26.1 - Types of Vectors

In the previous section, we saw direction cosines. In this  section, we will see types of vectors.

We need to learn about seven types of vectors:
1. Zero Vector
For a vector, if the terminal point is same as the initial point, then that vector is called a zero vector.
• A zero vector is denoted as $\small{\vec{0}}$
• A zero vector is also known as a null vector.
• A zero vector cannot be assigned any definite direction because, it has no magnitude.
• Alternatively, a zero vector can be regarded as a vector having any of the infinite possible directions.
• Vectors $\small{\vec{AA}, \vec{BB},~.~.~}$ are examples of zero vector.

2. Unit Vector
If the magnitude of a vector is 1 unit, then it is called a unit vector.
• Suppose that, we are given a vector $\small{\vec{a}}$. We can draw a unit vector which has the same direction as $\small{\vec{a}}$. Such a vector is denoted as $\small{\hat{a}}$

3. Coinitial Vectors
If two or more vectors have the same initial point, then they are called coinitial vectors.
• $\small{\vec{AB}}$ and $\small{\vec{AC}}$ are coinitial vectors

4. Collinear Vectors
If two or more vectors are parallel to the same line, then they are called collinear vectors.
• Collinear vectors need not lie along the same line.
• Collinear vectors can have only two possible directions. One direction exactly opposite to the other.

5. Equal Vectors
Two vectors are said to be equal if they have the same magnitude and direction.
• Equal vectors need not have the same initial point.
• If two vectors $\small{\vec{a}~\text{and}~\vec{b}}$ are equal, then we write $\small{\vec{a} = \vec{b}}$

6. Negative of a Vector
A vector is said to be the negative of another vector, if they have the same magnitude but opposite directions.
• $\small{\vec{AB}}$ is the negative of $\small{\vec{BA}}$. We write: $\small{\vec{BA} = -\vec{AB}}$

7. Free Vectors
Free vectors are those which can be moved around. The only condition is that, the magnitude and direction should not be altered.
• In this chapter, we will be dealing with free vectors only.

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Now we will see some solved examples

Solved example 26.1
Represent graphically a displacement of 40 km, 30^o west of south.
Solution:
1. An object undergoes a displacement of 40 km in a specified direction. We are asked to draw the displacement vector.

2. We setup the rectangular coordinate system in such a way that, the origin O of the coordinate system coincides with the initial position of the object. Also, the following directions should be clearly marked:
    ♦ +ve x direction towards east
    ♦ +ve y direction towards north
    ♦ −ve x direction towards west
    ♦ −ve y direction towards south
This is shown in fig.26.8 below:

Fig.26.8

3. Next step is to draw a dashed line which satisfies two conditions:
    ♦ The line must make 30^o with the south direction
    ♦ The line must be inclined towards west
This dashed line is shown in fig.a

4. Next step is to mark a point P on the dashed line in such a way that, OP represents 40 km. A suitable scale can be used based on the size of the paper available.

5. Now, the dashed portion between O and P should be changed to a continuous line. And an arrowhead can be placed at P. This is shown in fig.b
• Thus we get $\small{\vec{OP}}$ which is the displacement vector.

6. Scale of the drawing should be written on the top left corner.

Solved example 26.2
Classify the following measures as scalars and vectors.
(i) 5 seconds  (ii) 1000 cm³  (iii) 10 newton  (iv) 30 km/hr  (v) 10 g/cm³  (vi) 20 m/s towards north
Solution:
(i) Time has magnitude only. It has no direction. So 5 seconds is a scalar quantity 
(ii) Volume has magnitude only. It has no direction. So 1000 cm³ is a scalar quantity 
(iii) Force has both magnitude and direction. So 10 newton is a vector quantity. 
(iv) Speed has magnitude only. It has no direction. So 30 km/hr is a scalar quantity. 
(v) Density has magnitude only. It has no direction. So 10 g/cm³ is a scalar quantity. 
(vi) Velocity has both magnitude and direction. So "20 m/s towards north" is a vector quantity.

