Cauchy Eular Differential Equation

1. 𝑎0 𝑥
𝑛
𝑑 𝑛 𝑦
𝑑𝑥 𝑛
+ 𝑎. 𝑥 𝑛−1
𝑑 𝑛−1𝑦
𝑑𝑛−1
+ ⋯ + 𝑎 𝑛−1 𝑥
𝑑𝑦
𝑑𝑥
+ 𝑎 𝑛 𝑦 = 𝑓(𝑥)
CAUCHY-EULAR DIFFRENTIAL EQUATION
Cauchy–Euler equation s a linear homogeneous ordinary differential
equation with variable coefficients. It is sometimes referred to as
an equidimensional equation. Because of its particularly simple equidimensional
structure the differential equation can be solved explicitly.

2. 𝑎0 𝑥
𝑛
𝑑 𝑛 𝑦
𝑑𝑥 𝑛
+ 𝑎. 𝑥 𝑛−1
𝑑 𝑛−1𝑦
𝑑𝑛−1
+ ⋯ + 𝑎 𝑛−1 𝑥
𝑑𝑦
𝑑𝑥
+ 𝑎 𝑛 𝑦 = 𝑓(𝑥)
𝑎0 𝑥
𝑛
𝑑 𝑛 𝑦
𝑑𝑥 𝑛
+ 𝑎. 𝑥 𝑛−1
𝑑 𝑛−1𝑦
𝑑𝑛−1
+ ⋯ + 𝑎 𝑛−1 𝑥
𝑑𝑦
𝑑𝑥
+ 𝑎 𝑛 𝑦 = 𝑓(𝑥)
• This equation is called Cauchy Euler equation
• We can convert this equation into linear
• equation by substituting x= 𝒆 𝒕
= 𝒍𝒏𝒙 = 𝒕
•
1
𝑥
=
𝑑𝑡
𝑑𝑥
•
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑡
.
𝑑𝑡
𝑑𝑥
=
𝑑𝑦
𝑑𝑡
.
1
𝑥
• =>
𝑥𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑡
= ∆𝑦
𝑎0 𝑥
𝑛
𝑑 𝑛 𝑦
𝑑𝑥 𝑛
+ 𝑎. 𝑥 𝑛−1
𝑑 𝑛−1𝑦
𝑑𝑛−1
+ ⋯ + 𝑎 𝑛−1 𝑥
𝑑𝑦
𝑑𝑥
+ 𝑎 𝑛 𝑦 = 𝑓(𝑥)

3. •
𝑑𝑦
𝑑𝑥
=
1
𝑥
𝑑𝑦
𝑑𝑡
•
𝑑2 𝑦
𝑑𝑥2 =
1
𝑥
𝑑2
𝑑𝑡2 .
𝑑𝑡
𝑑𝑥
+
𝑑𝑦
𝑑𝑡
−1
𝑥2
•
𝑑2 𝑦
𝑑𝑥2 =
1
𝑥
𝑑2
𝑑𝑡2 .
1
𝑥
−
𝑑𝑦
𝑑𝑡
1
𝑥2
•
𝑑2 𝑦
𝑑𝑥2 =
1
𝑥2 (
𝑑2 𝑦
𝑑𝑥2 −
𝑑𝑦
𝑑𝑡
)
𝑥2 𝑑𝑦
𝑑𝑥2 = ( ∆2 − ∆)𝑦
∆ ∆ − 1 𝑦

𝑥3𝑑3
𝑦
𝑑𝑥3 =
∆ ∆ − 1 (∆ −

4. PROBLEM:
𝑥2
𝑑2 𝑦
𝑑𝑥2
+ 7𝑥
𝑑𝑦
𝑑𝑥
+ 5𝑦 = 𝑥5
Solution: x= 𝑒 𝑡
⇒ 𝑙𝑛𝑥 = 𝑡
So,
𝑑𝑡
𝑑𝑥
=
1
𝑥
Now we know the values already, therefore,
𝑥2 𝑑2 𝑦
𝑑𝑥2 =∆ ∆ − 1 𝑦
𝑥
𝑑𝑦
𝑑𝑥
= ∆∆𝑦 ∆

5. ⇒ ∆(∆ − 1)𝑦 + 7 ∆𝑦 + 5𝑦 = 𝑒5𝑡
⇒ (∆2 −∆ + 7∆ + 5)𝑦 = 𝑒5𝑥
(∆2= 6∆ + 5 )𝑦 = 𝑒5𝑥

6. Remaining Problem