Homogenous Differential Equations

1 Homogenous Equation

Suppose

a_n

	dny
	dxn

+ ⋯ + a₁

	dy
	dx

+ a₀ y ...= ...f ( x )

is an n^th order linear differential equation (l.d.e.). The equation is homogenous ...if ...f ( x ) = 0 , otherwise it is inhomogenous.

2 Second Order Homogenous Equations

a

	d2y
	dx2

+ b

	dy
	dx

+ c y ...= ...0 ..................a, b, c ...constants.

Try ...y = e^mx ...then ...a m² e^mx + b m e^mx + c e^mx ...= ...0 ......⇒ ......a m² + b m + c ...= ...0

m ...= ...

	–b ± √ b² – 4ac
	2a

In general there will be two roots for m , ...p and q say. The general solution of the equation is ...y ...= ...A e ^px + B e ^qx

2.1 Auxiliary Equations

The equation, ...a m² + b m + c ...= ...0 , ...is called the auxiliary equation or indicial equation.

Example 1

	d2y
	dx2

–

	dy
	dx

– 2 y ...= ...0

y = e^mx ...is a solution if ...m² – m – 2 ...= ...0 ...⇔ ...( m + 1 ) ( m – 2 ) ...= ...0
i.e. the general soultion is ...y ...= ...A e ^–x + B e ^2x

Example 2

	d2y
	dx2

– k² y ...= ...0

y = e^mx ...is a solution if ...m² – k² ...= ...0 ...⇔ ...m ...= ...± k .
The general soultion is ...y ...= ...A e ^{k x} + B e ^{– k x}

Example 3

	d2y
	dx2

+ k² y ...= ...0

y = e^mx ...is a solution if ...m² + k² ...= ...0 ...⇔ ...m ...= ...± i k . ...The general soultion is

y ...= ...A e ^{i k x} + B e ^{– i k x} ...= ...( A + B ) cos k x ...+ ...i ( A – B ) sin k x

Note that this is a more general version of the solution , we found earlier, (where k = 1). If the initial conditions require that the solution be real, then the general (real) solution is :

y ...= ...C cos k x ...+ ...D sin k x , ......C, D ∊ IR

where ...A = ^C/₂ – i ^D/₂ ...and ...B = ^C/₂ + i ^D/₂ . ...Of course this is an additional constraint on the constants.

2.2 Concurrent Roots

If ...b² = 4ac ...then the two roots will be equal: ...p = q.

So the 'general' solution becomes ...y ...= ...A e ^px + B e ^px ...= ...C e ^px , ...i.e. only one constant.

But in this case ...y = x e ^px ...is also a solution, since ...p = –b ⁄ 2a , ...and

y ...= ...x e ^px ......⇒ ......y' ...= ...x pe ^px + e ^px = ( p x + 1 ) e ^px

⇒ ............y'' ...= ...p e ^px + ( p x + 1 ) p e ^px ...= ...( p² x + 2 p ) e ^px

So substituting these values in the differential equation:

a y'' + b y' + c y ...= ...( ( a p² + b p + c ) x + 2 a p + b ) e ^px

but ...a p² + b p + c ...= ...0 , ...and ...2 a p ...= ...2 a × –b ⁄ 2 a ...= ...-b , ...proving that ...y = x e ^px ...is a solution.

The general solution is therefore: .........y ...= ...A e ^px + B x e ^px

Example 4

	d2y
	dx2

+ 2

	dy
	dx

+ y ...= ...0

The auxiliary equation is ...m² + 2m + 1 ...= ...0 , ...i.e. ...( m – 1 )² ...= ...0 ......⇒ ......p = q = 1

The general solution is therefore: .........y ...= ...A e ^x + B x e ^x

3 General Solution of Second Order Homogenous Equations

By dividing through by a, a second order equation can be written in the form

	d2y
	dx2

+ 2d

	dy
	dx

+ k y ...= ...0

If ...k > 0, ...and ...d ² < k ...then the roots of the auxiliary equation are imaginary, and the general solution is

y ...= ...A exp { ( –d + i √ k – d ² ...) x } ...+ ...B exp { ( –d – i √ k – d ² ...) x }

......= ...e^–dx [ A exp { i x √ k – d ² } ...+ ...B exp { – i x √ k – d ² } ]

......= ...e^–dx [ ( A + B ) cos ( x √ k – d ² ) ...+ ...( A – B ) i sin ( x √ k – d ² ) ]

In this case the general real valued solution is

y ...= ...e^–dx [ C cos ( x √ k – d ² ) ...+ ...D sin ( x √ k – d ² ) ]

......= ...e^–dx ...E ...sin ( x √ k – d ² ...+ ...φ ) , ......where ...E = √ C ² + D ² , ...φ = sin^–1 C / E

If ...k > 0, ...and ...d ² > k ...then the roots of the auxiliary equation are real, and the general solution is

y ...= ...A exp { ( –d + √ d ² – k ...) x } ...+ ...B exp { ( –d – √ d ² – k ...) x }

......= ...e^–dx [ A exp { x √ d ² – k } ...+ ...B exp { – x √ d ² – k } ]

If ...k > 0, ...and ...d ² = k ...then the roots are concurrent, and the general solution is

y ...= ...e^–dx [ A + Bx ]

4 Higher Order Homogenous Equations

We can generalize the above by stating that the equation

a_n

	dny
	dxn

+ ⋯ + a₁

	dy
	dx

+ a₀ y ...= ...0

will have a general solution ......A₁ exp ( m₁ x ) + ⋯ + A_n exp ( m_n x ) , ......where ...m₁, ⋯ m_n ...are the separate roots of the auxiliary equation

a₁ m₁² + ⋯ + a_n m_n² ...= ...0

If this equation does not have n separate roots, say for example ...m_i = m_j , ...then the general solution becomes:

A₁ exp ( m₁ x ) + ⋯ + A_i exp ( m_i x ) + ⋯ + A_j x exp ( m_i x ) + ⋯ + A_n exp ( m_n x )

and if ...m_i = m_j = m_k :

A₁ exp ( m₁ x ) + ⋯ + A_i exp ( m_i x ) + ⋯ + A_j x exp ( m_i x ) + ⋯ + A_k x² exp ( m_i x ) + ⋯ + A_n exp ( m_n x )

etcetera.

 

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