25.9 - Solved Examples on Solution of Homogeneous Differential Equations

In the previous section, we saw the method to solve homogeneous differential equations. We also saw some solved examples. In this section, we will see a few more solved examples. 

Solved example 25.52
Show that the differential equation $\boldsymbol{x \cos\left(\frac{y}{x} \right) \,\frac{dy}{dx}~=~y \cos \left(\frac{y}{x} \right)~+~x}$ is homogeneous and solve it.
Solution:
Part I: Showing that the given differential equation is homogeneous
1. We can rearrange the given differential equation as:
$\small{\frac{dy}{dx}~=~\frac{y \cos \left(\frac{y}{x} \right)~+~x}{x \cos\left(\frac{y}{x} \right)}}$

2. Let $\small{F(x,y)~=~\frac{y \cos \left(\frac{y}{x} \right)~+~x}{x \cos\left(\frac{y}{x} \right)}}$
• We need to show that, this is a homogeneous function of degree zero.

3. Dividing both numerator and denominator by x, we get:

$\small{F(x,y)~=~\frac{\left(\frac{y}{x} \right) \cos \left(\frac{y}{x} \right)~+~1}{\cos\left(\frac{y}{x} \right)}~=~x^0\left[g\left(\frac{y}{x} \right) \right]}$

• Here, $\boldsymbol{g\left(\frac{y}{x} \right) ~=~\frac{\left(\frac{y}{x} \right) \cos \left(\frac{y}{x} \right)~+~1}{\cos\left(\frac{y}{x} \right)}}$

4. 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 n is a natural number.
• So it is a homogeneous function.

• For this function, n = 0. So degree is zero.
• So the given differential equation is homogeneous.

Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

2. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

3. Substituting (1) and (2) in the given differential equation, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{\left(\frac{y}{x} \right) \cos \left(\frac{y}{x} \right)~+~1}{\cos\left(\frac{y}{x} \right)}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{\frac{\left(v \right) \cos \left(v \right)~+~1}{\cos\left(v \right)}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{v \cos \left(v \right)~+~1}{\cos\left(v \right)}~-~v}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{v \cos \left(v \right)~+~1~-~v \cos\left(v \right)}{\cos\left(v \right)}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{1}{\cos v}}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{\left[\cos v \right]dv}    & {~=~}    &{\frac{dx}{x}}    \\
\end{array}}$

4. So we have separated the variables. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\int{\left[\cos v \right]dv}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}    &{\sin v~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 3    }&{{\Rightarrow}}    &{\sin v}    & {~=~}&{\log \left|x \right|~+~\rm{C}_3}    \\
\end{array}}$

5. Replacing 'v', we will get the general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\sin v}    & {~=~}    &{\log \left|x \right|~+~\rm{C}_3}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{\sin\left(\frac{y}{x} \right)}    & {~=~}    &{\log \left|x \right|~+~\log \left|\rm{C} \right|}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{\sin\left(\frac{y}{x} \right)}    & {~=~}    &{\log \left|\rm{C}\,x \right|}    \\
\end{array}}$

Solved example 25.53
Show that the differential equation $\boldsymbol{x \, \frac{dy}{dx}~-~y~+~x \sin \left(\frac{y}{x} \right)~=~0}$ is homogeneous and solve it.
Solution:
Part I: Showing that the given differential equation is homogeneous

1. We can rearrange the given differential equation as:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x \, \frac{dy}{dx}~-~y~+~x \sin \left(\frac{y}{x} \right)}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{x \, \frac{dy}{dx}}    & {~=~}    &{y~-~x \sin \left(\frac{y}{x} \right)}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{ \frac{dy}{dx}}    & {~=~}    &{\frac{y~-~x \sin \left(\frac{y}{x} \right)}{x}}    \\
\end{array}}$

2. Let $\small{F(x,y)~=~\frac{y~-~x \sin \left(\frac{y}{x} \right)}{x}}$
• We need to show that, this is a homogeneous function of degree zero.

3. Dividing both numerator and denominator by x, we get:

$\small{F(x,y)~=~\frac{\left(\frac{y}{x} \right)~-~ \sin \left(\frac{y}{x} \right)}{1}~=~x^0\left[g\left(\frac{y}{x} \right) \right]}$

• Here, $\boldsymbol{g\left(\frac{y}{x} \right) ~=~\left(\frac{y}{x} \right)~-~ \sin \left(\frac{y}{x} \right)}$

4. 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 n is a natural number.
• So it is a homogeneous function.

• For this function, n = 0. So degree is zero.
• So the given differential equation is homogeneous.

Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

2. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

3. Substituting (1) and (2) in the given differential equation, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\left(\frac{y}{x} \right)~-~ \sin \left(\frac{y}{x} \right)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{v~-~\sin v}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{-\sin v}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left[\frac{1}{\sin v} \right]dv}    & {~=~}    &{-\frac{dx}{x}}    \\
\end{array}}$

4. So we have separated the variables. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\int{\left[\frac{1}{\sin v} \right]dv}}    & {~=~}    &{(-1)\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}    &{\log \left|\csc v~-~\cot v \right|~+~\rm{C}_1}    & {~=~}&{(-1)\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 3    }&{{\Rightarrow}}    &{\log \left|\csc v~-~\cot v \right|~+~\log \left|x \right|}    & {~=~}&{\rm{C}_2~-~\rm{C}_1}    \\
{~\color{magenta} 4    }&{{\Rightarrow}}    &{\log \left|\csc v~-~\cot v \right|~+~\log \left|x \right|}    & {~=~}&{\rm{C}_3}    \\
{~\color{magenta} 5    }&{{\Rightarrow}}    &{\log \left|(\csc v~-~\cot v)x  \right|}    & {~=~}&{\rm{C}_3}    \\
{~\color{magenta} 6    }&{{\Rightarrow}}    &{(\csc v~-~\cot v)x}    & {~=~}&{e^{\rm{C}_3}~=~\rm{C}}    \\
{~\color{magenta} 7    }&{{\Rightarrow}}    &{\frac{x(1~-~\cos v)}{\sin v}}    & {~=~}&{e^{\rm{C}_3}~=~\rm{C}}    \\
\end{array}}$

