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I have 2 nonhomogenous differential equations ($\alpha,\beta, c, d$ are constants and $z,y,z_2,y_2$ are functions of x)

1) $z_2'(x)-\alpha[z_2(x)-2z_2(x+c)-y_2(x+c)]=-2\alpha z(x+c)[2y(x+c)+z(x+c)]$

2) $z_2'(x)+\beta[y_2(x)-2y_2(x-d)-z_2(x-d)]=2\beta y(x-d)[2z(x-d)+y(x-d)] $

I need to solve for $z_2$ and $y_2$. From knowing and solving for $z$ and $y$, I know that they have the form $z=c_1+c_1'e^{rx}$ and $y=c_2+c_2'e^{rx}$. I know the values of $c_1$, $c_1'$, $c_2$ and $c_2'$ from the boundary condition of that problem (not stated here).

What I know:

A) It is a nonhomogeneous differential equation. The solution of the homogeneous part of 1) and 2) gives me the same characteristic equations as that for y and z.

B) I choose $z_2$ and $y_2$ to be of the form

$z_2=u_1e^{rx}+u_2e^{2rx}+u_3xe^{rx}+u_4x +u_5$ and $y_2=v_1e^{rx}+v_2e^{2rx}+v_3xe^{rx}+v_4x +v_5$

How do I solve for the exact values of u and v? I have two boundary conditions: $z_2(w-d_e)=1$ and $y_2(d_h)=1$

I was advised to take each solution of $z_2$ and $y_2$ and plug it into 1) and 2) and then solve them with the addition of the two boundary conditions (total 10+2 equations which will have redundancy) for the 10 unknown coeffecients, $u_{1...5}$,$v_{1...5}$ but I'm stuck. Can I remove the RHS of equation 1 and 2, now that I have included that form in the solution for $z_2$ and $y_2$ on the LHS of eq 1) and 2)? The RHS of the equations 1) and 2) become (after simplification):

For 1) $s_1+s_2e^{rx}+s_3e^{2rz}$

For 2) $t_1+t_2e^{rx}+t_3e^{2rz}$

Please help!

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