Let $u=v+\dfrac{y+1}{2}$ ,
Then $u_x=v_x$
$u_{xx}=v_{xx}$
$u_{xy}=v_{xy}$
$u_y=v_y+\dfrac{1}{2}$
$u_{yy}=v_{yy}$
$\therefore av_x+b\left(v_y+\dfrac{1}{2}\right)+\dfrac{\alpha}{2}v_{xx}+\beta v_{xy}+\dfrac{\gamma}{2}v_{yy}=0$
$\alpha v_{xx}+2\beta v_{xy}+\gamma v_{yy}+2av_x+2bv_y=-b$ with $v(x,-1)=0$ , $v(x,1)=0$
When $\beta=0$ , the homogeneous part of the PDE is separable.
Let $v(x,y)=X(x)Y(y)$ ,
Then $aX'(x)Y(y)+bX(x)Y'(y)+\dfrac{\alpha}{2}X''(x)Y(y)+\dfrac{\gamma}{2}X(x)Y''(y)=0$
$\dfrac{\alpha}{2}X''(x)Y(y)-aX'(x)Y(y)=-\dfrac{\gamma}{2}X(x)Y''(y)+bX(x)Y'(y)$
$\dfrac{(\alpha X''(x)+2aX'(x))Y(y)}{2}=-\dfrac{X(x)(\gamma Y''(y)+2bY'(y))}{2}$
$\dfrac{\alpha X''(x)+2aX'(x)}{X(x)}=-\dfrac{\gamma Y''(y)+2bY'(y)}{Y(y)}$
$\therefore$ Let $v(x,y)=\sum\limits_{n=0}^\infty C(x,n)e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}$ so that it automatically satisfies $v(x,-1)=0$ and $v(x,1)=0$ ,
Then $\sum\limits_{n=0}^\infty\alpha C_{xx}(x,n)e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}+\sum\limits_{n=0}^\infty\dfrac{b^2}{\gamma}C(x,n)e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}+\sum\limits_{n=0}^\infty b(2n+1)\pi C(x,n)e^{-\frac{by}{\gamma}}\sin\dfrac{(2n+1)\pi y}{2}-\sum\limits_{n=0}^\infty\dfrac{\gamma(2n+1)^2\pi^2}{4}C(x,n)e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}+\sum\limits_{n=0}^\infty2aC_x(x,n)e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}-\sum\limits_{n=0}^\infty\dfrac{2b^2}{\gamma}C(x,n)e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}-\sum\limits_{n=0}^\infty b(2n+1)\pi C(x,n)e^{-\frac{by}{\gamma}}\sin\dfrac{(2n+1)\pi y}{2}=-b$
$\sum\limits_{n=0}^\infty\left(\alpha C_{xx}(x,n)+2aC_x(x,n)-\dfrac{\gamma^2(2n+1)^2\pi^2+4b^2}{4\gamma}C(x,n)\right)e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}=-b$
$\sum\limits_{n=0}^\infty\left(\alpha C_{xx}(x,n)+2aC_x(x,n)-\dfrac{\gamma^2(2n+1)^2\pi^2+4b^2}{4\gamma}C(x,n)\right)\cos\dfrac{(2n+1)\pi y}{2}=-be^\frac{by}{\gamma}$
For $-1 , with reference to Wave equation with initial and boundary conditions - is this function right? ,
$\sum\limits_{n=0}^\infty\left(\alpha C_{xx}(x,n)+2aC_x(x,n)-\dfrac{\gamma^2(2n+1)^2\pi^2+4b^2}{4\gamma}C(x,n)\right)\cos\dfrac{(2n+1)\pi y}{2}=-\sum\limits_{n=0}^\infty\int_{-1}^1be^\frac{by}{\gamma}\cos\dfrac{(2n+1)\pi y}{2}dy~\cos\dfrac{(2n+1)\pi y}{2}$
$\sum\limits_{n=0}^\infty\left(\alpha C_{xx}(x,n)+2aC_x(x,n)-\dfrac{\gamma^2(2n+1)^2\pi^2+4b^2}{4\gamma}C(x,n)\right)\cos\dfrac{(2n+1)\pi y}{2}=-\sum\limits_{n=0}^\infty\dfrac{4b\gamma^2(-1)^n(2n+1)\pi\cosh\dfrac{b}{\gamma}}{\gamma^2(2n+1)^2\pi^2+4b^2}\cos\dfrac{(2n+1)\pi y}{2}$
$\therefore\alpha C_{xx}(x,n)+2aC_x(x,n)-\dfrac{\gamma^2(2n+1)^2\pi^2+4b^2}{4\gamma}C(x,n)=-\dfrac{4b\gamma^2(-1)^n(2n+1)\pi\cosh\dfrac{b}{\gamma}}{\gamma^2(2n+1)^2\pi^2+4b^2}$
$4\alpha\gamma C_{xx}(x,n)+8a\gamma C_x(x,n)-(\gamma^2(2n+1)^2\pi^2+4b^2)C(x,n)=-\dfrac{16b\gamma^3(-1)^n(2n+1)\pi\cosh\dfrac{b}{\gamma}}{\gamma^2(2n+1)^2\pi^2+4b^2}$
$C(x,n)=A(n)e^{-\frac{ax}{\alpha}}\cosh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}x}{2\alpha}+B(n)e^{-\frac{ax}{\alpha}}\sinh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}x}{2\alpha}+\dfrac{16b\gamma^3(-1)^n(2n+1)\pi\cosh\dfrac{b}{\gamma}}{(\gamma^2(2n+1)^2\pi^2+4b^2)^2}$