Solved example 26.3
In fig.26.9, which of the vectors are:
(i) Collinear  (ii) Equal  (iii) Coinitial

Fig.26.9

Solution:
1. We are given four vectors.
• Each of $\small{\vec{a}}$ and $\small{\vec{c}}$ has a small white perpendicular line. This line divides the vector into two equal parts. So these two vectors have a magnitude of two units.
• Each of $\small{\vec{b}}$ and $\small{\vec{d}}$ has two small white perpendicular lines. These lines divide the vectors into three equal parts. So these two vectors have a magnitude of three units.
2. We see that $\small{\vec{a}~\text{and}~\vec{c}}$ have the same direction. Also, both have a magnitude of 2 units. So they are equal vectors.
3. $\small{\vec{b}, \vec{c}~\text{and}~\vec{d}}$ have the same initial point. So they are coinitial vectors.
4. $\small{\vec{a}, \vec{c}~\text{and}~\vec{d}}$ are parallel. So they are collinear vectors

Solved example 26.4
Represent graphically a displacement of 40 km, 30^o east of north
Solution:

1. An object undergoes a displacement of 40 km in a specified direction. We are asked to draw the displacement vector.

2. We setup the rectangular coordinate system in such a way that, the origin O of the coordinate system coincides with the initial position of the object. Also, the following directions should be clearly marked:
    ♦ +ve x direction towards east
    ♦ +ve y direction towards north
    ♦ −ve x direction towards west
    ♦ −ve y direction towards south
This is shown in fig.26.10 below:

Fig.26.10

3. Next step is to draw a dashed line which satisfies two conditions:
    ♦ The line must make 30^o with the north direction
    ♦ The line must be inclined towards east
This dashed line is shown in fig.a

4. Next step is to mark a point P on the dashed line in such a way that, OP represents 40 km. A suitable scale can be used based on the size of the paper available.

5. Now, the dashed portion between O and P should be changed to a continuous line. And an arrowhead can be placed at P. This is shown in fig.b
• Thus we get $\small{\vec{OP}}$ which is the displacement vector.

6. Scale of the drawing should be written on the top left corner.

Solved example 26.5
In fig.26.11 (a square), identify the following vectors.
(i) Coinitial  (ii) Equal  (iii) Collinear but not equal

Fig.26.11

Solution:
1. We are given four vectors. They form the sides of a square. So all four of them have the same magnitude.
2. $\small{\vec{a}~\text{and}~\vec{d}}$ have the same initial point. So they are coinitial vectors.
3. $\small{\vec{b}~\text{and}~\vec{d}}$ have the same magnitude and direction. So they are equal vectors.
4. $\small{\vec{a}~\text{and}~\vec{c}}$ have the same magnitude but opposite directions. So they are not equal vectors. However, since they are parallel, they are collinear.

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The link below gives a few more solved examples:

Exercise 26.1

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In the next section, we will see addition of vectors.

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Chapter 26 - Vector Algebra

In the previous section, we completed a discussion on differential equations. In this chapter, we will see vector algebra.

Some basics about vectors can be written in 2 steps:
1. Consider the length of a room. We can write:
Length of the room is 3.2 m
• Length involves only one value, which is the magnitude.
• Quantities which involve only the magnitude, are called scalars.
• A scalar quantity can be represented by a real number and a proper unit.
• Some examples are:
    ♦ A mass of 32 kg
    ♦ An area of 90 sq.cm
    ♦ A speed of 7 m/s

2. A quantities having both magnitude and direction are called vectors.
• Some examples are:
    ♦ A car moving with a velocity. We can write: velocity is 35 km/h towards east. Here, velocity is a vector quantity.
    ♦ A force applied on a body. We can write: Force is 25 N along a line which makes 30 degrees with the x-axis. Here, force is a vector quantity.

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Let us see vectors in some more detail. It can be written in 8 steps:

1. Fig.26.1(a) below, shows a line $\small{l}$. It extents to infinity in both ways.

Fig.26.1

2. In fig.b, there is an arrowhead attached to the same line $\small{l}$. This arrowhead gives a sense of direction to the observer.

3. In fig.c the arrowhead is attached in the opposite direction. This arrowhead gives the sense of opposite direction to the observer.

4. In fig.d, two points A and B are marked on the same line $\small{l}$. AB is a line segment.
• Any line segment will have a definite length. That means, any line segment will have a magnitude.
• An arrowhead is also attached to the segment AB. This arrowhead gives the sense of direction.
• So the line segment AB in fig.d has both magnitude and direction. It is a vector.