5. Replacing 'v', we will get the general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\frac{x(1~-~\cos v)}{\sin v}}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{\frac{x\left[1~-~\cos \left(\frac{y}{x} \right) \right]}{\sin \left(\frac{y}{x} \right)}}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{x\left[1~-~\cos \left(\frac{y}{x} \right) \right]}    & {~=~}    &{\rm{C}\sin\left(\frac{y}{x} \right)}    \\
\end{array}}$

Solved example 25.54
Show that the differential equation $\boldsymbol{2y e^{x/y} dx~+~\left(y\,-\,2x e^{x/y} \right)dy~=~0}$ is homogeneous and find it's particular solution, given that, x = 0 when y = 1.
Solution:
Part I: Showing that the given differential equation is homogeneous

1. We can rearrange the given differential equation as:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2y e^{x/y} dx~+~\left(y\,-\,2x e^{x/y} \right)dy}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{2y e^{x/y} dx}    & {~=~}    &{(-1)\left(y\,-\,2x e^{x/y} \right)dy}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{2y e^{x/y} dx}    & {~=~}    &{\left(2x e^{x/y}\,-\,y \right)dy}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{dx}{dy}}    & {~=~}    &{\frac{2x e^{x/y}\,-\,y }{2y e^{x/y}}}    \\
\end{array}}$

• Here we write $\small{\frac{dx}{dy}}$ instead of $\small{\frac{dy}{dx}}$ because, the given differential equation involves $\small{x/y}$.

2. Let $\small{F(x,y)~=~\frac{2x e^{x/y}\,-\,y }{2y e^{x/y}}}$
• We need to show that, this is a homogeneous function of degree zero.

3. Dividing both numerator and denominator by y, we get:

$\small{F(x,y)~=~\frac{2(x/y) e^{x/y}\,-\,1 }{2 e^{x/y}}~=~y^0\left[h\left(\frac{x}{y} \right) \right]}$

• Here, $\boldsymbol{h\left(\frac{x}{y} \right) ~=~\frac{2(x/y) e^{x/y}\,-\,1 }{2 e^{x/y}}}$

4. We see that:
For the function $\small{F(x,y)}$, it is possible to write:
$\small{F(x,y)~=~y^n\left[h\left(\frac{x}{y} \right) \right]}$
   ♦ Where n is a natural number.
• So it is a homogeneous function.

• For this function, n = 0. So degree is zero.
• So the given differential equation is homogeneous.

Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{x}{y}~=~u}$
Which is same as: $\small{x~=~uy}$

2. Differentiating the above equation with respect to y, we get:
$\small{\frac{dx}{dy}~=~u~+~y \frac{du}{dy}}$

3. Substituting (1) and (2) in the given differential equation, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dx}{dy}}    & {~=~}    &{\frac{2(x/y) e^{x/y}\,-\,1 }{2 e^{x/y}}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{u~+~y \frac{du}{dy}}    & {~=~}    &{\frac{2(u) e^{u}\,-\,1 }{2 e^{u}}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{u~+~y \frac{du}{dy}}    & {~=~}    &{u~-~\frac{1 }{2 e^{u}}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{y \frac{du}{dy}}    & {~=~}    &{-\frac{1 }{2 e^{u}}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{2 e^u\,du}    & {~=~}    &{-\frac{1 }{y}dy}    \\
\end{array}}$

4. So we have separated the variables. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\int{\left[2 e^u \right]du}}    & {~=~}    &{(-1)\int{\left[\frac{1}{y} \right]dy}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}    &{2 e^u~+~\rm{C}_1}    & {~=~}&{(-1)\log \left|y \right|~+~\rm{C}_2}    \\
{~\color{magenta} 3    }&{{\Rightarrow}}    &{2 e^u~+~\log \left|y \right|}    & {~=~}&{\rm{C}_2~-~\rm{C}_1}    \\
{~\color{magenta} 4    }&{{\Rightarrow}}    &{2 e^u~+~\log \left|y \right|}    & {~=~}&{\rm{C}}    \\
\end{array}}$

5. Replacing 'u', we will get the general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{2 e^u~+~\log \left|y \right|}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{2 e^{(x/y)}~+~\log \left|y \right|}    & {~=~}    &{\rm{C}}    \\
\end{array}}$

Part III: To find the particular solution
1. Given that, x = 0 when y = 1.

2. Substituting in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{2 e^{(x/y)}~+~\log \left|y \right|}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{2 e^{(0/1)}~+~\log \left(1 \right)}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{2 e^{(0)}~+~0}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}&{2 (1)~+~0}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    5    }&{{\Rightarrow}}&{2}    & {~=~}    &{\rm{C}}    \\
\end{array}}$

3. So based on the result from part II, we get the particular solution as:

$\boldsymbol{2 e^{(x/y)}~+~\log \left|y \right|~=~2}$

Solved example 25.55
Show that the family of curves for which the slope of the tangent at any point (x,y) on it is $\boldsymbol{\frac{x^2 + y^2}{2xy}}$, is  given by $\boldsymbol{x^2 ~-~y^2~=~cx}$.
Solution:
Part I: Finding the differential equation
1. We know that, the slope is same as the derivative $\small{\frac{dy}{dx}}$.

