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.

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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.

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25.6 - Solved Examples on Variables Separable

In the previous section, we saw how to solve first order first degree differential equations with variables separable. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 25.36
Find the particular solution of the differential equation $\cos\left(\frac{dy}{dx} \right)~=~a~~(a \in R)$ given that y = 2 when x = 0.
Solution:
1. The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos\left(\frac{dy}{dx} \right)}    & {~=~}    &{a~~(a \in R)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{\cos^{-1}a}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{dy}    & {~=~}    &{\cos^{-1}a\,dx}    \\
\end{array}}$

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[1\right]dy}}    & {~=~}    &{\int{\left[\cos^{-1}a \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{y~+~\rm{C}_1}    & {~=~}    &{x~\cos^{-1}a~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}&{y}    & {~=~}    &{x~\cos^{-1}a~+~\rm{C}}    \\
\end{array}}$

3. So the general solution is:
$\small{y~=~x~\cos^{-1}a~+~\rm{C}}$

4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• We want that particular solution in which y becomes 2 when x = 0

• So substituting x = 0 and y = 2 in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2}    & {~=~}    &{(0)~\cos^{-1}a~+~\rm{C}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{2}    & {~=~}    &{0~+~\rm{C}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{2}    & {~=~}    &{\rm{C}}    \\
\end{array}}$

5. So the particular solution is:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{x~\cos^{-1}a~+~2}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{x~\cos^{-1}a}    & {~=~}    &{y~-~2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\cos^{-1}a}    & {~=~}    &{\frac{y-2}{x}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\cos\left(\frac{y-2}{x} \right)}    & {~=~}    &{a}    \\
\end{array}}$ 

Solved example 25.37
Find the equation of the curve passing through the point (1,1) whose differential equation is $x dy~=~(2x^2 + 1) dx~(x \ne 0)$.
Solution:
1. The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{dy}    & {~=~}    &{\left[\frac{2x^2 + 1}{x} \right]dx}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{dy}    & {~=~}    &{\left[2x + \frac{1}{x} \right] dx}    \\
\end{array}}$

2. Integrating both sides, we get:

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

3. So the general solution is:
$\small{y~=~x^2~+~\log \left|x \right|~+~\rm{C}}$

4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• We want that particular solution in which y becomes 1 when x = 1

• So substituting x = 1 and y = 1 in the general solution, we get:

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

5. So the particular solution is:

$\small{y~=~x^2~+~\log \left|x \right|}$

6. The red curve in fig.25.19 below is the graph of this particular solution. Note that, the point (1,1) falls on the red curve only.

Fig.25.19

    ♦ For the yellow curve, C = 2  
    ♦ For the red curve, C = 0
    ♦ For the green curve, C = -0.5

Solved example 25.38
Find the equation of the curve passing through the point (0,0) and whose differential equation is $y'~=~e^x\,\sin x$.
Solution:
1. The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y'}    & {~=~}    &{e^x\,\sin x}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{e^x\,\sin x}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{dy}    & {~=~}    &{\left[e^x\,\sin x \right]dx}    \\
\end{array}}$

2. Integrating both sides, we get:

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

3. So the general solution is:
$\small{y~=~\frac{1}{2}e^x (\sin x ~-~\cos x)~+~\rm{C}}$

4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• We want that particular solution in which y becomes zero when x = zero

• So substituting x = 0 and y = 0 in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{\frac{1}{2}e^x (\sin x ~-~\cos x)~+~\rm{C}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{0}    & {~=~}    &{\frac{1}{2}e^0 (\sin 0 ~-~\cos 0)~+~\rm{C}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}        &{0}    & {~=~}    &{\frac{1}{2}(1) (0 ~-~1)~+~\rm{C}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}        &{0}    & {~=~}    &{\frac{1}{2}(1) (-1)~+~\rm{C}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}        &{\rm{C}}    & {~=~}    &{\frac{1}{2}}    \\
\end{array}}$

5. So the particular solution is:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{\frac{1}{2}e^x (\sin x ~-~\cos x)~+~\frac{1}{2}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{2y}    & {~=~}    &{e^x (\sin x ~-~\cos x)~+~1}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}        &{2y~-~1}    & {~=~}    &{e^x (\sin x ~-~\cos x)}    \\
\end{array}}$

Solved example 25.39
For the differential equation $xy \frac{dy}{dx}~=~(x+2)(y+2)$, find the solution curve passing through the point (1,−1).
Solution:
1. The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{xy\frac{dy}{dx}}    & {~=~}    &{(x+2)(y+2)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{y}{y+2}\,dy}    & {~=~}    &{\frac{x+2}{x}\,dx}    \\
\end{array}}$

2. Integrating both sides, we get:

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

3. So the general solution is:
$\small{x~-~y~+~\log \left[\left(y+2 \right)^2 \,\left(x \right)^2\right]~=~\rm{C}}$

4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• We want that particular solution in which y becomes −1 when x = 1

• So substituting x = 1 and y = −1 in the general solution, we get:

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

5. So the particular solution is:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x~-~y~+~\log \left[\left(y+2 \right)^2 \,\left(x \right)^2\right]}    & {~=~}    &{2}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{y-x+2}    & {~=~}    &{\log \left[\left(y+2 \right)^2 \,\left(x \right)^2\right]}    \\
\end{array}}$

