Cauchy – Euler Homogeneous Differential
Equations
21UMT2T431: Ordinary Differential Equations And Real Analysis – I
Dr. Soya Mathew
Assistant Professor in Mathematics
Department of Physical Sciences
Kristu Jayanti College (Autonomous)
Bengaluru – 560077, India
Cauchy – Euler Differential Equations
A Linear Differential Equation of the form
𝒙𝒏 𝒅𝒏𝒚
𝒅𝒙𝒏 + 𝒂𝟏𝒙𝒏−𝟏 𝒅𝒏−𝟏𝒚
𝒅𝒙𝒏−𝟏 + ⋯ + 𝒂𝒏𝒚 = 𝒇 𝒙
where 𝑎1, 𝑎2, … 𝑎𝑛 are constants and 𝑓 𝑥 is a known function of the 𝑥
is called a called Cauchy – Euler Homogeneous Differential Equation of order 𝑛.
Working Rule
Given 𝑥𝑛
𝐷𝑛
+ 𝑎1𝑥𝑛−1
𝐷𝑛−1
+ ⋯ + 𝑎𝑛 𝑦 = 𝑓 𝑥 where 𝐷 =
𝑑
𝑑𝑥
Put 𝑥 = 𝑒𝑧
⟹ 𝑧 = log 𝑥
Then 𝑥𝐷 = 𝐷1 where 𝐷1 =
𝑑
𝑑𝑧
𝑥2𝐷2 = 𝐷1(𝐷1 − 1)
𝑥3
𝐷3
= 𝐷1(𝐷1 − 1)(𝐷1 − 2) and so on.
Hence the equation becomes, 𝐹 𝐷1 𝑦 = 𝑓1 𝑧
where 𝑓1(𝑧) is the value of 𝑓(𝑥) in terms of 𝑧.
Problem: Solve 𝟒𝒙𝟐 𝒅𝟐𝒚
𝒅𝒙𝟐 + 𝟒𝒙
𝒅𝒚
𝒅𝒙
− 𝒚 = 𝟒𝒙𝟐
Solution:
Given 4𝑥2
𝑦′′
+ 4𝑥𝑦′
− 𝑦 = 4𝑥2
⟹ (4𝑥2
𝐷2
+ 4𝑥𝐷 − 1)𝑦 = 4𝑥2
… … 1 where 𝐷 =
𝑑
𝑑𝑥
This is a second order Cauchy – Euler equation.
Put 𝑥 = 𝑒𝑧
⟹ 𝑧 = log 𝑥
Then 𝑥𝐷 = 𝐷1, 𝑥2
𝐷2
= 𝐷1 𝐷1 − 1 , where 𝐷1 =
𝑑
𝑑𝑧
Problem: Solve 𝟒𝒙𝟐 𝒅𝟐𝒚
𝒅𝒙𝟐 + 𝟒𝒙
𝒅𝒚
𝒅𝒙
− 𝒚 = 𝟒𝒙𝟐
(Cont…)
1 ⟹ 4𝐷1 𝐷1 − 1 + 4𝐷1 − 1 𝑦 = 4 𝑒2𝑧
⟹ 4𝐷1
2
− 4𝐷1 + 4𝐷1 − 1 𝑦 = 4 𝑒2𝑧
⟹ 4𝐷1
2
− 1 𝑦 = 4 𝑒2𝑧
Auxiliary Equation (AE) is 4𝑚2
− 1 = 0
⟹ 2𝑚 + 1 2𝑚 − 1 = 0
Problem: Solve 𝟒𝒙𝟐 𝒅𝟐𝒚
𝒅𝒙𝟐 + 𝟒𝒙
𝒅𝒚
𝒅𝒙
− 𝒚 = 𝟒𝒙𝟐
(Cont…)
⟹ 𝑚 = −
1
2
,
1
2
The complementary function is 𝑐1𝑒−
𝑧
2 + 𝑐2𝑒
𝑧
2.
⟹ 𝑐1𝑒−
1
2
log 𝑥
+ 𝑐2𝑒
1
2
log 𝑥
⟹
𝑐1
𝑥
+ 𝑐2 𝑥 , since 𝑒log 𝑥
= 𝑥
Problem: Solve 𝟒𝒙𝟐 𝒅𝟐𝒚
𝒅𝒙𝟐 + 𝟒𝒙
𝒅𝒚
𝒅𝒙
− 𝒚 = 𝟒𝒙𝟐
(Cont…)
Now, Particular integral,
𝑃. 𝐼 =
1
4𝐷1
2 − 1
4 𝑒2𝑧
⟹ 𝑃. 𝐼 = 4
1
15
𝑒2𝑧
⟹ 𝑃. 𝐼 =
4
15
𝑥2
Problem: Solve 𝟒𝒙𝟐 𝒅𝟐𝒚
𝒅𝒙𝟐 + 𝟒𝒙
𝒅𝒚
𝒅𝒙
− 𝒚 = 𝟒𝒙𝟐
(Cont…)
Hence the complete solution is
𝑦 = 𝐶. 𝐹 + 𝑃. 𝐼
⟹ 𝑦 =
𝑐1
𝑥
+ 𝑐2 𝑥 +
4
15
𝑥2
Cauchy Euler DE.pptx

Cauchy Euler DE.pptx

  • 1.
