Solution of Differential Equations of First Order and First Degree

Homogeneous Differential Equations

A first-order ordinary differential equation in the form:

𝑀(𝑥, 𝑦)𝑑𝑥 + 𝑁(𝑥, 𝑦)𝑑𝑦 = 0

is a homogeneous type if both functions 𝑀(𝑥, 𝑦)and 𝑁(𝑥, 𝑦)are homogeneous functions of the
same degree 𝑛. A homogeneous differential equation of first order and first degree can be written
in the form

𝑑𝑦 𝑦
= 𝑓( )
𝑑𝑥 𝑥

Working rule for solving homogeneous equations

Let the given equation be homogeneous. Then, by definition, the given equation can be put in the
form

𝑑𝑦 𝑦
= 𝑓 ( ) … … … (1)
𝑑𝑥 𝑥

To solve (1), let 𝑦⁄𝑥 = 𝑣 ⇒ 𝑦 = 𝑣𝑥 … … … (2)

Differentiating (2)w.r.to 𝑥,

𝑑𝑦 𝑑𝑣
=𝑣+𝑥 … … … (3)
𝑑𝑥 𝑑𝑥

Using (2) and (3) in (1), we obtain,

𝑑𝑣
𝑣+𝑥 = 𝑓(𝑣)
𝑑𝑥

𝑑𝑣
⇒𝑥 = 𝑓(𝑣) − 𝑣
𝑑𝑥

𝑑𝑥 𝑑𝑣
⇒ =
𝑥 𝑓(𝑣) − 𝑣

1 Solution of Homogeneous Differential Equations

So the equation is reduced to an equation in which variables are separated. Finally, by integrating
we can easily obtain the solution.

Example 01: Solve

(𝑥 2 + 𝑦 2 )𝑑𝑥 + 2𝑥𝑦𝑑𝑦 = 0

Solution: We have,

(𝑥 2 + 𝑦 2 )𝑑𝑥 = −2𝑥𝑦𝑑𝑦

𝑦2
𝑑𝑦 (𝑥 2 + 𝑦 2 ) 𝑑𝑦 𝑥 2 (1 + 𝑥 2 )
⇒ =− ⇒ =−
𝑑𝑥 2𝑥𝑦 𝑑𝑥 2𝑥𝑦

𝑦2 𝑦 2
𝑑𝑦 𝑥 (1 + 𝑥 2 ) 𝑑𝑦 {1 + (𝑥 ) }
⇒ =− ⇒ =− 𝑦 … … … (1)
𝑑𝑥 2𝑦 𝑑𝑥 2. 𝑥

To solve (1), let 𝑦⁄𝑥 = 𝑣 ⇒ 𝑦 = 𝑣𝑥 … … … (2)

Differentiating (2)w.r.to 𝑥,

𝑑𝑦 𝑑𝑣
=𝑣+𝑥 … … … (3)
𝑑𝑥 𝑑𝑥

Using (2) and (3) in (1), we obtain,

𝑑𝑣 {1 + (𝑣)2 } 𝑑𝑣 (1 + 𝑣 2 )
𝑣+𝑥 =− ⇒𝑥 =− −𝑣
𝑑𝑥 2. 𝑣 𝑑𝑥 2𝑣
𝑑𝑣 −1 − 𝑣 2 − 2𝑣 2 𝑑𝑣 −1 − 3𝑣 2
⇒𝑥 = ⇒𝑥 =
𝑑𝑥 2𝑣 𝑑𝑥 2𝑣
2𝑣 𝑑𝑣 𝑑𝑥
⇒− 2
=
1 + 3𝑣 𝑥
Integrating both sides,

1 6𝑣 𝑑𝑣 𝑑𝑥 1
⇒− ∫ 2
= ∫ ⇒ − ln(1 + 3𝑣 2 ) + ln 𝐶 = ln 𝑥
3 1 + 3𝑣 𝑥 3
1 1
⇒ ln 𝐶 = ln 𝑥 + ln(1 + 3𝑣 2 )3 ⇒ ln 𝐶 = ln 𝑥 (1 + 3𝑣 2 )3
1
𝑦2 3
⇒ 𝐶 = 𝑥 (1 + 3 2 )
𝑥

2 Solution of Homogeneous Differential Equations

which is the required general solution.

Example 02: Solve

(𝑥 3 + 3𝑥𝑦 2 )𝑑𝑥 + (𝑦 3 + 3𝑥 2 𝑦)𝑑𝑦 = 0
Solution: We have,
(𝑥 3 + 3𝑥𝑦 2 )𝑑𝑥 + (𝑦 3 + 3𝑥 2 𝑦)𝑑𝑦 = 0

𝑑𝑦 𝑥 3 + 3𝑥𝑦 2
⇒ =− 3
𝑑𝑥 𝑦 + 3𝑥 2 𝑦
𝑦 2
𝑑𝑦 1+3( ⁄𝑥)
⇒ =− 𝑦 3 𝑦 … … … (1)
𝑑𝑥 ( ⁄𝑥) +3( ⁄𝑥)

