25.10 - Linear Differential Equations

In the previous section, we completed a discussion on the method to solve homogeneous differential equations. In this section, we will see linear differential equations.

Some basic details can be written in 3 steps:

Type I:
1. Consider the differential equation $\small{\frac{dy}{dx}~+~\rm{P}y~=~\rm{Q}}$

2. Suppose that P and Q satisfy any one of the following three conditions:
(i) Both P and Q are constants
(ii) Both P and Q are functions of x
(iii) P is a constant and Q is a function of x

3. Then the equation written in (1), is a first order linear differential equation.

Let us see some examples:

• $\small{\frac{dy}{dx}~+~y~=~\sin x}$
   ♦ Condition  (iii) is satisfied

• $\small{\frac{dy}{dx}~+~\left(\frac{1}{x} \right)y~=~e^x}$
   ♦ Condition  (ii) is satisfied

• $\small{\frac{dy}{dx}~+~\left(\frac{1}{x\,\log x} \right)y~=~\frac{1}{x}}$
   ♦ Condition  (ii) is satisfied

Type II:
1. Consider the differential equation $\small{\frac{dx}{dy}~+~\rm{P}_1 x~=~\rm{Q}_1}$

2. Suppose that P₁ and Q₁ satisfy any one of the following three conditions:
(i) Both P₁ and Q₁ are constants
(ii) Both P₁ and Q₁ are functions of y
(iii) P₁ is a constant and Q₁ is a function of y

3. Then the equation written in (1), is a first order linear differential equation.

Let us see some examples:

• $\small{\frac{dx}{dy}~+~x~=~\cos y}$
   ♦ Condition  (iii) is satisfied

• $\small{\frac{dx}{dy}~+~\left(\frac{-2}{y} \right)x~=~y^2\;e^y}$
   ♦ Condition  (ii) is satisfied

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So now we know what a first order linear differential equation is. Let us try to solve such equations:

First we will solve type I:
1. We have: $\small{\frac{dy}{dx}~+~\rm{P}y~=~\rm{Q}}$

2. Multiply both sides by a function of x, say $\small{g(x)}$.
• We get:
$\small{g(x)\,\frac{dy}{dx}~+~\rm{P}\,g(x)\,y~=~{\rm{Q}}\,g(x)}$

3. The function $\small{g(x)}$ should be such that:
$\small{{\rm{Q}}\,g(x)~=~\frac{d}{dx}\left[g(x)\,y \right]}$

4. Based on (2) and (3), we get:

$\small{\begin{array}{ll}{~\color{magenta}    1    }  &{{}}    &{g(x)\,\frac{dy}{dx}~+~{\rm{P}}\,g(x)\,y}    & {~=~}    &{{\rm{Q}}\,g(x)}    \\
{~\color{magenta} 2    }    &{{\Rightarrow}}    &{g(x)\,\frac{dy}{dx}~+~{\rm{P}}\,g(x)\,y}    & {~=~}    &{\frac{d}{dx}\left[g(x)\,y \right]}    \\
{~\color{magenta}    3}    &{{\Rightarrow}}    &{g(x)\,\frac{dy}{dx}~+~{\rm{P}}\,g(x)\,y}    & {~=~}    &{g(x)\,\frac{dy}{dx}~+~y\;g'(x)}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{{\rm{P}}\,g(x)}    & {~=~}    &{g'(x)}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{{\rm{P}}}    & {~=~}    &{\frac{g'(x)}{g(x)}}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{\int{\left[{\rm{P}} \right]dx}}    & {~=~}  &{\int{\left[\frac{g'(x)}{g(x)} \right]dx}}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}    &{\int{\left[{\rm{P}} \right]dx}}    & {~=~}  &{\log\left[g(x) \right]}    \\
{~\color{magenta}    8    }    &{{\Rightarrow}}    &{g(x)}    & {~=~}  &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}    \\
\end{array}}$

5. So we have obtained $\small{g(x)}$.
• It is called the integrating factor (abbreviated as I.F).
• Now the equation in (2) becomes:

$\small{\begin{array}{ll}{~\color{magenta}    1    }  &{{}}    &{g(x)\,\frac{dy}{dx}~+~{\rm{P}}\,g(x)\,y}    & {~=~}    &{{\rm{Q}}\,g(x)}    \\
{~\color{magenta} 2    }    &{{\Rightarrow}}    &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}\,\frac{dy}{dx}~+~{\rm{P}}\,e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}\,y}    & {~=~}    &{{\rm{Q}}\,e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}    \\
{~\color{magenta}    3}    &{{\Rightarrow}}    &{\frac{d}{dx}\left[y\,e^{\left(\int{\left[{\rm{P}} \right]dx} \right)} \right]}    & {~=~}    &{{\rm{Q}}\,e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{y\,e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}    & {~=~}    &{\int{\left[{\rm{Q}}\,e^{\left(\int{\left[{\rm{P}} \right]dx} \right)} \right]dx}}    \\
\end{array}}$

• From this, we can write y.

