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Euler-Cauchy Equation

Muhammad Nadeem

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Differential

Equations

ODE PDE

Second orderFirst order

LinearExact

Separable

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NonhomogeneousHomogeneous

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Second Order Homogeneous linear

equation: Euler-Cauchy Equation0

2=++ byyaxyx (1)

m

xy =Let its solution isSo (1) will be

=m

0)1(122

=++ mmm

bxamxxxmmx

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Or

It is called characteristic or auxiliary equation. It will have two roots say

(2)

2

4)1()1(2

4)1()1(2

2

2

1baamandbaam =+=

These roots can be real and distinct, real double, or complex conjugate

0)1(2

=++ bmam

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Case I

Real and Distinct Roots

If pm =1 and qm =2So there will be two linearly independent solutions of equation 1

p=

q=

And General solutionqp

xcxcy 21 +=

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Example

Solution

Find the G.S. of the ODE 022

=+ yyxyx

mxy =

0}2)1({ =+m

xmmm

Let its solution is

So (1) will be 02)1(122

=+

mmm

xxmxxmmx

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21 =m 22 =mSo general solution

2

2

2

1

+= xcxcy

022 =m

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Example

Solution

Find the G.S. of the ODE

mxy =

0}312)1(4{ =++m

xmmm

Let its solution is

So (1) will be 0312)1(4122

=++

mmm

xxmxxmmx

031242

=++ yyxyx

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2/11 =m 2/32 =m

So general solution2/3

2

2/1

1

+= xcxcy

0384 2 =++ mm

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Case II

Real Double RootIf 2/)1(21 amm ==

So there is only one solution2/)1(

1

axy

=

The second linearly independent solution (needed for basis)can be found by Reduction of Order. Let

Not G.S.!

=

2y

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112 yuyuy += 1112 2 yuyuyuy ++=

Since is solution of Eq. (1) so

0)()2(111111

2=+++++ buyyuyuaxyuyuyux

2

y

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0)()2( 1112

111

2=+++++ byyaxyxuayyxxuyux

2/)1(

1

xaxy

=Since is solution of eq. (1) so

0111

2=++ byyaxyx 1112 yayyx =+and

So 0112

=+ xyuyux

0Since 2 =

2/)1(

2)(ln

a

xxy

=

And General solution of equation (1) will be

( ) 2/)1(212211 lna

xxccycycy

+=+=

Or xu ln=Hence second lineally independent solution of eq. (1) is

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Example

Solution

Find the G.S. of the ODE

mxy =

0}97)1({ =++m

xmmm

Let its solution is

So (1) will be 097)1(

122

=++

mmm

xxmxxmmx

0972

=++ yyxyx

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321 == mm

So general solution3

21 )ln(

+= xxccy

0962 =++ mm

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Example

Solution

Find the G.S. of the ODE

mxy =

0}2524)1(4{ =++m

xmmm

Let its solution is

So (1) will be 02524)1(4

122

=++

mmm

xxmxxmmx

0252442

=++ yyxyx

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2/521 == mm

So general solution2/5

21 )ln(

+= xxccy

025204 2 =++ mm

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Case III

Complex Conjugate RootsIf iqpmiqpm =+= 21 ,

Now we can find real solutions as

ln1 pxiqpiqpm +

21 ,mm

xxSo two complex solutions

And complex general solution iqpiqp xcxcy + += 21

xqxxx

y pmm

lncos2

21

1

=+

=

)lnsinln(cosln2 xqixqxexxx pxiqpiqp

m===

xqxi

xxy p

mm

lnsin2

21

2

=

=

So real general solution is

( )xqcxqcxy p lnsinlncos 21 +=Muhammad Nadeem

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Example

Solution

Find the G.S. of the ODE

mxy =

0}1)1({ =++m

xmmm

Let its solution is

So (1) will be 0)1(

122

=++

mmm

xxmxxmmx

02

=++ yyxyx

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im =1

So general solution

012

=+m

im =2

( )xqcxqcxy p lnsinlncos 21 +=

( )xcxcy lnsinlncos 21 +=

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Example

Solution

Find the G.S. of the ODE

mxy =

0}137)1({ =++m

xmmm

Let its solution is

So (1) will be 0137)1(

122

=++

mmm

xxmxxmmx

01372

=++ yyxyx

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im 231 +=

So general solution

01362

=++ mm

im 232 =

( )xqcxqcxy p lnsinlncos 21 +=

( )xcxcxy ln2sinln2cos 213

+=