To solve (1), let 𝑦⁄𝑥 = 𝑣 ⇒ 𝑦 = 𝑣𝑥 … … … (2)

Differentiating (2)w.r.to 𝑥,

𝑑𝑦 𝑑𝑣
=𝑣+𝑥 … … … (3)
𝑑𝑥 𝑑𝑥
𝑑𝑣 1+3𝑣 2
Using (2) and (3) in (1), we obtain, 𝑣 + 𝑥 𝑑𝑥 = − 𝑣3 +3𝑣

𝑑𝑣 1 + 3𝑣 2 𝑣 4 + 6𝑣 2 + 1
⇒𝑥 =− 3 −𝑣 =−
𝑑𝑥 𝑣 + 3𝑣 𝑣 3 + 3𝑣

𝑑𝑥 4𝑣 3 + 12𝑣
⇒4 =− 4 𝑑𝑣
𝑥 𝑣 + 6𝑣 2 + 1

Integrating, 4𝑙𝑜𝑔𝑥 = −𝑙𝑜𝑔(𝑣 4 + 6𝑣 2 + 1) + 𝑙𝑜𝑔𝑐

⇒ 𝑙𝑜𝑔𝑥 4 = 𝑙𝑜𝑔[𝑐/(𝑣 4 + 6𝑣 2 + 1)]

⇒ 𝑥 4 (𝑣 4 + 6𝑣 2 + 1) = 𝑐

⇒ 𝑦 4 + 6𝑥 2 𝑦 2 + 𝑥 4 = 𝑐, as 𝑦⁄𝑥 = 𝑣

Therefore, 𝑦 4 + 6𝑥 2 𝑦 2 + 𝑥 4 = 𝑐 is the required solution.

Example 03: Solve

𝑥𝑐𝑜𝑠(𝑦⁄𝑥)(𝑦𝑑𝑥 + 𝑥𝑑𝑦) = 𝑦𝑠𝑖𝑛(𝑦⁄𝑥)(𝑥𝑑𝑦 − 𝑦𝑑𝑥)

3 Solution of Homogeneous Differential Equations

Solution: We have,

𝑥𝑐𝑜𝑠(𝑦⁄𝑥)(𝑦𝑑𝑥 + 𝑥𝑑𝑦) = 𝑦𝑠𝑖𝑛(𝑦⁄𝑥)(𝑥𝑑𝑦 − 𝑦𝑑𝑥)

⇒ 𝑥𝑐𝑜𝑠(𝑦⁄𝑥 )𝑦𝑑𝑥 + 𝑥𝑐𝑜𝑠(𝑦⁄𝑥)𝑥𝑑𝑦 = 𝑦𝑠𝑖𝑛(𝑦⁄𝑥)𝑥𝑑𝑦 − 𝑦𝑠𝑖𝑛(𝑦⁄𝑥)𝑦𝑑𝑥

⇒ 𝑥𝑐𝑜𝑠(𝑦⁄𝑥 )𝑦𝑑𝑥 + 𝑦𝑠𝑖𝑛(𝑦⁄𝑥)𝑦𝑑𝑥 = 𝑦𝑠𝑖𝑛(𝑦⁄𝑥)𝑥𝑑𝑦 − 𝑥𝑐𝑜𝑠(𝑦⁄𝑥)𝑥𝑑𝑦

⇒ {𝑥𝑐𝑜𝑠(𝑦⁄𝑥) + 𝑦𝑠𝑖𝑛(𝑦⁄𝑥)}𝑦𝑑𝑥 = {𝑦𝑠𝑖𝑛(𝑦⁄𝑥) − 𝑥𝑐𝑜𝑠(𝑦⁄𝑥)}𝑥𝑑𝑦

𝑦{𝑥𝑐𝑜𝑠(𝑦⁄𝑥) + 𝑦𝑠𝑖𝑛(𝑦⁄𝑥)} 𝑑𝑦
⇒ =
𝑥{𝑦𝑠𝑖𝑛(𝑦⁄𝑥) − 𝑥𝑐𝑜𝑠(𝑦⁄𝑥)} 𝑑𝑥
𝑦
𝑑𝑦 𝑦𝑥 {𝑐𝑜𝑠(𝑦⁄𝑥) + 𝑥 𝑠𝑖𝑛(𝑦⁄𝑥)}
⇒ =
𝑑𝑥 𝑥 2 {𝑦 𝑠𝑖𝑛(𝑦⁄𝑥) − 𝑐𝑜𝑠(𝑦⁄𝑥)}
𝑥

𝑦 𝑦
𝑑𝑦 𝑥 {𝑐𝑜𝑠(𝑦⁄𝑥) + 𝑥 𝑠𝑖𝑛(𝑦⁄𝑥)}
⇒ = 𝑦 … … … (1)
𝑑𝑥 {𝑥 𝑠𝑖𝑛(𝑦⁄𝑥) − 𝑐𝑜𝑠(𝑦⁄𝑥)}

To solve(1), let 𝑦⁄𝑥 = 𝑣 ⇒ 𝑦 = 𝑣𝑥 … … … (2)