◼ Remarks:
• The L.H.S of [(2) magenta color], is the derivative of $y\,e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}$, w.r.t x.
• In [(4) magenta color], we integrate both sides of [(3) magenta color]

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The above five steps can be rewritten in an "easy to remember" form:
1. Write the given differential equation in the form:
$\small{\frac{dy}{dx}~+~{\rm{P}}y~=~\rm{Q}}$

• P and Q must  satisfy any one of the following three conditions:
(i) Both P and Q are constants
(ii) Both P and Q are functions of x
(iii) P is a constant and Q is a function of x

2. Find I.F:
I.F =$\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$

3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$

4. Do the integration in the R.H.S, and add the constant of integration.

5. Multiply the above result in (4), by the reciprocal of the I.F

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Next, we will solve type II.
We can directly write the "easy to remember" five steps:

1. Write the given differential equation in the form:
$\small{\frac{dx}{dy}~+~{\rm{P}_1}y~=~\rm{Q}_1}$

• P₁ and Q₁ must satisfy any one of the following three conditions:
(i) Both P₁ and Q₁ are constants
(ii) Both P₁ and Q₁ are functions of y
(iii) P₁ is a constant and Q₁ is a function of y

2. Find I.F:
I.F =$\small{e^{\left(\int{\left[{\rm{P}_1} \right]dy} \right)}}$

3. Write the equation with independent x:
$\small{x\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q_1(I.F)}} \right]dy}}$

4. Do the integration in the R.H.S, and add the constant of integration.

5. Multiply the above result in (4), by the reciprocal of the I.F

---

Now we will see some solved examples:

Solved example 25.56
Find the general solution of the differential equation $\boldsymbol{\frac{dy}{dx}~+~3y~=~e^{-2x}}$.
Solution:
1. The given differential equation is:
$\small{\frac{dy}{dx}~+~3y~=~e^{-2x}}$

• Here P = 3 and Q = $\small{e^{-2x}}$

• P and Q must  satisfy any one of the following three conditions:
(i) Both P and Q are constants
(ii) Both P and Q are functions of x
(iii) P is a constant and Q is a function of x


• Here, P is a constant and Q is a function of x. So it is a first order linear differential equation.

2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$

$\small{\begin{array}{ll}{~\color{magenta}    1    }  &{{}}    &{\int{\left[{\rm{P}} \right]dx}}    & {~=~}    &{\int{\left[3 \right]dx}~=~ 3x}    \\
{~\color{magenta} 2    }    &{{\Rightarrow}}    &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}    & {~=~}    &{e^{\left(3x \right)}}    \\
\end{array}}$

3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$

We get: $\small{y\left[e^{\left(3x \right)} \right]~=~\int{\left[e^{-2x}~\left(e^{\left(3x \right)} \right) \right]dx}~=~\int{\left[e^{x} \right]dx}}$

4. Do the integration in the R.H.S, and add the constant of integration.

We get: $\small{e^x~+~\rm{C}}$

5. Multiply the above result in (4), by the reciprocal of the I.F

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

Solved example 25.57
Find the general solution of the differential equation $\boldsymbol{\frac{dy}{dx}~+~\frac{y}{x}~=~x^2}$.
Solution:
1. The given differential equation is:
$\small{\frac{dy}{dx}~+~\frac{y}{x}~=~x^2}$

• Here P = 1/x and Q = x²

• P and Q must  satisfy any one of the following three conditions:
(i) Both P and Q are constants
(ii) Both P and Q are functions of x
(iii) P is a constant and Q is a function of x


• Here, both P and Q are functions of x. So it is a first order linear differential equation.

2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$  

$\small{\begin{array}{ll}{~\color{magenta}    1    }  &{{}}    &{\int{\left[{\rm{P}} \right]dx}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}~=~ \log x}    \\
{~\color{magenta} 2    }    &{{\Rightarrow}}    &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}    & {~=~}    &{e^{\left(\log x \right)}}    \\
{~\color{magenta} 3    }    &{{\Rightarrow}}    &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}    & {~=~}    &{x~~\left({\rm{since}}~~e^{\log u}~=~u \right)}    \\
\end{array}}$

3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$

We get: $\small{y\left[x \right]~=~\int{\left[x^2~\left(x \right) \right]dx}~=~\int{\left[x^3 \right]dx}}$

4. Do the integration in the R.H.S, and add the constant of integration.

We get: $\small{\frac{x^4}{4}~+~\rm{C}}$

• The reader must write all the steps involved in this integration.