2. So based on the given information, we can write:
$\small{\frac{dy}{dx}~=~\frac{x^2 + y^2}{2xy}}$
• This is the required differential equation.

Part II: Showing that the differential equation is homogeneous

1. We have the  differential equation:
$\small{\frac{dy}{dx}~=~\frac{x^2 + y^2}{2xy}}$

2. Let $\small{F(x,y)~=~\frac{x^2 + y^2}{2xy}}$
• We need to show that, this is a homogeneous function of degree zero.

3. Dividing both numerator and denominator by $\small{x^2}$, we get:

$\small{F(x,y)~=~\frac{1 + \frac{y^2}{x^2}}{2\left(\frac{y}{x} \right)}~=~x^0\left[g\left(\frac{y}{x} \right) \right]}$

• Here, $\boldsymbol{g\left(\frac{y}{x} \right) ~=~\frac{1 + \left(\frac{y}{x} \right)^2}{2\left(\frac{y}{x} \right)}}$

4. 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 n is a natural number.
• So it is a homogeneous function.

• For this function, n = 0. So degree is zero.

• So the given differential equation is homogeneous.

Part III: Solving the differential equation
1. We make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

2. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

3. Substituting (1) and (2) in the given differential equation, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{1 + \left(\frac{y}{x} \right)^2}{2\left(\frac{y}{x} \right)}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{\frac{1 + \left(v \right)^2}{2\left(v \right)}~=~\frac{1}{2v}~+~\frac{v}{2}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{1}{2v}~-~\frac{v}{2}~=~\frac{1}{2}\left[\frac{1}{v}~-~v \right]}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{ \frac{dv}{\frac{1}{v}~-~v }}    & {~=~}    &{\frac{dx}{2x}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{\left[ \frac{1}{\frac{1}{v}~-~v } \right]dv}    & {~=~}    &{\left[ \frac{1}{2x } \right]dx}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{\left[\frac{v}{1~-~v^2 } \right]dv}    & {~=~}    &{\left[ \frac{1}{2x } \right]dx}    \\
\end{array}}$

4. So we have separated the variables. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\int{\left[\frac{v}{1~-~v^2 } \right]dv}}    & {~=~}    &{\int{\left[\frac{1}{2x} \right]dx}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}    &{(-1)\left(\frac{1}{2} \right)\log \left|(1 - v^2) \right|~+~\rm{C}_1}    & {~=~}&{\left(\frac{1}{2} \right)\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}    &{(-1)\log \left|(1 - v^2) \right|~+~2\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~2\rm{C}_2}    \\
{~\color{magenta} 4    }&{{\Rightarrow}}    &{\log \left|x \right|~+~\log \left|(1 - v^2) \right|}    & {~=~}&{2\rm{C}_1~-~2\rm{C}_2}    \\
{~\color{magenta} 5    }&{{\Rightarrow}}    &{\log \left|x(1 - v^2) \right|}    & {~=~}&{\rm{C}_3}    \\
{~\color{magenta} 6    }&{{\Rightarrow}}    &{\log \left|x(1 - v^2) \right|}    & {~=~}&{\log \left|\rm{C}_4 \right|}    \\
{~\color{magenta} 7    }&{{\Rightarrow}}    &{x(1 - v^2)}    & {~=~}&{\pm\rm{C}_4}    \\
\end{array}}$

5. Replacing 'v', we will get the general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{x(1 - v^2)}    & {~=~}    &{\pm \rm{C}_4}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{x\left(1 - \frac{y^2}{x^2}\right) }    & {~=~}    &{\pm \rm{C}_4}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{\frac{x^2~-~y^2}{x}}    & {~=~}    &{\pm \rm{C}_4}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}&{x^2~-~y^2}    & {~=~}    &{{\rm{C}}\,x}    \\
\end{array}}$

---

The link below gives a few more solved examples:

Exercise 9.5

---

In the next section, we will see linear differential equations.

Previous

Contents

Next 

Copyright©2025 Higher secondary mathematics.blogspot.com

 

25.8 - Solution of Homogeneous Differential Equations

In the previous section, we saw homogeneous functions and their degrees. We also saw homogeneous differential equation. In this section, we will see the method to solve such differential equations.

The method can be written in 5 steps:

1. We want to solve the homogeneous differential equation

$\boldsymbol{\frac{dy}{dx}~=~F(x,y)}$

• Since it is a homogeneous differential equation, we can write: $\small{\frac{dy}{dx}~=~F(x,y)~=~x^0\left[g\left(\frac{y}{x} \right) \right]~=~g\left(\frac{y}{x} \right)}$

• In short, we can write: $\small{\frac{dy}{dx}~=~g\left(\frac{y}{x} \right)}$

• Note that, if we are given the homogeneous differential equation $\boldsymbol{\frac{dx}{dy}~=~F(x,y)}$, we must write:
$\small{\frac{dx}{dy}~=~h\left(\frac{x}{y} \right)}$

2. Next, we make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

3. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

4. Substituting (2) and (3) in (1), we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{g(v)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{g(v)~-~v}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{ \frac{dv}{g(v)~-~v}}    & {~=~}    &{\frac{dx}{x}}    \\
\end{array}}$

5. So we have separated the variables. Integrating both sides, we will get the general solution. In that general solution, we must substitute $\small{\frac{y}{x}~~\text{in the place of}~~v}$

Let us see some solved examples:

Solved example 25.45
Show that the differential equation $\boldsymbol{y'~=~\frac{x+y}{x}}$ is homogeneous and solve it.
Solution:
Part I: Showing that the given differential equation is homogeneous
1. We can rearrange the given differential equation as:
$\small{\frac{dy}{dx}~=~\frac{x+y}{x}}$

