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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