5. Multiply the above result in (4), by the reciprocal of the I.F

$\small{\begin{array}{ll}{~\color{magenta}    1    }  &{{}}    &{y}    & {~=~}    &{\left[\frac{x^4}{4}~+~\rm{C} \right]\frac{1}{x}}    \\
{~\color{magenta} 2    }    &{{\Rightarrow}}    &{xy}    & {~=~}    &{\frac{x^4}{4}~+~{\rm{C}}}    \\
\end{array}}$

Solved example 25.58
Find the general solution of the differential equation $\boldsymbol{x\frac{dy}{dx}~+~2y~=~x^2~~(x\ne 0)}$.
Solution:
1. The given differential equation is:
$\small{x\frac{dy}{dx}~+~2y~=~x^2~~(x\ne 0)}$
Dividing both sides by x, we get:
$\small{\frac{dy}{dx}~+~\left(\frac{2}{x} \right)y~=~x~~(x\ne 0)}$
• Here P = 2/x and Q = x

• P and Q must  satisfy any one of the following three conditions:
(i) Both P and Q are constants
(ii) Both P and Q are functions of x
(iii) P is a constant and Q is a function of x


• Both P and Q are functions of x. So it is a first order linear differential equation.

2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$  

$\small{\begin{array}{ll}{~\color{magenta}    1    }  &{{}}    &{\int{\left[{\rm{P}} \right]dx}}    & {~=~}    &{\int{\left[\frac{2}{x} \right]dx}~=~2 \log x}    \\
{~\color{magenta} 2    }    &{{\Rightarrow}}    &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}    & {~=~}    &{e^{\left(2 \log x \right)}~=~e^{\left( \log x^2 \right)}}    \\
{~\color{magenta} 3    }    &{{\Rightarrow}}    &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}    & {~=~}    &{x^2~~\left({\rm{since}}~~e^{\log u}~=~u \right)}    \\
\end{array}}$

3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$

We get: $\small{y\left[x^2 \right]~=~\int{\left[x~\left(x^2 \right) \right]dx}}$

4. Do the integration in the R.H.S, and add the constant of integration.

We get: $\small{\frac{x^4}{4}~+~\rm{C}}$

• The reader must write all the steps involved in this integration.

5. Multiply the above result in (4), by the reciprocal of the I.F

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

Solved example 25.59
Find the general solution of the differential equation $\boldsymbol{x\frac{dy}{dx}~+~2y~=~x^2\,\log x}$.
Solution:
1. The given differential equation is:
$\small{x\frac{dy}{dx}~+~2y~=~x^2\,\log x}$

Dividing both sides by x, we get:

$\small{\frac{dy}{dx}~+~\left(\frac{2}{x} \right)y~=~x\,\log x}$

• Here P = $\small{\frac{2}{x}}$ and Q = $\small{x\,\log x}$

• P and Q must  satisfy any one of the following three conditions:
(i) Both P and Q are constants
(ii) Both P and Q are functions of x
(iii) P is a constant and Q is a function of x


• Here, both P and Q are functions of x. So it is a first order linear differential equation.

2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$  

$\small{\begin{array}{ll}{~\color{magenta}    1    }  &{{}}    &{\int{\left[{\rm{P}} \right]dx}}    & {~=~}    &{\int{\left[\frac{2}{x} \right]dx}~=~ 2\,\log x~=~\log\left(x^2 \right)}    \\
{~\color{magenta} 2    }    &{{\Rightarrow}}    &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}    & {~=~}    &{e^{\log\left(x^2 \right)}~=~x^2}    \\
\end{array}}$

3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$

We get: $\small{y\left[x^2 \right]~=~\int{\left[x \log x~\left(x^2 \right) \right]dx}~=~\int{\left[x^3~\times~\log x \right]dx}}$

4. Do the integration in the R.H.S, and add the constant of integration.

We get: $\small{\frac{x^4 \log \left|x \right|}{4}~-~\frac{x^4}{16}~+~\rm{C}}$

5. Multiply the above result in (4), by the reciprocal of the I.F

$\small{\begin{array}{ll}{~\color{magenta}    1    }  &{{}}    &{y}    & {~=~}    &{\left[\frac{x^4 \log \left|x \right|}{4}~-~\frac{x^4}{16}~+~\rm{C} \right]\frac{1}{x^2}}    \\
{~\color{magenta} 2    }    &{{\Rightarrow}}    &{y}    & {~=~}    &{\frac{x^2 \log \left|x \right|}{4}~-~\frac{x^2}{16}~+~{\rm{C}}x^{-2}}    \\
\end{array}}$