$\therefore u(x,y)=\sum\limits_{n=0}^\infty A(n)e^{-\frac{ax}{\alpha}-\frac{by}{\gamma}}\cosh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}x}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}+\sum\limits_{n=0}^\infty B(n)e^{-\frac{ax}{\alpha}-\frac{by}{\gamma}}\sinh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}x}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}+\sum\limits_{n=0}^\infty\dfrac{16b\gamma^3(-1)^n(2n+1)\pi\cosh\dfrac{b}{\gamma}}{(\gamma^2(2n+1)^2\pi^2+4b^2)^2}e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}+\dfrac{y+1}{2}$
$u(-\varepsilon,y)=0$ , $u(\varepsilon,y)=0$ :
$\begin{cases}\sum\limits_{n=0}^\infty A(n)e^{\frac{a\varepsilon}{\alpha}-\frac{by}{\gamma}}\cosh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}\varepsilon}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}-\sum\limits_{n=0}^\infty B(n)e^{\frac{a\varepsilon}{\alpha}-\frac{by}{\gamma}}\sinh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}\varepsilon}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}+\sum\limits_{n=0}^\infty\dfrac{16b\gamma^3(-1)^n(2n+1)\pi\cosh\dfrac{b}{\gamma}}{(\gamma^2(2n+1)^2\pi^2+4b^2)^2}e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}+\dfrac{y+1}{2}=0\\\sum\limits_{n=0}^\infty A(n)e^{-\frac{a\varepsilon}{\alpha}-\frac{by}{\gamma}}\cosh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}\varepsilon}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}+\sum\limits_{n=0}^\infty B(n)e^{-\frac{a\varepsilon}{\alpha}-\frac{by}{\gamma}}\sinh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}\varepsilon}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}+\sum\limits_{n=0}^\infty\dfrac{16b\gamma^3(-1)^n(2n+1)\pi\cosh\dfrac{b}{\gamma}}{(\gamma^2(2n+1)^2\pi^2+4b^2)^2}e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}+\dfrac{y+1}{2}=0\end{cases}$
$\begin{cases}\sum\limits_{n=0}^\infty A(n)e^{\frac{a\varepsilon}{\alpha}}\cosh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}\varepsilon}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}-\sum\limits_{n=0}^\infty B(n)e^{\frac{a\varepsilon}{\alpha}}\sinh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}\varepsilon}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}=-\sum\limits_{n=0}^\infty\dfrac{16b\gamma^3(-1)^n(2n+1)\pi\cosh\dfrac{b}{\gamma}}{(\gamma^2(2n+1)^2\pi^2+4b^2)^2}\cos\dfrac{(2n+1)\pi y}{2}-\dfrac{y+1}{2}e^\frac{by}{\gamma}\\\sum\limits_{n=0}^\infty A(n)e^{-\frac{a\varepsilon}{\alpha}}\cosh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}\varepsilon}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}+\sum\limits_{n=0}^\infty B(n)e^{-\frac{a\varepsilon}{\alpha}}\sinh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}\varepsilon}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}=-\sum\limits_{n=0}^\infty\dfrac{16b\gamma^3(-1)^n(2n+1)\pi\cosh\dfrac{b}{\gamma}}{(\gamma^2(2n+1)^2\pi^2+4b^2)^2}\cos\dfrac{(2n+1)\pi y}{2}-\dfrac{y+1}{2}e^\frac{by}{\gamma}\end{cases}$