differential equation including $g^{\prime}(y)$ and $g(-y)$

Question

I am looking for $g(y)$ in the following differential equation:

$g^{\prime}(y)-g(-y)\big[-\lambda+\lambda^2(\tau/2-y)\big] = 0$

Answer 1

Differentiating your equation gives $$g''(y)+g'(-y) \left[- \lambda+\lambda^2(\tau/2-y) \right]-g(-y)\left[ -\lambda^2 \right] \tag{1} $$

Using the original equation we can express $g(-y)$ in terms of $g'(y)$ as follows:

$$g(-y)=\frac{g'(y)}{- \lambda+\lambda^2(\tau/2-y)} \tag{2} $$

Moreover, taking $y \mapsto -y$ in the original equation gives

$$g'(-y)-g(y) \left[ -\lambda+\lambda^2(\tau/2+y) \right]=0 \tag{3} $$

which allows us to express $g'(-y)$ in terms of $g(y)$ as follows $$g'(-y)=g(y) \left[ -\lambda+\lambda^2(\tau/2+y) \right] \tag{4} $$

Substituting (2) and (4) in (1) gives an ODE for $g(y)$:

$$g''(y)+g(y) \left[ -\lambda+\lambda^2(\tau/2+y) \right] \left[- \lambda+\lambda^2(\tau/2-y) \right]-\frac{g'(y)}{- \lambda+\lambda^2(\tau/2-y)}\left[ -\lambda^2 \right] =0 $$ which you can try and solve.