5. The vector in fig.d can be denoted as: $\small{\vec{AB}}$ or as $\small{\vec{a}}$.
    ♦ $\small{\vec{AB}}$ is read as “vector AB”
    ♦ $\small{\vec{a}}$ is read as “vector a”

6. $\small{\vec{AB}}$ starts at A. So the point A is called the initial point of $\small{\vec{AB}}$

7. $\small{\vec{AB}}$ ends at B. So the point B is called the terminal point of $\small{\vec{AB}}$.

8. The distance between points A and B is called the magnitude of $\small{\vec{AB}}$. It is denoted as $\small{\left|{\vec{AB}}\right|}$
Note that, a length cannot be −ve. So we never write: $\small{\left|{\vec{AB}}\right| < 0}$

Position Vector

This can be explained in 4 steps:
1. In class 11, we have seen the three dimensional right handed rectangular coordinate system. Details here.
In that coordinate system, consider a point P. This is shown in fig.26.2 below:

Fig.26.2

    ♦ The coordinates of origin O are (0,0,0)
    ♦ The coordinates of P are (x,y,z)

2. In such a situation, we can draw $\small{\vec{OP}}$
• For $\small{\vec{OP}}$, O is the initial point and P is the terminal point.
• This $\small{\vec{OP}}$ is called the position vector of point P with respect to O.

3. Based on the discussions that we had in class 11, we are able to find the distance OP. Details here.
We get: $\small{OP = \sqrt{x^2 + y^2 + z^2}}$.
• So we can write:  $\small{\left|{\vec{OP}}\right| = \sqrt{x^2 + y^2 + z^2}}$

4. Position vector is an important concept in science and engineering. For points A, B, C, . . . , we denote the position vectors as:
$\small{\vec{a},~\vec{b},~\vec{c},~.~.~.~,}$.
This is shown in fig.26.3 below:

Fig.26.3

Direction Cosines

There are three direction cosines. First we will see the direction cosine related to the x-axis. It can be written in 4 steps:

1.  Fig.26.4 below shows a point P(x,y,z).

Fig.26.4

• The position vector $\small{\vec{OP}}$ makes an angle $\small{\alpha}$ with the +ve direction of the x-axis.
• For better clarity, the observer is looking at the coordinate system in the direction marked as I in fig.26.5 below.

Fig.26.5
Top view of rectangular coordinate system

2. A perpendicular is dropped from P onto the xy-plane. Foot of this perpendicular is marked as P₁.
• From P₁, a perpendicular is dropped onto the x-axis. Foot of this perpendicular is marked as A.

3. Consider the plane containing the three point P, P₁ and A.
• This plane will be perpendicular to the x-axis because, AP₁ is perpendicular to the x-axis.
• Since this plane is perpendicular to the x-axis, any line in the plane will be perpendicular to the x-axis.
• So AP is perpendicular to the x-axis.
• Therefore, $\small{\angle{OAP} = 90^\circ}$

4. So triangle OAP is a right triangle with OP as hypotenuse. We get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos \alpha}    & {~=~}    &{\frac{OA}{\left|\vec{OP} \right|} = \frac{x}{\left|\vec{OP} \right|}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{x}    & {~=~}    &{\cos \alpha\,\left|\vec{OP} \right|}
\\ \end{array}}$

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Next we will see the direction cosine related to the y-axis. It can be written in 4 steps:

1.  Fig.26.6 below shows the same point P(x,y,z).

Fig.26.6

• The position vector $\small{\vec{OP}}$ makes an angle $\small{\beta}$ with the +ve direction of the y-axis.
• For better clarity, the observer is looking at the coordinate system in the direction marked as II in fig.26.5 above.

2. We have the same point P₁ as before.
• This time, the perpendicular from P₁, is dropped onto the y-axis. Foot of this perpendicular is marked as B.