Solved example 25.60
Find the general solution of the differential equation $\boldsymbol{x\,\log x \frac{dy}{dx}~+~y~=~\left(\frac{2}{x} \right)\log x}$.
Solution:
1. The given differential equation is:
$\small{x\,\log x \frac{dy}{dx}~+~y~=~\left(\frac{2}{x} \right)\log x}$

Dividing both sides by x log x, we get:

$\small{\frac{dy}{dx}~+~\left(\frac{1}{x\,\log x} \right)y~=~\frac{2}{x^2}}$

• Here P = $\small{\frac{1}{x\,\log x}}$ and Q = $\small{\frac{2}{x^2}}$

• P and Q must  satisfy any one of the following three conditions:
(i) Both P and Q are constants
(ii) Both P and Q are functions of x
(iii) P is a constant and Q is a function of x


• Here, both P and Q are functions of x. So it is a first order linear differential equation.

2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$  

$\small{\begin{array}{ll}{~\color{magenta}    1    }  &{{}}    &{\int{\left[{\rm{P}} \right]dx}}    & {~=~}    &{\int{\left[\frac{1}{x\,\log x} \right]dx}~=~\log \left|\log \left|x \right| \right|}    \\
{~\color{magenta} 2    }    &{{\Rightarrow}}    &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}    & {~=~}    &{e^{\log \left|\log \left|x \right| \right|}~=~\log \left|x \right|}    \\
\end{array}}$

3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$

We get: $\small{y\left[\log x \right]~=~\int{\left[\frac{2}{x^2}~\left(\log \left|x \right| \right) \right]dx}}$

4. Do the integration in the R.H.S, and add the constant of integration.

We get: $\small{\frac{-2 \log \left|x \right|~-~2}{x}~+~\rm{C}}$

• The reader must write all the steps involved in this integration.

5. Multiply the above result in (4), by the reciprocal of the I.F

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

Solved example 25.61
For the differential equation $\boldsymbol{\left(1 +x^2 \right) \frac{dy}{dx}~+~2xy~=~\frac{1}{1+x^2}}$; y = 0 when x = 1, find the particular solution satisfying the given conditions.
Solution:
1. The given differential equation is:
$\small{\left(1 +x^2 \right) \frac{dy}{dx}~+~2xy~=~\frac{1}{1+x^2}}$

Dividing both sides by $\small{\left(1 +x^2 \right)}$, we get:

$\small{\frac{dy}{dx}~+~\left(\frac{2x}{1 +x^2 } \right)y~=~\frac{1}{\left(1 +x^2 \right)^2}}$

• Here P = $\small{\frac{2x}{1 +x^2 }}$ and Q = $\small{\frac{1}{\left(1 +x^2 \right)^2}}$

• P and Q must  satisfy any one of the following three conditions:
(i) Both P and Q are constants
(ii) Both P and Q are functions of x
(iii) P is a constant and Q is a function of x


• Here, both P and Q are functions of x. So it is a first order linear differential equation.

2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$  

$\small{\begin{array}{ll}{~\color{magenta}    1    }  &{{}}    &{\int{\left[{\rm{P}} \right]dx}}    & {~=~}    &{\int{\left[\frac{2x}{1 +x^2 } \right]dx}~=~\log (1+x^2)}    \\
{~\color{magenta} 2    }    &{{\Rightarrow}}    &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}    & {~=~}    &{e^{\log (1+x^2)}~=~1 + x^2}    \\
\end{array}}$

3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$

We get: $\small{y\left[1 + x^2 \right]~=~\int{\left[\frac{1}{\left(1 +x^2 \right)^2}~\left(1 + x^2 \right) \right]dx}~=~\int{\left[\frac{1}{1 +x^2} \right]dx}}$

4. Do the integration in the R.H.S, and add the constant of integration.

We get: $\small{\tan^{-1}x~+~\rm{C}}$

5. Multiply the above result in (4), by the reciprocal of the I.F

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

6. Now we can find the particular solution.
• Given that: y = 0 when x = 1
• Substituting in the general solution, we get:

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

• So the particular solution is:

$\small{y\,\left(1 + x^2 \right)~=~\tan^{-1}x~-~\frac{\pi}{4}}$

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

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25.9 - Solved Examples on Solution of Homogeneous Differential Equations

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

1. We can rearrange the given differential equation as:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

1. We can rearrange the given differential equation as:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

2. Substituting in the general solution, we get:

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

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

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

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

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

Part II: Showing that the differential equation is homogeneous

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

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

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

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

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

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

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

• So the given differential equation is homogeneous.

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

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

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

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

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

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

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

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

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The link below gives a few more solved examples:

Exercise 9.5

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

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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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