2. Let $\small{F(x,y)~=~\frac{x+y}{x}}$
• We need to show that, this is a homogeneous function of degree zero.

(i) We can rearrange $\small{F(x,y)}$ as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{F(x,y)}    & {~=~}    &{\frac{x+y}{x}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{F(x,y)}    & {~=~}    &{x\left[\frac{1 ~+~ \frac{y}{x}}{x} \right]}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}        &{\left[\frac{x}{x} \right]\left[1 ~+~ \frac{y}{x} \right]}    \\
{~\color{magenta}    4    }    &{}    &{}    & {~=~}        &{\left[1 \right]\left[1 ~+~ \frac{y}{x} \right]}    \\
{~\color{magenta}    5    }    &{}    &{}    & {~=~}        &{x^0\left[1 ~+~ \frac{y}{x} \right]}    \\
{~\color{magenta}    6    }    &{}    &{}    & {~=~}    &{x^0\left[g\left(\frac{y}{x} \right) \right]}    \\
\end{array}}$

Here, $\boldsymbol{g\left(\frac{y}{x} \right)~=~1 ~+~ \frac{y}{x}}$

(ii) 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.
• Therefore, the given differential equation is homogeneous.

Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

2. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

3. Substituting (1) and (2) in the given differential equation, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{x+y}{x}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{\frac{x+(vx)}{x}~=~1+v}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{1+v~-~v}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{1}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{\left[1 \right]dv}    & {~=~}    &{\left[\frac{1}{x} \right]dx}    \\
\end{array}}$

4. So we have separated the variables. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\int{\left[1 \right]dv}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}    &{v~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 3    }&{{\Rightarrow}}    &{v}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2~-~\rm{C}_1}    \\
{~\color{magenta} 4    }&{{\Rightarrow}}    &{v}    & {~=~}&{\log \left|x \right|~+~\rm{C}_3}    \\
\end{array}}$

5. Replacing 'v', we will get the general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{v}    & {~=~}    &{\log \left|x \right|~+~\rm{C}_3}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{\frac{y}{x}}    & {~=~}    &{\log \left|x \right|~+~\rm{C}_3}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{y}    & {~=~}    &{x\log \left|x \right|~+~x \rm{C}_3}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}&{y}    & {~=~}    &{x\log \left|x \right|~+~{\rm{C}}\,x}    \\
\end{array}}$

Solved example 25.46
Show that the differential equation $\boldsymbol{(x-y)dy~-~(x+y)dx~=~0}$ is homogeneous and solve it.
Solution:
Part I: Showing that the given differential equation is homogeneous

1. We can rearrange the given differential equation as:
$\small{\frac{dy}{dx}~=~\frac{x+y}{x-y}}$

2. Let $\small{F(x,y)~=~\frac{x+y}{x-y}}$
• We need to show that, this is a homogeneous function of degree zero.

(i) We can rearrange $\small{F(x,y)}$ as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{F(x,y)}    & {~=~}    &{\frac{x+y}{x-y}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{F(x,y)}    & {~=~}    &{\left[\frac{x}{x} \right]\left[\frac{1 ~+~ \frac{y}{x}}{1 ~-~ \frac{y}{x}} \right]}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}        &{\left[1 \right]\left[\frac{1 ~+~ \frac{y}{x}}{1 ~-~ \frac{y}{x}} \right]}    \\
{~\color{magenta}    4    }    &{}    &{}    & {~=~}        &{x^0\left[\frac{1 ~+~ \frac{y}{x}}{1 ~-~ \frac{y}{x}} \right]}    \\
{~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{x^0\left[g\left(\frac{y}{x} \right) \right]}    \\
\end{array}}$

Here, $\boldsymbol{g\left(\frac{y}{x} \right)~=~\frac{1 ~+~ \frac{y}{x}}{1 ~-~ \frac{y}{x}}}$

(ii) 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.
• Therefore, the given differential equation is homogeneous.

Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

2. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

3. Substituting (1) and (2) in the given differential equation, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{1 ~+~ \frac{y}{x}}{1 ~-~ \frac{y}{x}}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{\frac{1+v}{1-v}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{1+v}{1-v}~-~v~=~\frac{1+v-v+v^2}{1-v}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{1+v^2}{1-v}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{\left[\frac{1-v}{1+v^2} \right]dv}    & {~=~}    &{\left[\frac{1}{x} \right]dx}    \\
\end{array}}$

4. So we have separated the variables. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\int{\left[\frac{1-v}{1+v^2} \right]dv}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}    &{\int{\left[\frac{1}{1+v^2} \right]dv}~-~\int{\left[\frac{v}{1+v^2} \right]dv}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}    &{\tan^{-1}v~-~\frac{1}{2}\log \left|1+v^2 \right|~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 4    }&{{\Rightarrow}}    &{\tan^{-1}v~-~\frac{1}{2}\log \left|1+v^2 \right|}    & {~=~}&{\log \left|x \right|~+~\rm{C}_3}    \\
{~\color{magenta} 5    }&{{\Rightarrow}}    &{2 \tan^{-1}v~-~\log \left|1+v^2 \right|}    & {~=~}&{2\log \left|x \right|~+~2\rm{C}_3}    \\
{~\color{magenta} 6    }&{{\Rightarrow}}    &{2 \tan^{-1}v}    & {~=~}&{\log \left|1+v^2 \right|~+~2\log \left|x \right|~+~\rm{C}_4}    \\
{~\color{magenta} 7    }&{{\Rightarrow}}    &{2 \tan^{-1}v}    & {~=~}&{\log \left|\left(1+v^2 \right)x^2 \right|~+~\rm{C}_4}    \\
\end{array}}$