    Cauchy – EulerHomogeneous Differential Equations 21UMT2T431: Ordinary Differential Equations And Real Analysis – I Dr. Soya Mathew Assistant Professor in Mathematics Department of Physical Sciences Kristu Jayanti College (Autonomous) Bengaluru – 560077, India
  • 2.
    Cauchy – EulerDifferential Equations A Linear Differential Equation of the form 𝒙𝒏 𝒅𝒏𝒚 𝒅𝒙𝒏 + 𝒂𝟏𝒙𝒏−𝟏 𝒅𝒏−𝟏𝒚 𝒅𝒙𝒏−𝟏 + ⋯ + 𝒂𝒏𝒚 = 𝒇 𝒙 where 𝑎1, 𝑎2, … 𝑎𝑛 are constants and 𝑓 𝑥 is a known function of the 𝑥 is called a called Cauchy – Euler Homogeneous Differential Equation of order 𝑛.
  • 3.
    Working Rule Given 𝑥𝑛 𝐷𝑛 +𝑎1𝑥𝑛−1 𝐷𝑛−1 + ⋯ + 𝑎𝑛 𝑦 = 𝑓 𝑥 where 𝐷 = 𝑑 𝑑𝑥 Put 𝑥 = 𝑒𝑧 ⟹ 𝑧 = log 𝑥 Then 𝑥𝐷 = 𝐷1 where 𝐷1 = 𝑑 𝑑𝑧 𝑥2𝐷2 = 𝐷1(𝐷1 − 1) 𝑥3 𝐷3 = 𝐷1(𝐷1 − 1)(𝐷1 − 2) and so on. Hence the equation becomes, 𝐹 𝐷1 𝑦 = 𝑓1 𝑧 where 𝑓1(𝑧) is the value of 𝑓(𝑥) in terms of 𝑧.
  • 4.
    Problem: Solve 𝟒𝒙𝟐𝒅𝟐𝒚 𝒅𝒙𝟐 + 𝟒𝒙 𝒅𝒚 𝒅𝒙 − 𝒚 = 𝟒𝒙𝟐 Solution: Given 4𝑥2 𝑦′′ + 4𝑥𝑦′ − 𝑦 = 4𝑥2 ⟹ (4𝑥2 𝐷2 + 4𝑥𝐷 − 1)𝑦 = 4𝑥2 … … 1 where 𝐷 = 𝑑 𝑑𝑥 This is a second order Cauchy – Euler equation. Put 𝑥 = 𝑒𝑧 ⟹ 𝑧 = log 𝑥 Then 𝑥𝐷 = 𝐷1, 𝑥2 𝐷2 = 𝐷1 𝐷1 − 1 , where 𝐷1 = 𝑑 𝑑𝑧
  • 5.
    Problem: Solve 𝟒𝒙𝟐𝒅𝟐𝒚 𝒅𝒙𝟐 + 𝟒𝒙 𝒅𝒚 𝒅𝒙 − 𝒚 = 𝟒𝒙𝟐 (Cont…) 1 ⟹ 4𝐷1 𝐷1 − 1 + 4𝐷1 − 1 𝑦 = 4 𝑒2𝑧 ⟹ 4𝐷1 2 − 4𝐷1 + 4𝐷1 − 1 𝑦 = 4 𝑒2𝑧 ⟹ 4𝐷1 2 − 1 𝑦 = 4 𝑒2𝑧 Auxiliary Equation (AE) is 4𝑚2 − 1 = 0 ⟹ 2𝑚 + 1 2𝑚 − 1 = 0
  • 6.
    Problem: Solve 𝟒𝒙𝟐𝒅𝟐𝒚 𝒅𝒙𝟐 + 𝟒𝒙 𝒅𝒚 𝒅𝒙 − 𝒚 = 𝟒𝒙𝟐 (Cont…) ⟹ 𝑚 = − 1 2 , 1 2 The complementary function is 𝑐1𝑒− 𝑧 2 + 𝑐2𝑒 𝑧 2. ⟹ 𝑐1𝑒− 1 2 log 𝑥 + 𝑐2𝑒 1 2 log 𝑥 ⟹ 𝑐1 𝑥 + 𝑐2 𝑥 , since 𝑒log 𝑥 = 𝑥
  • 7.
    Problem: Solve 𝟒𝒙𝟐𝒅𝟐𝒚 𝒅𝒙𝟐 + 𝟒𝒙 𝒅𝒚 𝒅𝒙 − 𝒚 = 𝟒𝒙𝟐 (Cont…) Now, Particular integral, 𝑃. 𝐼 = 1 4𝐷1 2 − 1 4 𝑒2𝑧 ⟹ 𝑃. 𝐼 = 4 1 15 𝑒2𝑧 ⟹ 𝑃. 𝐼 = 4 15 𝑥2
  • 8.
    Problem: Solve 𝟒𝒙𝟐𝒅𝟐𝒚 𝒅𝒙𝟐 + 𝟒𝒙 𝒅𝒚 𝒅𝒙 − 𝒚 = 𝟒𝒙𝟐 (Cont…) Hence the complete solution is 𝑦 = 𝐶. 𝐹 + 𝑃. 𝐼 ⟹ 𝑦 = 𝑐1 𝑥 + 𝑐2 𝑥 + 4 15 𝑥2