1. We can rearrange $\small{F(x,y)}$ as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{F(x,y)} & {~=~} &{\frac{x^2 + y^2}{2xy}} \\
{~\color{magenta} 2 } &{\Rightarrow} &{F(x,y)} & {~=~} &{x^2\left[\frac{1 ~+~ \frac{y^2}{x^2}}{2xy} \right]} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{\left[\frac{x^2}{x^2} \right]\left[\frac{1 ~+~ \frac{y^2}{x^2}}{\frac{2xy}{x^2}} \right]} \\
{~\color{magenta} 4 } &{} &{} & {~=~} &{\left[\frac{x^2}{x^2} \right]\left[\frac{1 ~+~ \frac{y^2}{x^2}}{\frac{2y}{x}} \right]} \\
{~\color{magenta} 5 } &{} &{} & {~=~} &{x^0\left[\frac{1 ~+~ \left(\frac{y}{x} \right)^2}{2\left(\frac{y}{x} \right)} \right]} \\
{~\color{magenta} 6 } &{} &{} & {~=~} &{x^0\left[g\left(\frac{y}{x} \right) \right]} \\
\end{array}}$
Here, $\boldsymbol{g\left(\frac{y}{x} \right)~=~\frac{1 ~+~ \left(\frac{y}{x} \right)^2}{2\left(\frac{y}{x} \right)}}$
2. We see that:
For the function $\small{F(x,y)}$, it is possible to write:
$\small{F(x,y)~=~x^n\left[g \left(\frac{y}{x} \right) \right]}$
♦ Where $\small{n}$ is a natural number.
• So it is a homogeneous function.
• For this function, $\small{n~=~0}$. So degree is zero.
We have defined homogeneous function of degree n. Now we can define homogeneous differential equation. It can be written in two steps.
1. Consider the differential equation:
$\boldsymbol{\frac{dy}{dx}~=~F(x,y)}$
2. If $\small{F(x,y)}$ is a homogeneous function of degree zero, then the differential equation in (1) is said to be a homogeneous differential equation.
In the next section, we will see the method to solve homogeneous differential equations.
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