• The reader may write all steps related to the integration in [(2) magenta color]

5. Replacing 'v', we will get the general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{2 \tan^{-1}v}    & {~=~}    &{\log \left|\left(1+v^2 \right)x^2 \right|~+~\rm{C}_4}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{2 \tan^{-1}\left(\frac{y}{x} \right)}    & {~=~}    &{\log \left|\left(1+\frac{y^2}{x^2} \right)x^2 \right|~+~\rm{C}_4}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{2 \tan^{-1}\left(\frac{y}{x} \right)}    & {~=~}    &{\log \left|x^2 + y^2 \right|~+~\rm{C}_4}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}&{ \tan^{-1}\left(\frac{y}{x} \right)}    & {~=~}    &{\frac{1}{2}\log \left|x^2 + y^2 \right|~+~\left(\frac{1}{2} \right)\rm{C}_4}    \\
{~\color{magenta}    5    }&{{\Rightarrow}}&{ \tan^{-1}\left(\frac{y}{x} \right)}    & {~=~}    &{\frac{1}{2}\log \left|x^2 + y^2 \right|~+~\rm{C}}    \\
\end{array}}$

Solved example 25.47
Show that the differential equation $\boldsymbol{(x-y)\frac{dy}{dx}~=~x+2y}$ is homogeneous and solve it.
Solution:
Part I: Showing that the given differential equation is homogeneous
1. We can rearrange the given differential equation as:
$\small{\frac{dy}{dx}~=~\frac{x+2y}{x-y}}$

2. Let $\small{F(x,y)~=~\frac{x+2y}{x-y}}$
• We need to show that, this is a homogeneous function of degree zero.
• We already showed it in solved example 25.43 of the previous section. So the given differential equation is homogeneous.

Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

2. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

3. Substituting (1) and (2) in the given differential equation, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{x+2y}{x-y}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{\frac{x+2(vx)}{x-vx}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{x+2(vx)}{x-vx}~-~v}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{x~+~2vx~-~vx~+~v^2x}{x-vx}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{x~+~vx~+~v^2x}{x-vx}~=~\frac{1~+~v~+~v^2}{1-v}}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{\left[\frac{1-v}{1~+~v~+~v^2} \right]dv}    & {~=~}    &{\frac{dx}{x}}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}    &{\left[\frac{v-1}{1~+~v~+~v^2} \right]dv}    & {~=~}    &{\left[\frac{-1}{x} \right]dx}    \\
\end{array}}$

4. So we have separated the variables. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\int{\left[\frac{v-1}{1~+~v~+~v^2} \right]dv}}    & {~=~}    &{\int{\left[\frac{-1}{x} \right]dx}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{\int{\left[\frac{2v + 1 - 3}{2(1~+~v~+~v^2)} \right]dv}}    & {~=~}    &{\int{\left[\frac{-1}{x} \right]dx}}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{\frac{1}{2} \int{\left[\frac{2v + 1}{1~+~v~+~v^2} \right]dv}~-~\frac{3}{2} \int{\left[\frac{1}{1~+~v~+~v^2} \right]dv}}    & {~=~}    &{\int{\left[\frac{-1}{x} \right]dx}}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}    &{\frac{1}{2} \log \left| v^2 + v+ 1  \right|~-~\left(\frac{3}{2} \right)\frac{2}{\sqrt 3}\tan^{-1}\left(\frac{2v+1}{\sqrt 3} \right)~+~\rm{C}_1}    & {~=~}&{-\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 5    }&{{\Rightarrow}}    &{\frac{1}{2} \log \left|v^2 + v+ 1 \right|~+~\log \left|x \right|}    & {~=~}&{\left(\frac{3}{2} \right)\frac{2}{\sqrt 3}\tan^{-1}\left(\frac{2v+1}{\sqrt 3} \right)~+~\rm{C}_2~-~\rm{C}_1}    \\
{~\color{magenta} 6    }&{{\Rightarrow}}    &{\frac{1}{2} \log \left|v^2 + v+ 1  \right|~+~\log \left|x \right|}    & {~=~}&{\sqrt 3 \tan^{-1}\left(\frac{2v+1}{\sqrt 3} \right)~+~\rm{C}_3}    \\
\end{array}}$

• The reader may write all steps related to the integration in [(3) magenta color]

5. Replacing 'v', we will get the general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\frac{1}{2} \log \left|v^2 + v+ 1  \right|~+~\log \left|x \right|}    & {~=~}    &{\sqrt 3 \tan^{-1}\left(\frac{2v+1}{\sqrt 3} \right)~+~\rm{C}_3}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{\log \left|v^2 + v+ 1  \right|~+~2 \log \left|x \right|}    & {~=~}    &{2 \sqrt 3 \tan^{-1}\left(\frac{2v+1}{\sqrt 3} \right)~+~2 \rm{C}_3}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{\log \left|\frac{y^2}{x^2} + \frac{y}{x}+ 1 \right|~+~\log \left|x^2 \right|}    & {~=~}    &{2 \sqrt 3 \tan^{-1}\left(\frac{2y+x}{\sqrt 3 \,x} \right)~+~2 \rm{C}_3}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}&{\log \left|\left(\frac{y^2}{x^2} + \frac{y}{x}+ 1 \right)x^2  \right|}    & {~=~}    &{2 \sqrt 3 \tan^{-1}\left(\frac{2y+x}{\sqrt 3 \,x} \right)~+~2 \rm{C}_3}    \\
{~\color{magenta}    5    }&{{\Rightarrow}}&{\log \left|y^2 + xy + x^2 \right|}    & {~=~}    &{2 \sqrt 3 \tan^{-1}\left(\frac{2y+x}{\sqrt 3 \,x} \right)~+~ \rm{C}}    \\
\end{array}}$