3. Consider the plane containing the three point P, P₁ and B.
• This plane will be perpendicular to the y-axis because, BP₁ is perpendicular to the y-axis.
• Since this plane is perpendicular to the y-axis, any line in the plane will be perpendicular to the y-axis.
• So BP is perpendicular to the y-axis.
• Therefore, $\small{\angle{OBP} = 90^\circ}$

4. So triangle OBP is a right triangle with OP as hypotenuse. We get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos \beta}    & {~=~}    &{\frac{OB}{\left|\vec{OP} \right|} = \frac{y}{\left|\vec{OP} \right|}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{y}    & {~=~}    &{\cos \beta\,\left|\vec{OP} \right|}
\\ \end{array}}$

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Finally we will see the direction cosine related to the z-axis. It can be written in 3 steps:

1.  Fig.26.7 below shows the same point P(x,y,z).

Fig.26.7

• The position vector $\small{\vec{OP}}$ makes an angle $\small{\gamma}$ with the +ve direction of the z-axis.
• For better clarity, the observer is looking at the coordinate system in the direction marked as I in fig.26.5 above.

2. This time, we do not need the point P₁.
• This time, we directly drop the perpendicular from P onto the z-axis. Foot of this perpendicular is marked as C.

3. Since PC is perpendicular to the z-axis, triangle OPC is a right triangle with OP as hypotenuse. We get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos \gamma}    & {~=~}    &{\frac{OC}{\left|\vec{OP} \right|} = \frac{z}{\left|\vec{OP} \right|}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{z}    & {~=~}    &{\cos \gamma\,\left|\vec{OP} \right|}
\\ \end{array}}$

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Now we can write the definitions. It can be written in 7 steps:

1. The angles $\small{\alpha, \beta~\text{and}~\gamma}$ that we saw above, are angles made with the +ve direction of the x, y and z axes respectively.
• They are called direction angles of the position vector $\small{\vec{OP}}$

2. We can take cosines of the above angles. We get:
$\small{\cos\alpha, \cos\beta~\text{and}~\cos\gamma}$
• These are called direction cosines of the position vector $\small{\vec{OP}}$

3. Usually we use short forms for denoting the direction cosines.
    ♦ $\small{\cos\alpha}$ is denoted by $\small{l}$
    ♦ $\small{\cos\beta}$ is denoted by $\small{m}$
    ♦ $\small{\cos\gamma}$ is denoted by $\small{n}$

4. There are short forms for the position vector also:
    ♦ $\small{\vec{OP}}$ is denoted by $\small{\vec{r}}$
    ♦ $\small{\left|\vec{OP} \right|}$ is denoted by $\small{r}$ 

5. Based on the above short forms, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x}    & {~=~}    &{\cos \alpha\,\left|\vec{OP} \right|}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{x}    & {~=~}    &{lr}
\\ {~\color{magenta}    3    }    &{}    &{y}    & {~=~}    &{\cos \beta\,\left|\vec{OP} \right|}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{y}    & {~=~}    &{mr}
\\ {~\color{magenta}    5    }    &{}    &{z}    & {~=~}    &{\cos \gamma\,\left|\vec{OP} \right|}
\\ {~\color{magenta}    6    }    &{\Rightarrow}    &{z}    & {~=~}    &{nr}
\\ \end{array}}$

6. So the coordinates (x,y,z) of the point P, can be written as: $\small{(lr,mr,nr)}$

7. $\small{lr,mr~\text{and}~nr}$ are mere numbers.
• They are called direction ratios of $\small{\vec{r}}$. We will see more details about this in later sections.

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In the next section, we will see types of vectors.

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25.14 - Miscellaneous Examples (2) on Differential Equations

In the previous section, we saw some miscellaneous examples on differential equations. In this section, we will see a few more miscellaneous examples.

Solved example 25.89
Find a particular solution of the differential equation ${(x-y)(dx+dy)~=~dx - dy}$, given that y = −1 when x = 0.
Solution:
1. The variables can be separated and the differential equation can be rewritten:

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

2. Put $\small{x-y~=~t}$
Differentiating w.r.t x, we get:

Integrating both sides, we get:

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

• Substituting this in the result in (1), we get:

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

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

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

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

4. Replacing 't', we will get the general solution:

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

5. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• We want that particular solution in which y becomes −1 when x = 0

• So substituting x = 0 and y = −1 in the general solution, we get:

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

6. So the particular solution is:
$\small{\log\left( \left|x-y \right| \right)~=~x + y + \rm{C}}$

Solved example 25.90
Find a particular solution of the differential equation ${(x+1)\frac{dy}{dx}~=~2 e^{-y}~-~1}$, given that y = 0 when x = 0.
Solution:
1. The variables can be separated and the differential equation can be rewritten:

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

2. Integrating both sides, we get:

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

3. So the general solution is:

$\small{(x+1)\left({e}^{y} - 2 \right)~=~\rm{C}}$

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

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

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

5. So the particular solution is:

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

Solved example 25.91
Find a particular solution of the differential equation ${\log\left(\frac{dy}{dx} \right)~=~3x + 4y}$, given that y = 0 when x = 0.
Solution:
1. The variables can be separated and the differential equation can be rewritten:

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

2. Integrating both sides, we get:

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

3. So the general solution is:

$\small{\frac{\mathrm{e}^{-4y}}{4}~+~\frac{e^{3x}}{3}~=~\rm{C}}$

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

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

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{\mathrm{e}^{-4y}}{4}~+~\frac{e^{3x}}{3}}    & {~=~}&{\rm{C}}    \\
{~\color{magenta} 2    }    &{{\Rightarrow}}    &{\frac{\mathrm{e}^{0}}{4}~+~\frac{e^{0}}{3}}    & {~=~}&{\rm{C}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{1}{4}~+~\frac{1}{3}}    & {~=~}&{\rm{C}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}        &{\rm{C}}    & {~=~}    &{\frac{7}{12}}    \\
\end{array}}$

5. So the particular solution is:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{\mathrm{e}^{-4y}}{4}~+~\frac{e^{3x}}{3}}    & {~=~}&{\frac{7}{12}}    \\
{~\color{magenta} 2    }    &{{\Rightarrow}}    &{3 e^{-4x}~+~4 e^{3x}}    & {~=~}&{7}    \\
\end{array}}$

Solved example 25.92
Solve the differential equation
$\small{(x dy - y dx)y\,\sin\left(\frac{y}{x} \right)~=~(y dx + x dy)x\,\cos\left(\frac{y}{x} \right)}$.
Solution:
Part I: Showing that the given differential equation is homogeneous

1. We can rearrange the given differential equation as:

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

2. Let $\small{F(x,y)~=~\frac{xy\,\cos\left(\frac{y}{x} \right)~+~y^2\,\sin\left(\frac{y}{x} \right) }{xy\,\sin\left(\frac{y}{x} \right)~-~x^2\,\cos\left(\frac{y}{x} \right) }}$

• We need to show that, this is a homogeneous function of degree zero.

(i) We can rearrange F(x,y) as shown below:

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

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

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

• So it is a homogeneous function.

• For this function, 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:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{\left(\frac{y}{x} \right)\cos\left(\frac{y}{x} \right)~+~\left(\frac{y}{x} \right)^2\sin\left(\frac{y}{x} \right) }{\left(\frac{y}{x} \right)\sin\left(\frac{y}{x} \right)~-~\cos\left(\frac{y}{x} \right) }}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{v~+~x \frac{dv}{dx}}    & {~=~}    &{\frac{(v)\cos\left(v \right)~+~\left(v \right)^2\sin\left(v \right) }{\left(v \right)\sin\left(v \right)~-~\cos\left(v \right) }}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{(v)\cos\left(v \right)~+~\left(v \right)^2\sin\left(v \right) }{\left(v \right)\sin\left(v \right)~-~\cos\left(v \right) }~-~v}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{(v)\cos\left(v \right)~+~\left(v \right)^2\sin\left(v \right)~-~\left(v \right)^2\sin\left(v \right)~+~(v)\cos(v) }{\left(v \right)\sin\left(v \right)~-~\cos\left(v \right) }}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{(v)\cos\left(v \right)~+~(v)\cos(v) }{\left(v \right)\sin\left(v \right)~-~\cos\left(v \right) }}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{x \frac{dv}{dx}}    & {~=~}    &{\frac{2(v)\cos\left(v \right)}{\left(v \right)\sin\left(v \right)~-~\cos\left(v \right) }}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}    &{\left[\frac{\left(v \right)\sin\left(v \right)~-~\cos\left(v \right)}{2(v)\cos(v)} \right]dv}    & {~=~}    &{\left[\frac{1}{x} \right]dx}    \\
{~\color{magenta}    8    }    &{{\Rightarrow}}    &{\left[\frac{\tan(v)}{2}~-~\frac{1}{2(v)} \right]dv}    & {~=~}    &{\left[\frac{1}{x} \right]dx}    \\
\end{array}}$

