How to solve a differential equation of nonhomogeneous fourth order real variable complex function with Mathematica? [closed]

Question

Recently, I was got in trouble in solving a fourth order differential equation with Mathematica. And it is written like: $$ -a^2\zeta+\frac{1}{b^2}(\zeta''''+4ia\zeta'''-6a^2\zeta''-4ia^3\zeta'+a^4\zeta)-\frac{c}{d}[-\zeta'-ia\zeta+(1-x)(\zeta''+2ia\zeta'-a^2\zeta)]-\frac{1}{d}(-2f(a)\sqrt{\frac{1-x}{x}}\zeta'+2a^2\sqrt{x(1-x)}\zeta) = a^2x-\frac{1}{b^2}(-4ia^3+a^4x)+\frac{c}{d}[-1-ia x+(1-x)(2ia-a^2 x)] $$ and $$ f(a)=\frac{H{_1^{(2)}}(a/2)}{H{_1^{(2)}}(a/2)+iH{_0^{(2)}}(a/2)} $$ The boundary condition are $$ \zeta(0)=0 $$ $$ \zeta''(0)+2ia\zeta'(0)=-2ia $$ $$ \zeta''(1)+2ia\zeta'(1)-a^2\zeta(1)=-2ia+a^2 $$ $$ \zeta'''(1)+3ia\zeta''(1)-3a^2\zeta'(1)-ia^3\zeta(1)=3a^2+ia^3 $$ where $H{_i^{(2)}}(a/2)$ are Hankel functions of the 2nd kind and $i$th order; $a$, $b$, $c$ and $d$ are known real numbers; $\zeta$ is an unknown complex function of real variable $x$. I really want to figure it out, so I would much appreciate anyone who can do me a favor. Thanks in advance.

Here are my related codes with Dsolve and HankelH2 commands:

a = 8*Pi; b = 5; c = 2.74*10^-3; d = 0.3;  
f = HankelH2[1, a/2]/(HankelH2[1, a/2] + I*HankelH2[0, a/2]);  
eq = -a^2*y + 1/b^2*(y'''' + I*4*a*y''' - 6*a^2*y'' - I*4*a^3*y') - 
    c/d*(-y' - I*a*y + (1 - x)*(y'' + I*2*a*y' - a^2*y)) - 
    1/d*(-2*f*Sqrt[(1 - x)/x]*y' + 2*a^2*Sqrt[(1 - x)*x]*y) == 
    a^2*x - 1/b^2*(-I*4*a^3 + a^4*x) + 
    c/d*(-1 - I*a*x + (1 - x)*(I*2*a - a^2*x));  
DSolve[{eq, y'[0] == 0, y''[0] + I*2*a*y'[0] == -I*2*a, 
    y''[1] + I*2*a*y'[1] - a^2*y[1] == -I*2*a + a^2, 
    y'''[1] + I*3*a*y''[1] - 3*a^2*y'[1] - I*a^3*y[1] == 3*a^2 + I*a^3},
    y[x], x]  

After implementing the codes, I got no answer.

Answer 1

This is not solution but it might suggest that it won't be trivial to find one.

Let us call

eqn = {-a^2 f[x] + 
1/b^2 (f''''[x] + 4 I a f'''[x] + 6  a^2 f''[x] - 4 I a^3 f'[x] + 
   a^4 f[x]) - c/d (-f'[x] - I a f[x] + (1 - x) (f''[x] + 2 I a f'[x] - a^2 f[x])) - 
1/d (-2 F[a] Sqrt[(1 - x)/x] f'[x] + 2 a^2 Sqrt[x (1 - x)] f[x]) == 
 h (a^2 x - 1/b^2 (-4 I a^3 + a^4 x) +  c/d (-1 - I a x + (1 - x) (2 I a - a^2 x))),
f[0] == 0, f''[0] + 2 I a f'[0] == -2  I a,
f''[1] + 2 I a f'[1] - a^2 f[1] == -2 I a + a^2 ,
f'''[1] + 3 I a f''[1] - 3 a^2 f[1] - I a^3 f[1] == 3 a^2 + I a^3}

Let us now simplify the problem: put F to zero make the equation homogeneous and replace variable by numbers

eqn2 = eqn /. 
F[a] -> 0  HankelH2[1, 
    a/2]/(HankelH2[1, a/2] + I HankelH2[0, a/2]) /. 
 h -> 0 /. {a -> 1 , b -> 2, c -> 3, d -> 4}

This does not yield a solution

DSolve[eqn2, f[x], x]

(* returns the same *) while the suggestion of Pragabhava

eqn3 = eqn2 /. f -> Function[x, -x + \[Nu][x]] // Simplify;
DSolve[eqn3, \[Nu][x], x]

does not seem to produce the expected miracle.

So in short, one needs to simplify the problem further and find a solution unless you are happy to find numerical solutions?