Solved example 25.48
Show that the differential equation $\boldsymbol{(x^2 + xy)dy~=~(x^2 + y^2)dx}$ is homogeneous and solve it.
Solution:
Part I: Showing that the given differential equation is homogeneous

1. We can rearrange the given differential equation as:
$\small{\frac{dy}{dx}~=~\frac{x^2+y^2}{x^2+xy}}$

2. Let $\small{F(x,y)~=~\frac{x^2+y^2}{x^2+xy}}$
• We need to show that, this is a homogeneous function of degree zero.

(i) We can rearrange $\small{F(x,y)}$ as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{F(x,y)}    & {~=~}    &{\frac{x^2+y^2}{x^2+xy}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{F(x,y)}    & {~=~}    &{\left[\frac{x^2}{x^2} \right]\left[\frac{1 ~+~ \frac{y^2}{x^2}}{1 ~+~ \frac{y}{x}} \right]}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}        &{\left[1 \right]\left[\frac{1 ~+~ \frac{y^2}{x^2}}{1 ~+~ \frac{y}{x}} \right]}    \\
{~\color{magenta}    4    }    &{}    &{}    & {~=~}        &{x^0\left[\frac{1 ~+~ \left(\frac{y}{x} \right)^2}{1 ~+~ \frac{y}{x}} \right]}    \\
{~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{x^0\left[g\left(\frac{y}{x} \right) \right]}    \\
\end{array}}$

Here, $\boldsymbol{g\left(\frac{y}{x} \right)~=~\frac{1 ~+~ \left(\frac{y}{x} \right)^2}{1 ~+~ \frac{y}{x}}}$

(ii) 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.
• Therefore, the given differential equation is homogeneous.

Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

2. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

3. Substituting (1) and (2) in the given differential equation, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{1 ~+~ \left(\frac{y}{x} \right)^2}{1 ~+~ \frac{y}{x}}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{\frac{1+v^2}{1+v}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{1+v^2~}{1+v}~-~v~=~\frac{1+v^2-v-v^2}{1+v}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{1-v}{1+v}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{\left[\frac{1+v}{1-v} \right]dv}    & {~=~}    &{\left[\frac{1}{x} \right]dx}    \\
\end{array}}$

4. So we have separated the variables. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\int{\left[\frac{1+v}{1-v} \right]dv}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}    &{\int{\left[\frac{1}{1-v} \right]dv}~+~\int{\left[\frac{v}{1-v} \right]dv}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}    &{(-1)\log \left|1-v \right|~+~\left[-v~-~\log \left|1-v \right| \right]~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 4    }&{{\Rightarrow}}    &{-\log \left|1-v \right|~-~v~-~\log \left|1-v \right| ~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 5    }&{{\Rightarrow}}    &{-2\log \left|1-v \right|~-~v ~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 6    }&{{\Rightarrow}}    &{\log \left|x \right|~+~2\log \left|1-v \right|~+~v}    & {~=~}&{\rm{C}_1~-~\rm{C}_2}    \\
{~\color{magenta} 7    }&{{\Rightarrow}}    &{\log \left|x(1-v)^2 \right|~+~v}    & {~=~}&{\rm{C}_3}    \\
\end{array}}$

• The reader may write all steps related to the integration in [(2) magenta color]

5. Replacing 'v', we will get the general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\log \left|x(1-v)^2 \right|~+~v}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{\log \left|x\left[1-\left(\frac{y}{x} \right) \right]^2 \right|~+~\frac{y}{x}}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{\log \left|x\left[\frac{x-y}{x} \right]^2 \right|~+~\frac{y}{x}}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}&{\log \left|\frac{(x-y)^2}{x} \right|~+~\frac{y}{x}}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    5    }&{{\Rightarrow}}&{\log \left|\frac{(x-y)^2}{x} \right|}    & {~=~}    &{\log \left|\rm{C}\right|~-~\frac{y}{x}}    \\
{~\color{magenta}    6    }&{{\Rightarrow}}&{\log \left|\frac{(x-y)^2}{x} \right|~-~\log \left|\rm{C}\right|}    & {~=~}    &{~-~\frac{y}{x}}    \\
{~\color{magenta}    7    }&{{\Rightarrow}}&{\log \left|\frac{(x-y)^2}{{\rm{C}}x} \right|}    & {~=~}    &{~-~\frac{y}{x}}    \\
{~\color{magenta}    8    }&{{\Rightarrow}}&{\frac{(x-y)^2}{{\rm{C}}x}}    & {~=~}    &{e^{-\frac{y}{x}}}    \\
{~\color{magenta}    9    }&{{\Rightarrow}}&{(x-y)^2}    & {~=~}    &{{\rm{C}}x\,e^{-\frac{y}{x}}}    \\
\end{array}}$

Solved example 25.49
Show that the differential equation $\boldsymbol{(x^2 - y^2)dx~+~2xy\,dy~=~0}$ is homogeneous and solve it.
Solution:
Part I: Showing that the given differential equation is homogeneous

1. We can rearrange the given differential equation as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{(x^2 - y^2)dx~+~2xy\,dy}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{2xy\,dy}    & {~=~}    &{(-1)(x^2 - y^2)dx}    \\
{~\color{magenta}    3    }    &{\Rightarrow}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{(-1)(x^2 - y^2)}{2xy}}    \\
\end{array}}$

2. Let $\small{F(x,y)~=~\frac{(-1)(x^2 - y^2)}{2xy}}$
• We need to show that, this is a homogeneous function of degree zero.