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

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

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

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

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}&{\frac{\sec v}{v x^2}}    & {~=~}    &{\rm{C}_4}    \\
{~\color{magenta}    2    }&{{\Rightarrow}}&{\frac{\sec(y/x)}{(y/x)x^2}}    & {~=~}    &{\rm{C}_4}    \\
{~\color{magenta}    3    }&{{\Rightarrow}}&{\frac{\sec(y/x)}{xy}}    & {~=~}    &{\rm{C}_4}    \\
{~\color{magenta}    4    }&{{\Rightarrow}}&{\sec(y/x)}    & {~=~}    &{{\rm{C}}xy}    \\
\end{array}}$

Solved example 25.93
Verify that the function $\small{y ~=~c_1\,e^{ax}\,\cos(bx)~+~c_2\,e^{ax}\,\sin(bx)}$, where $\small{c_1,~c_2}$ are arbitrary constants is a solution of the differential equation
$\small{\frac{dy}{dx}~-~2a\,\frac{dy}{dx}~+~\left(a^2 + b^2 \right)y~=~0}$.
Solution:
1. Find the derivatives from the given solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{c_1\,e^{ax}\,\cos(bx)~+~c_2\,e^{ax}\,\sin(bx)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{{\frac{dy}{dx}}}    & {~=~}    &{c_1\left[e^{ax}(-\sin bx)(b)+(\cos bx)e^{ax}(a) \right]}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{+~c_2\left[e^{ax}(\cos bx)(b)+(\sin bx)e^{ax}(a) \right]}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{{\frac{dy}{dx}}}    & {~=~}    &{e^{ax}(\sin bx)\left[c_2\,a~-~c_1\,b \right]}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{+e^{ax}(\cos bx)\left[c_2\,b~+~c_1\,a \right]}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{{\frac{d^2y}{dx^2}}}    & {~=~}    &{\left[c_2\,a~-~c_1\,b \right]\left[e^{ax}(\cos bx)(b)+(\sin bx)e^{ax}(a) \right]}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{+\left[c_2\,b~+~c_1\,a \right]\left[e^{ax}(-\sin bx)(b)+(\cos bx)e^{ax}(a) \right]}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{{\frac{d^2y}{dx^2}}}    & {~=~}    &{e^{ax}(\sin bx)\Big[\left[c_2\,a~-~c_1\,b \right]a~-~\left[c_2\,b~+~c_1\,a \right]b \Big] }    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{+~e^{ax}(\cos bx)\Big[\left[c_2\,a~-~c_1\,b \right]b~+~\left[c_2\,b~+~c_1\,a \right]a \Big]}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{{\frac{d^2y}{dx^2}}}    & {~=~}    &{e^{ax}(\sin bx)\Big[c_2\left(a^2 - b^2 \right)~-~2c_1\,ab \Big] }    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{+~e^{ax}(\cos bx)\Big[2c_2\,ab~+~c_1\left(a^2 - b^2 \right) \Big] }    \\
\end{array}}$

2. Let us write $\small{2a\,\frac{dy}{dx}}$:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{e^{ax}(\sin bx)\left[c_2\,a~-~c_1\,b \right]}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{+e^{ax}(\cos bx)\left[c_2\,b~+~c_1\,a \right]}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{{2a\,\frac{dy}{dx}}}    & {~=~}    &{e^{ax}(\sin bx)\left[2c_2\,a^2~-~2c_1\,ab \right]}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{+e^{ax}(\cos bx)\left[2c_2\,ab~+~2c_1\,a^2 \right]}    \\
\end{array}}$

3. Let us write $\small{\left(a^2~+~b^2 \right)y}$:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{c_1\,e^{ax}\,\cos(bx)~+~c_2\,e^{ax}\,\sin(bx)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(a^2~+~b^2 \right)y}    & {~=~}    &{c_1\,a^2\,e^{ax}\,\cos(bx)~+~c_2\,a^2\,e^{ax}\,\sin(bx)}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{+c_1\,b^2\,e^{ax}\,\cos(bx)~+~c_2\,b^2\,e^{ax}\,\sin(bx)}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(a^2~+~b^2 \right)y}    & {~=~}    &{e^{ax}(\sin bx)\left[c_2\,a^2~+~c_2\,b^2 \right]}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{+e^{ax}(\cos bx)\left[c_1\,a^2~+~c_1\,b^2 \right]}    \\
\end{array}}$

