Solving Linear Second Order ODEs for which a Symmetry of the Form [xi=0, eta=F(x)] Can Be Found - Maple Programming Help

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Solving Linear Second Order ODEs for which a Symmetry of the Form [xi=0, eta=F(x)] Can Be Found

 

Description

Examples

Description

• 

All second order linear ODEs have symmetries of the form [xi=0, eta=F(x)]. Actually, F(x) is always a solution of the related homogeneous ODE. There is no general scheme for determining F(x); see dsolve,linear).

• 

When a symmetry of the form [xi=0, eta=F(x)] is found, this information is enough to integrate the homogeneous ODE (see Murphy's book, p. 88).

• 

In the case of nonhomogeneous ODEs, you can do the following:

  

1) look for F(x) as a symmetry of the homogeneous ODE;

  

2) solve the homogeneous ODE using this information;

  

3) set each of _C1 and _C2 equal to 0 and 1 in the answer of the previous step, in order to obtain the two linearly independent solutions of the homogeneous ODE;

  

4) use these two independent solutions of the homogeneous ODE to build the general solution to the nonhomogeneous ODE (see Bluman and Kumei, Symmetries and Differential Equations, p. 132 and ?dsolve,references).

Examples

withDEtools,odeadvisor,symgen

odeadvisor,symgen

(1)

ode1ⅆ2ⅆx2yx=aⅆⅆxyx1yxlnxx2ayxlnxx

ode1:=ⅆ2ⅆx2yx=aⅆⅆxyxyxlnxx2ayxlnxx

(2)

odeadvisorode1

_2nd_order,_with_linear_symmetries,_2nd_order,_linear,_with_symmetry_[0,F(x)]

(3)

dsolveode1

yx=∫ⅇaxlnx2ⅆx_C1+_C2lnx

(4)

A nonhomogeneous ODE example

ode2ⅆ2ⅆx2yx=ⅆ2ⅆx2FxyxFx+Hx

ode2:=ⅆ2ⅆx2yx=ⅆ2ⅆx2FxyxFx+Hx

(5)

dsolveode2,yx

yx=∫1Fx2ⅆxFx_C2+Fx_C1+Fx∫HxFxⅆx∫1Fx2ⅆx∫∫1Fx2ⅆxFxHxⅆx

(6)

A nonhomogeneous example step by step

ode3ⅆ2ⅆx2yxyx=Fx

ode3:=ⅆ2ⅆx2yxyx=Fx

(7)

homogeneous_odeⅆ2ⅆx2yxyx

homogeneous_ode:=ⅆ2ⅆx2yxyx

(8)

Steps 1) and 2) mentioned above

ans_hdsolvehomogeneous_ode

ans_h:=yx=_C1ⅇx+_C2ⅇx

(9)

Step 3): two independent solutions for the homogeneous_ode

sol_1rhssubs_C1=1,_C2=0,ans_h

sol_1:=ⅇx

(10)

sol_2rhssubs_C1=0,_C2=1,ans_h

sol_2:=ⅇx

(11)

Step 4): a procedure for the general solution to the original nonhomogeneous ODE (ode[3]) is given by

ANSs1,s2,F→yx=_C1s2+_C2s1+s2∫Fs1Ws1,s2ⅆxs1∫Fs2Ws1,s2ⅆx

ANS:=s1,s2,F→yx=_C1s2+_C2s1+s2∫Fs1Ws1,s2ⅆxs1∫Fs2Ws1,s2ⅆx

(12)

where s1 and s2 are the linearly independent solutions of the homogeneous ode (sol_1 and sol_2 above), F is the nonhomogeneous term (here represented by F(x)), and W is the Wronskian, in turn given by

Ws1,s2→simplifys1xs2s2xs1

W:=s1,s2→simplifys1xs2s2xs1

(13)

from which the answer to the nonhomogeneous ODE follows

ansANSsol_1,sol_2,Fx

ans:=yx=_C1ⅇx+_C2ⅇx+ⅇx∫12Fxⅇxⅆxⅇx∫12Fxⅇxⅆx

(14)

odetestans,ode3

0

(15)

See Also

DEtools

odeadvisor

dsolve,Lie

quadrature

missing

reducible

linear_ODEs

exact_linear

exact_nonlinear

sym_Fx

linear_sym

Bessel

Painleve

Halm

Gegenbauer

Duffing

ellipsoidal

elliptic

erf

Emden

Jacobi

Hermite

Lagerstrom

Laguerre

Liouville

Lienard

Van_der_Pol

Titchmarsh

odeadvisor,types

 


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