(i) We can rearrange $\small{F(x,y)}$ as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{F(x,y)}    & {~=~}    &{\frac{(-1)(x^2 - y^2)}{2xy}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{F(x,y)}    & {~=~}    &{\left[\frac{x^2}{x^2} \right]\left[\frac{1 ~-~ \frac{y^2}{x^2}}{(-2)\frac{y}{x}} \right]}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}        &{\left[1 \right]\left[\frac{1 ~-~ \frac{y^2}{x^2}}{(-2) \frac{y}{x}} \right]}    \\
{~\color{magenta}    4    }    &{}    &{}    & {~=~}        &{x^0\left[\frac{1 ~-~ \left(\frac{y}{x} \right)^2}{(-2) \frac{y}{x}} \right]}    \\
{~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{x^0\left[g\left(\frac{y}{x} \right) \right]}    \\
\end{array}}$

Here, $\boldsymbol{g\left(\frac{y}{x} \right)~=~\frac{1 ~-~ \left(\frac{y}{x} \right)^2}{(-2) \frac{y}{x}}}$

(ii) 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.
• Therefore, the given differential equation is homogeneous.

Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

2. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

3. Substituting (1) and (2) in the given differential equation, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{1 ~-~ \left(\frac{y}{x} \right)^2}{(-2) \frac{y}{x}}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{\frac{1-v^2}{(-2)v}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{1-v^2~}{(-2)v}~-~v~=~\frac{1-v^2+2v^2}{(-2)v}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{1+v^2}{(-2)v}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{\left[\frac{(-2)v}{1+v^2} \right]dv}    & {~=~}    &{\left[\frac{1}{x} \right]dx}    \\
\end{array}}$

4. So we have separated the variables. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\int{\left[\frac{(-2)v}{1+v^2} \right]dv}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}    &{(-2)\left(\frac{1}{2} \right)\log \left|1+v^2 \right|~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 3    }&{{\Rightarrow}}    &{-\log \left|1+v^2 \right| ~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 4    }&{{\Rightarrow}}    &{\log \left|x \right|~+~\log \left|1+v^2 \right|}    & {~=~}&{\rm{C}_1~-~\rm{C}_2~=~\rm{C}_3}    \\
\end{array}}$

• The reader may write all steps related to the integration in [(1) magenta color]

5. Replacing 'v', we will get the general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\log \left|x \right|~+~\log \left|1+v^2 \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{\log \left|x\left[1+\left(\frac{y}{x} \right)^2 \right] \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{\log \left|x\left[\frac{x^2 + y^2}{x^2} \right] \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}&{\log \left|\frac{x^2 + y^2}{x} \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    5    }&{{\Rightarrow}}&{\frac{x^2 + y^2}{x}}    & {~=~}    &{e^{\rm{C}_3}~=~\rm{C}}    \\
{~\color{magenta}    6    }&{{\Rightarrow}}&{x^2~+~y^2}    & {~=~}    &{{\rm{C}}x}    \\
\end{array}}$

Solved example 25.50
Show that the differential equation $\boldsymbol{x^2\,\frac{dy}{dx}~=~x^2~-~2y^2~+~xy}$ is homogeneous and solve it.
Solution:
Part I: Showing that the given differential equation is homogeneous

1. We can rearrange the given differential equation as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2\,\frac{dy}{dx}}    & {~=~}    &{x^2~-~2y^2~+~xy}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{\frac{dy}{dx}}    & {~=~}    &{1~-~\frac{2y^2}{x^2}~+~\frac{y}{x}}    \\
\end{array}}$

2. Let $\small{F(x,y)~=~1~-~\frac{2y^2}{x^2}~+~\frac{y}{x}}$
• We need to show that, this is a homogeneous function of degree zero.

(i) We can rearrange $\small{F(x,y)}$ as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{F(x,y)}    & {~=~}    &{1~-~\frac{2y^2}{x^2}~+~\frac{y}{x}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{F(x,y)}    & {~=~}    &{1~-~ 2\left(\frac{y}{x} \right)^2~+~\frac{y}{x}}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}        &{\left[1 \right]\left[1~-~ 2\left(\frac{y}{x} \right)^2~+~\frac{y}{x} \right]}    \\
{~\color{magenta}    4    }    &{}    &{}    & {~=~}        &{x^0\left[1~-~ 2\left(\frac{y}{x} \right)^2~+~\frac{y}{x} \right]}    \\
{~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{x^0\left[g\left(\frac{y}{x} \right) \right]}    \\
\end{array}}$

Here, $\boldsymbol{g\left(\frac{y}{x} \right)~=~1~-~ 2\left(\frac{y}{x} \right)^2~+~\frac{y}{x}}$

(ii) 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.
• Therefore, the given differential equation is homogeneous.

Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

2. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

3. Substituting (1) and (2) in the given differential equation, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{1~-~ 2\left(\frac{y}{x} \right)^2~+~\frac{y}{x}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{1~-~ 2\left(v \right)^2~+~v}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{1 - 2 v^2}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left[\frac{1}{1 - 2v^2} \right]dv}    & {~=~}    &{\left[\frac{1}{x} \right]dx}    \\
\end{array}}$

4. So we have separated the variables. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\int{\left[\frac{1}{1 - 2v^2} \right]dv}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}    &{\left(\frac{1}{2 \sqrt 2} \right)\log \left| \frac{2v + \sqrt 2}{2v - \sqrt 2}  \right|~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 3    }&{{\Rightarrow}}    &{\left(\frac{1}{2 \sqrt 2} \right)\log \left| \frac{2v + \sqrt 2}{2v - \sqrt 2}  \right|~-~\log \left|x \right|}    & {~=~}&{\rm{C}_3}    \\
\end{array}}$