4. Let us write all three terms of the differential equation:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{d^2y}{dx^2}}    & {~=~}    &{e^{ax}(\sin bx)\Big[c_2\left(a^2 - b^2 \right)~-~2c_1\,ab \Big]}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{+~e^{ax}(\cos bx)\Big[2c_2\,ab~+~c_1\left(a^2 - b^2 \right) \Big] }    \\
{~\color{magenta}    2    }    &{{}}    &{-2a\,\frac{dy}{dx}}    & {~=~}    &{-e^{ax}(\sin bx)\left[2c_2\,a^2~-~2c_1\,ab \right]}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{-e^{ax}(\cos bx)\left[2c_2\,ab~+~2c_1\,a^2 \right]}    \\
{~\color{magenta}    3    }    &{{}}    &{\left(a^2~+~b^2 \right)y}    & {~=~}    &{e^{ax}(\sin bx)\left[c_2\,a^2~+~c_2\,b^2 \right]}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{+e^{ax}(\cos bx)\left[c_1\,a^2~+~c_1\,b^2 \right]}    \\
\end{array}}$

4. Let us number the similar terms:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{d^2y}{dx^2}}    & {~=~}    &{e^{ax}(\sin bx)\Big[c_2\left(a^2(I) - b^2(II) \right)~-~2c_1\,ab(III) \Big]}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{+~e^{ax}(\cos bx)\Big[2c_2\,ab(IV)~+~c_1\left(a^2(V) - b^2(VI) \right) \Big] }    \\
{~\color{magenta}    2    }    &{{}}    &{-2a\,\frac{dy}{dx}}    & {~=~}    &{-e^{ax}(\sin bx)\left[2c_2\,a^2(I)~-~2c_1\,ab(III) \right]}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{-e^{ax}(\cos bx)\left[2c_2\,ab(IV)~+~2c_1\,a^2(V) \right]}    \\
{~\color{magenta}    3    }    &{{}}    &{\left(a^2~+~b^2 \right)y}    & {~=~}    &{e^{ax}(\sin bx)\left[c_2\,a^2(I)~+~c_2\,b^2(II) \right]}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{+e^{ax}(\cos bx)\left[c_1\,a^2(V)~+~c_1\,b^2(VI) \right]}    \\
\end{array}}$

5. We see that, the terms with the same numbers, cancel each other. So the net sum is zero. That means, L.H.S is equal to R.H.S.

Solved example 25.94
Solve the differential equation
$\small{(\tan^{-1}y~-~x)dy~=~(1~+~y^2)dx}$.
Solution:
1. The given differential equation is:
$\small{(\tan^{-1}y~-~x)dy~=~(1~+~y^2)dx}$

This can be rearranged as: $\small{\frac{dy}{dx}~=~\frac{1~+~y^2}{\tan^{-1}y~-~x}}$

It is not a linear form. So we rearrange in another way: $\small{\frac{dx}{dy}~=~\frac{\tan^{-1}y~-~x}{1~+~y^2}}$

$\small{\Rightarrow \frac{dx}{dy}~=~\frac{\tan^{-1}y}{1~+~y^2}~-~\frac{x}{1~+~y^2}}$

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

• Here P₁ = $\small{\frac{1}{1~+~y^2}}$ and Q1 = $\small{\frac{\tan^{-1}y}{1~+~y^2}}$

• 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

• Here, both P₁ and Q₁ are functions of y. So it is a first order linear differential equation.

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

$\small{\begin{array}{ll}{~\color{magenta}    1    }  &{{}}    &{\int{\left[{\rm{P}_1} \right]dy}}    & {~=~}    &{\int{\left[\frac{1}{1~+~y^2} \right]dy}~=~\tan^{-1}y}    \\
{~\color{magenta} 2    }    &{{\Rightarrow}}    &{e^{\left(\int{\left[{\rm{P}_1} \right]dx} \right)}}    & {~=~}    &{e^{\left(\tan^{-1}y \right)}}    \\
\end{array}}$

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

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

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

We get: $\small{e^{\left(\tan^{-1}y \right)}\left[\tan^{-1}y~-~1 \right]~+~\rm{C}}$

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

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

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The link below gives a few more miscellaneous examples:

Miscellaneous Exercise

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In the next chapter, we will see Vector Algebra.

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