Differentiating (2)w.r.to 𝑥,

𝑑𝑦 𝑑𝑣
=𝑣+𝑥 … … … (3)
𝑑𝑥 𝑑𝑥

Using (2) and (3) in (1), we obtain,

𝑑𝑣 𝑣(𝑐𝑜𝑠𝑣 + 𝑣𝑠𝑖𝑛𝑣)
𝑣+𝑥 =
𝑑𝑥 𝑣𝑠𝑖𝑛𝑣 − 𝑐𝑜𝑠𝑣
𝑑𝑣 𝑣(𝑐𝑜𝑠𝑣 + 𝑣𝑠𝑖𝑛𝑣)
⇒𝑥 = −𝑣
𝑑𝑥 𝑣𝑠𝑖𝑛𝑣 − 𝑐𝑜𝑠𝑣
𝑑𝑣 𝑣𝑐𝑜𝑠𝑣 + 𝑣 2 𝑠𝑖𝑛𝑣 − 𝑣 2 𝑠𝑖𝑛𝑣 + 𝑣𝑐𝑜𝑠𝑣
⇒𝑥 =
𝑑𝑥 𝑣𝑠𝑖𝑛𝑣 − 𝑐𝑜𝑠𝑣
𝑑𝑣 2𝑣𝑐𝑜𝑠𝑣
⇒𝑥 =
𝑑𝑥 𝑣𝑠𝑖𝑛𝑣 − 𝑐𝑜𝑠𝑣
𝑣𝑠𝑖𝑛𝑣 − 𝑐𝑜𝑠𝑣 2𝑑𝑥
⇒ 𝑑𝑣 =
𝑣𝑐𝑜𝑠𝑣 𝑥
𝑣𝑠𝑖𝑛𝑣 𝑐𝑜𝑠𝑣 2𝑑𝑥
⇒( − ) 𝑑𝑣 =
𝑣𝑐𝑜𝑠𝑣 𝑣𝑐𝑜𝑠𝑣 𝑥

4 Solution of Homogeneous Differential Equations

 1 2𝑑𝑥
⇒ (𝑡𝑎𝑛𝑣 − ) 𝑑𝑣 =
𝑣 𝑥
Integrating both sides,

1 𝑑𝑥
∫ 𝑡𝑎𝑛𝑣𝑑𝑣 − ∫ 𝑑𝑣 = 2 ∫
𝑣 𝑥

⇒ 𝑙𝑛𝑠𝑒𝑐𝑣 − 𝑙𝑛𝑣 + 𝑙𝑛𝐶 = 2𝑙𝑛𝑥

𝐶. 𝑠𝑒𝑐𝑣
⇒ 𝑙𝑛 = 𝑙𝑛𝑥 2
𝑣
𝐶. 𝑠𝑒𝑐𝑣
⇒ = 𝑥2
𝑣
⇒ 𝐶. 𝑠𝑒𝑐𝑣 = 𝑥 2 𝑣

⇒ 𝐶. 𝑠𝑒𝑐(𝑦⁄𝑥) = 𝑥 2 . 𝑦⁄𝑥

1
⇒𝐶 = 𝑥𝑦
𝑐𝑜𝑠(𝑦⁄𝑥)

⇒ 𝑥𝑦𝑐𝑜𝑠(𝑦⁄𝑥) = 𝐶

which is the required general solution.

Exercise from homogeneous equations:

1. Solve the following differential equation:

𝑑𝑦 √𝑥 2 − 𝑦 2 + 𝑦
=
𝑑𝑥 𝑥
𝑦
𝐴𝑛𝑠𝑤𝑒𝑟: 𝑠𝑖𝑛−1 = 𝑙𝑛𝑥 + 𝐶
𝑥
2. Solve
𝑑𝑦 𝑦 𝑦
= + 𝑡𝑎𝑛
𝑑𝑥 𝑥 𝑥

𝑦
𝐴𝑛𝑠𝑤𝑒𝑟: 𝑥 = 𝑐𝑠𝑖𝑛( ⁄𝑥).

3. Solve:
𝑑𝑦
(𝑥 2 + 𝑦 2 ) = 𝑥𝑦
𝑑𝑥
𝑥2
𝐴𝑛𝑠𝑤𝑒𝑟: − 2 + ln 𝑦 = 𝐶
2𝑦

5 Solution of Homogeneous Differential Equations

4. Solve
𝑥𝑑𝑦 − 𝑦𝑑𝑥 = √𝑥 2 + 𝑦 2 𝑑𝑥
𝐴𝑛𝑠𝑤𝑒𝑟: 𝑥 2 𝐶 = 𝑦 + √𝑥 2 + 𝑦 2

5. Solve (𝑥 + 𝑦)𝑑𝑦 + (𝑥 − 𝑦)𝑑𝑥 = 0

𝑦 1
Answer: tan−1 (𝑥 ) + 2 ln(𝑥 2 + 𝑦 2 ) = 𝑐.

6 Solution of Homogeneous Differential Equations