Solving differential equation

Question

I want to solve the following differential equation

DSolve[y''[x] == (λ y[x])/Sqrt[-1 + x], y[x],x]

But do not know how to actually solve it. Any suggestion?

Answer 1

Mathematica can also solve this differential equation if you help it with the choice of variable substitution:

Clear[y,λ,x,z,u,v]
eqn = y''[x] == λ y[x]/Sqrt[-1 + x]
(* ==> (y^\[Prime]\[Prime])[x] == (λ y[x])/Sqrt[-1 + x] *)

u[x_] := Sqrt[x - 1]
y[x_] := z[u[x]]
eqn2 = Simplify[eqn /. x -> InverseFunction[u][v], v > 0]
(*
==> 4 v^2 λ z[v] + Derivative[1][z][v] == 
 v (z^\[Prime]\[Prime])[v]
*)

zSolution[v_] = z[v] /. First[DSolve[eqn2, z[v], v]]
(*
==> AiryAiPrime[2^(2/3) v λ^(1/3)] C[1] + 
 AiryBiPrime[2^(2/3) v λ^(1/3)] C[2]
*)

ySolution[x_] = zSolution[u[x]]
(*
==> AiryAiPrime[2^(2/3) Sqrt[-1 + x] λ^(1/3)] C[1] + 
 AiryBiPrime[2^(2/3) Sqrt[-1 + x] λ^(1/3)] C[2]
*)

Block[{y}, Simplify[eqn /. y -> ySolution]]
(* ==> True *)

All I did here was to define a new function z through the substitution of variable $u = \sqrt{x-1}$. The original differential equation eqn is then transformed into the new one, eqn2, which can be solved analytically with DSolve. Finally, I also check that the general solution obeys the original equation when substituting back the original independent variable x.

Edit: an alternate substitution

To get Bessel functions as solutions, instead of derivatives of Airy functions, the only change to the above would be to choose as our new variable $u = (x-1)^{(3/4)}$. This is shown here by repeating the same steps as before with that single change:

Clear[y, λ, x, z, u, v]
eqn = y''[x] == λ y[x]/Sqrt[-1 + x];
u[x_] := (x - 1)^(3/4)
y[x_] := z[u[x]]
eqn2 = Simplify[eqn /. x -> InverseFunction[u][v], v > 0];
zSolution[v_] = z[v] /. First[DSolve[eqn2, z[v], v]];
ySolution[x_] = zSolution[u[x]];   
Block[{y = ySolution}, FullSimplify[eqn]]
(* ==> True *)

TeXForm[ySolution[x]]

$$c_1 \sqrt{x-1} J_{\frac{2}{3}}\left(-\frac{4}{3 } i (x-1)^{3/4} \sqrt{\lambda }\right)+c_2 \sqrt{x-1} Y_{\frac{2}{3}}\left(-\frac{4}{3 } i (x-1)^{3/4} \sqrt{\lambda }\right)$$

This is an alternate general solution.

Answer 2

A simpler direct solution is the following:

Ignoring the -1 for a moment, i.e. replacing Sqrt[x-1] by Sqrt[x], Mathematica solves the resulting ODE easily

sol = DSolve[z''[x] == \[Lambda] z[x]/Sqrt[x], z[x], x]

(*
Out[1]= {{z[x] -> (2/3)^(2/3) Sqrt[x] \[Lambda]^(1/3)
      BesselI[-(2/3), 4/3 x^(3/4) Sqrt[\[Lambda]]] C[1] Gamma[1/3] + ((-2)^(
     2/3) Sqrt[x] \[Lambda]^(1/3)
      BesselI[2/3, 4/3 x^(3/4) Sqrt[\[Lambda]]] C[2] Gamma[5/3])/3^(2/3)}}
*)

Now we restore the -1 and get

y[x_] = z[x] /. sol[[1]] /. x -> x - 1

(*
Out[2]= (2/3)^(2/3) Sqrt[-1 + x] \[Lambda]^(1/3)
   BesselI[-(2/3), 4/3 (-1 + x)^(3/4) Sqrt[\[Lambda]]] C[1] Gamma[1/
   3] + ((-2)^(2/3) Sqrt[-1 + x] \[Lambda]^(1/3)
   BesselI[2/3, 4/3 (-1 + x)^(3/4) Sqrt[\[Lambda]]] C[2] Gamma[5/3])/3^(2/3)
*)

Finally we check that y[x] really solves the original ODE

y''[x]/(\[Lambda] y[x]) // FullSimplify

(*
Out[3]= 1/Sqrt[-1 + x]
*)

Check ok. Done.

Answer 3

This is a Bessel type ODE. So it has solutions in terms of special Bessel functions. This can be done by hand using Frobenius series method if needed. You might want to do this to verify the solution below.

Here is the solution from Maple and I converted it to Mathematica syntax. The solution contains 2 arbitrary constants of integrations and use BesselI and BesselY both implemented in Mathematica. So direct translation was only needed

Translate to Mathematica:

Clear[x, c1, c2, lam]
sol1 = Sqrt[x - 1] BesselJ[2/3, 4/3 Sqrt[-lam] (x - 1)^(3/4)];
sol2 = Sqrt[x - 1] BesselY[2/3, 4/3 Sqrt[-lam] (x - 1)^(3/4)];
fullSolution = c1 sol1 + c2 sol2

Answer 4

for reference here is the direct numerical approach. Obviously you need to specify lambda and initial conditions and a domain.

 \[Lambda] = 1;
 x0 = 1 + 1*^-4;
 x1 = 4;
 ySolution[x_] = y[x] /. First@NDSolve[y''[x] == (\[Lambda] y[x])/
   Sqrt[-1 + x] && y[x0] == 0 && y'[x0] == 1 , y[x],
      {x, x0, x1}]
 Plot[ySolution[x], {x, x0, x1}]

I just guessed x==1 is an important point. The equation has a singularity but the solution is well behaved at x==1 , which we can see by setting an initial condition at some small epsilon.

aside, here is how you match those initial conditions using Dr WH analytic solution:

 \[Lambda] = 1
 sol = DSolve[z''[x] == \[Lambda] z[x]/Sqrt[x], z[x], x];
 y[x_] = z[x] /. sol[[1]] /. x -> x - 1 

 Limit[y[x], x -> 1] -> C[1]  (* so C[1] must be zero *)

some handwaving later we find:

 C[2] -> real * (1+I Sqrt[3] )

produces a real solution that we can differentiate to match the y'[1] condition.. sol we have:

 yprime1=1
 ySolution[x_] = 
    y[x] /. {C[1] -> 0 , 
     C[2] -> - yprime1 Gamma[2/3]/Gamma[5/3]/2/(2/3)^(1/3) (1 + I Sqrt[3])}

 Plot[ySolution[x], {x, 1, 4}]

(* same plot *)