• The reader may write all steps related to the integration in [(1) magenta color]

5. Replacing 'v', we will get the general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\left(\frac{1}{2 \sqrt 2} \right)\log \left| \frac{2v + \sqrt 2}{2v - \sqrt 2}  \right|~-~\log \left|x \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{\left(\frac{1}{2 \sqrt 2} \right)\log \left| \frac{2(y/x) + \sqrt 2}{2(y/x) - \sqrt 2}  \right|~-~\log \left|x \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{\left(\frac{1}{2 \sqrt 2} \right)\log \left| \frac{\sqrt 2(y/x) + 1}{\sqrt 2(y/x) - 1}  \right|~-~\log \left|x \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}&{\left(\frac{1}{2 \sqrt 2} \right)\log \left| \frac{\sqrt 2 \,y ~+~ x}{\sqrt 2 \,y ~-~ x}  \right|~-~\log \left|x \right|}    & {~=~}    &{\rm{C}}    \\
\end{array}}$

Solved example 25.51
Show that the differential equation $\boldsymbol{x dy ~-~y dx~=~\sqrt{x^2 + y^2}\,dx}$ is homogeneous and solve it.
Solution:
Part I: Showing that the given differential equation is homogeneous

1. We can rearrange the given differential equation as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x dy ~-~y dx}    & {~=~}    &{\sqrt{x^2 + y^2}\,dx}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{x dy }    & {~=~}    &{\sqrt{x^2 + y^2}\,dx~+~y dx}    \\
{~\color{magenta}    3    }    &{\Rightarrow}    &{x dy }    & {~=~}    &{\left[\sqrt{x^2 + y^2}~+~y \right]dx}    \\
{~\color{magenta}    4    }    &{\Rightarrow}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{\sqrt{x^2 + y^2}~+~y}{x}}    \\
\end{array}}$

2. Let $\small{F(x,y)~=~\frac{\sqrt{x^2 + y^2}~+~y}{x}}$
• We need to show that, this is a homogeneous function of degree zero.

(i) We can rearrange $\small{F(x,y)}$ as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{F(x,y)}    & {~=~}    &{\frac{\sqrt{x^2 + y^2}~+~y}{x}}    \\
{~\color{magenta}    2    }    &{\Rightarrow}    &{F(\lambda x,\lambda y)}    & {~=~}    &{\frac{\sqrt{(\lambda x)^2 + (\lambda y)^2}~+~\lambda y}{\lambda x}}    \\
{~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{\lambda \sqrt{x^2 + y^2}~+~\lambda y}{\lambda x}}    \\
{~\color{magenta}    4    }    &{}    &{}    & {~=~}        &{\frac{\sqrt{x^2 + y^2}~+~y}{x}}    \\
{~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{\lambda^0\left[F(x,y) \right]}    \\
\end{array}}$

(ii) We see that:
For the function $\small{F(x,y)}$, it is possible to write:
$\small{F(\lambda x,\lambda y)~=~\lambda^n\left[F(x,y) \right]}$
   ♦ Where $\small{n}$ is a natural number.

• So it is a homogeneous function.

• For this function, $\small{n~=~0}$. So degree is zero.
• Therefore, the given differential equation is homogeneous.

Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{y}{x}~=~v}$
Which is same as: $\small{y~=~vx}$

2. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$

3. Substituting (1) and (2) in the given differential equation, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{\sqrt{x^2 + y^2}~+~y}{x}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{\frac{\sqrt{x^2 + v^2\,x^2}~+~vx}{x}~=~\sqrt{1+v^2}~+~v}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\sqrt{1 + v^2}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left[\frac{1}{\sqrt{1 + v^2}} \right]dv}    & {~=~}    &{\left[\frac{1}{x} \right]dx}    \\
\end{array}}$

4. So we have separated the variables. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\int{\left[\frac{1}{\sqrt{1 + v^2}} \right]dv}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}    &{\log \left|\sqrt{1 + v^2}~+~v \right|~+~\rm{C}_1}    & {~=~}&{\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta} 3    }&{{\Rightarrow}}    &{\log \left|\sqrt{1 + v^2}~+~v \right|~-~\log \left|x \right|}    & {~=~}&{\rm{C}_3}    \\
\end{array}}$

• The reader may write all steps related to the integration in [(1) magenta color]

5. Replacing 'v', we will get the general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\log \left|\sqrt{1 + v^2}~+~v \right|~-~\log \left|x \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{\log \left|\sqrt{1 + (y/x)^2}~+~(y/x) \right|~-~\log \left|x \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{\log \left|\frac{\sqrt{x^2 + y^2}~+~y}{x} \right|~-~\log \left|x \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}&{\log \left|\frac{\sqrt{x^2 + y^2}~+~y}{x^2} \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    5    }&{{\Rightarrow}}&{\frac{\sqrt{x^2 + y^2}~+~y}{x^2}}    & {~=~}    &{e^{\rm{C}_3}}    \\
{~\color{magenta}    6    }&{{\Rightarrow}}&{\frac{\sqrt{x^2 + y^2}~+~y}{x^2}}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    7    }&{{\Rightarrow}}&{\sqrt{x^2 + y^2}~+~y}    & {~=~}    &{{\rm{C}}x^2}    \\
\end{array}}$

---

In the next section, we will see a few more solved examples.

Previous

Contents

Next 

Copyright©2025 Higher secondary mathematics.blogspot.com

25.7 - Homogeneous Differential Equations

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.

Previous

Contents

Next 

Copyright©2025 Higher secondary mathematics.blogspot.com