Solved example 25.40
Find the equation of a curve passing through the point (−2,3), given that the slope of the tangent to the curve at any point (x,y) is $\frac{2x}{y^2}$.
Solution:
1. We know that, slope of the tangent is same as the derivative. So we can write the differential equation:
$\frac{dy}{dx}~=~\frac{2x}{y^2}$
• The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{2x}{y^2}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{y^2\,dy}    & {~=~}    &{2x\,dx}    \\
\end{array}}$

2. Integrating both sides, we get:

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

3. So the general solution is:
$\small{x^2~-~\frac{y^3}{3}~=~\rm{C}}$

4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• We want that particular solution in which y becomes 3 when x = −2

• So substituting x = −2 and y = 3 in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2~-~\frac{y^3}{3}}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{(-2)^2~-~\frac{(3)^3}{3}}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}        &{4~-~9}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}        &{\rm{C}}    & {~=~}    &{-5}    \\
\end{array}}$

5. So the particular solution is:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2~-~\frac{y^3}{3}}    & {~=~}    &{-5}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{\frac{y^3}{3}}    & {~=~}    &{x^2~+~5}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}        &{y^3}    & {~=~}    &{3x^2~+~15}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}        &{y}    & {~=~}    &{\left(3x^2~+~15 \right)^{\frac{1}{3}}}    \\
\end{array}}$

Solved example 25.41
Find the equation of a curve passing through the point (0,−2), given that at any point (x,y) on the curve, the product of the slope of it's tangent and y-coordinate of the point is equal to the x-coordinate of the point.
Solution:
1. We know that, slope of the tangent is same as the derivative. So we can write the differential equation:
$y\,\frac{dy}{dx}~=~x$
• The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y\,\frac{dy}{dx}}    & {~=~}    &{x}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{y\,dy}    & {~=~}    &{x\,dx}    \\
\end{array}}$

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[y\right]dy}}    & {~=~}    &{\int{\left[x \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{y^2}{2}~+~\rm{C}_1}    & {~=~}    &{\frac{x^2}{2}~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}&{\frac{x^2}{2}~-~\frac{y^2}{2}}    & {~=~}    &{\rm{C}_1~-~\rm{C}_2}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}&{x^2~-~y^2}    & {~=~}    &{2\rm{C}_1~-~2\rm{C}_2}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}&{x^2~-~y^2}    & {~=~}    &{\rm{C}}    \\
\end{array}}$

3. So the general solution is:
$\small{x^2~-~y^2~=~\rm{C}}$

4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• We want that particular solution in which y becomes −2 when x = 0

• So substituting x = 0 and y = −2 in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2~-~y^2}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{(0)^2~-~(-2)^2}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}        &{0~-~4}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}        &{\rm{C}}    & {~=~}    &{-4}    \\
\end{array}}$

5. So the particular solution is:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2~-~y^2}    & {~=~}    &{-4}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{y^2~-~x^2}    & {~=~}    &{4}    \\
\end{array}}$

Solved example 25.42
In a bank, principal increases continuously at the rate of 5% per year. In how many years Rs 1000 double itself?
Solution:
1. Let P be the principal amount (in Rs), at any time t (in years).
• Then from the given data, we can write:
$\frac{dP}{dt}~=~P\left(\frac{5}{100} \right)~=~0.05P$
• The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dP}{dt}}    & {~=~}    &{0.05P}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dP}{P}}    & {~=~}    &{0.05\,dt}    \\
\end{array}}$

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{1}{P} \right]dP}}    & {~=~}    &{\int{\left[0.05 \right]dt}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\log P~+~\rm{C}_1}    & {~=~}    &{0.05t~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}&{\log P}    & {~=~}    &{0.05t~+~\rm{C}_2~-~\rm{C}_1}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}&{\log P}    & {~=~}    &{0.05t~+~\rm{C}_3}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}&{P}    & {~=~}    &{e^{0.05t + \rm{C}_3}~=~e^{0.05t}.\,e^{\rm{C}_3}}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}&{P}    & {~=~}    &{e^{0.05t}.\,\rm{C}}    \\
\end{array}}$

3. So the general solution is:
$\small{P~=~\rm{C}\,e^{0.05t}}$

4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• At the instant when 1000 Rs is deposited, time is zero.
• We want that particular solution in which P becomes 1000 when t = 0

• So substituting t = 0 and P = 1000 in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{P}    & {~=~}    &{\rm{C}\,e^{0.05t}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{1000}    & {~=~}    &{\rm{C}\,e^{0.05(0)}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}        &{1000}    & {~=~}    &{\rm{C}\,e^0}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}        &{1000}    & {~=~}    &{\rm{C}(1)}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}        &{\rm{C}}    & {~=~}    &{1000}    \\
\end{array}}$

5. So the particular solution is:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{P}    & {~=~}    &{\rm{C}\,e^{0.05t}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{P}    & {~=~}    &{1000\,e^{0.05t}}    \\
\end{array}}$

6. We want the time when Rs. 1000/- become Rs. 2000/-
So based on the particular solution, we can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{P}    & {~=~}    &{\rm{C}\,e^{0.05t}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{2000}    & {~=~}    &{1000\,e^{0.05t}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}        &{2}    & {~=~}    &{e^{0.05t}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}        &{\log 2}    & {~=~}    &{0.05t}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}        &{\log 2}    & {~=~}    &{(1/20)t}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}        &{t}    & {~=~}    &{20 \log_e 2}    \\
\end{array}}$

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The link below gives a few more solved examples:

Exercise 25.4

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

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