Solving Linear Second Order ODEs for which a Symmetry of the Form [xi=0, eta=F(x)] Can Be Found
Description
Examples
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).
with⁡DEtools,odeadvisor,symgen
odeadvisor,symgen
ode1≔diff⁡y⁡x,x,x=a⁢diff⁡y⁡x,x−1ln⁡x⁢y⁡xx2−aln⁡x⁢y⁡xx
ode1≔ⅆ2ⅆx2y⁡x=a⁢ⅆⅆxy⁡x−y⁡xln⁡x⁢x2−a⁢y⁡xln⁡x⁢x
odeadvisor⁡ode1
_2nd_order,_with_linear_symmetries,_2nd_order,_linear,_with_symmetry_[0,F(x)]
dsolve⁡ode1
y⁡x=∫ⅇa⁢xln⁡x2ⅆx⁢c__1+c__2⁢ln⁡x
A nonhomogeneous ODE example
ode2≔diff⁡y⁡x,x,x=diff⁡F⁡x,x,x⁢y⁡xF⁡x+H⁡x
ode2≔ⅆ2ⅆx2y⁡x=ⅆ2ⅆx2F⁡x⁢y⁡xF⁡x+H⁡x
dsolve⁡ode2,y⁡x
y⁡x=∫1F⁡x2ⅆx⁢F⁡x⁢c__2+F⁡x⁢c__1+F⁡x⁢∫H⁡x⁢F⁡xⅆx⁢∫1F⁡x2ⅆx−∫∫1F⁡x2ⅆx⁢F⁡x⁢H⁡xⅆx
A nonhomogeneous example step by step
ode3≔diff⁡y⁡x,`$`⁡x,2−y⁡x=F⁡x
ode3≔ⅆ2ⅆx2y⁡x−y⁡x=F⁡x
homogeneous_ode≔diff⁡y⁡x,`$`⁡x,2−y⁡x
homogeneous_ode≔ⅆ2ⅆx2y⁡x−y⁡x
Steps 1) and 2) mentioned above
ans_h≔dsolve⁡homogeneous_ode
ans_h≔y⁡x=c__1⁢ⅇ−x+c__2⁢ⅇx
Step 3): two independent solutions for the homogeneous_ode
sol_1≔rhs⁡subs⁡_C1=1,_C2=0,ans_h
sol_1≔ⅇ−x
sol_2≔rhs⁡subs⁡_C1=0,_C2=1,ans_h
sol_2≔ⅇx
Step 4): a procedure for the general solution to the original nonhomogeneous ODE (ode[3]) is given by
ANS≔s1,s2,F↦y⁡x=_C1⋅s2+_C2⋅s1+s2⋅int⁡F⋅s1W⁡s1,s2,x−s1⋅int⁡F⋅s2W⁡s1,s2,x
ANS≔s1,s2,F↦y⁡x=c__1⋅s2+c__2⋅s1+s2⋅∫F⋅s1W⁡s1,s2ⅆx−s1⋅∫F⋅s2W⁡s1,s2ⅆx
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
W≔s1,s2→simplify⁡s1⁢∂∂x⁢s2−s2⁢∂∂x⁢s1
from which the answer to the nonhomogeneous ODE follows
ans≔ANS⁡sol_1,sol_2,F⁡x
ans≔y⁡x=c__1⁢ⅇx+c__2⁢ⅇ−x+ⅇx⁢∫F⁡x⁢ⅇ−x2ⅆx−ⅇ−x⁢∫F⁡x⁢ⅇx2ⅆx
odetest⁡ans,ode3
0
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
Download Help Document