I'm trying to solve the following integral equation:

Integrate[-k/2 Exp[-k Abs[x-z]] f[x], {x,0,d}] == a f[z]

Differentiating twice with respect to z should yield the following differential equation if i'm not mistaken:

a*f''[z] == k^2 (a+1) f[z]

Or

f[z] = C[1] Exp[Sqrt[1+1/a] k z] + C[2] Exp[-Sqrt[1+1/a] k z]

However, when I specify this function as the input, i.e.

f[z_]:= Exp[Sqrt[1+1/a] k z] + Exp[-Sqrt[1+1/a] k z]

Integrate[-k/2 Exp[-k Abs[x-z]] f[x], {x,0,d}, Assumptions->{z \[Element] Reals, 
    z > 0, z < d, k \[Element] Reals, k > 0, d \[Element] Reals} ]

I dont get a*f[z] back, nor anything similar. Is there anything I am missing?

I'm trying to solve the following integral equation:

Integrate[-k/2 Exp[-k Abs[x-z]] f[x], {x,0,d}] == a f[z]

Differentiating twice with respect to z should yield the following differential equation if i'm not mistaken (using Leibniz integral rule):

a f[z] == Integrate[-k/2 Exp[k (x-z)] f[x], {x,0,z}] + 
    Integrate[-k/2 Exp[-k (x-z)] f[x], {x,z,d}]

a f'[z] == (-k/2 f[z] + Integrate[k^2/2 Exp[k (x-z)] f[x], {x,0,z}]) + 
    (k/2 f[z] + Integrate[-k^2/2 Exp[-k (x-z)] f[x], {x,z,d}]

a f''[z] == (k^2/2 f[z] + Integrate[-k^3/2 Exp[k (x-z)] f[x], {x,0,z}] +     
    (k^2/2 f[z] + Integrate[-k^3/2 Exp[-k (x-z)] f[x], {x,z,d}]

a*f''[z] == k^2 (a+1) f[z]

Or

f[z] = C[1] Exp[Sqrt[1+1/a] k z] + C[2] Exp[-Sqrt[1+1/a] k z]

However, when I specify this function as the input, i.e.

f[z_]:= Exp[Sqrt[1+1/a] k z] + Exp[-Sqrt[1+1/a] k z]

Integrate[-k/2 Exp[-k Abs[x-z]] f[x], {x,0,d}, Assumptions->{z \[Element] Reals, 
    z > 0, z < d, k \[Element] Reals, k > 0, d \[Element] Reals} ]

I dont get a*f[z] back, nor anything similar. Is there anything I am missing?