I have an equation $$ \tag 1 \frac{\partial^{2} y}{\partial t^{2}} - \frac{\partial^{2}y}{\partial z^{2}} + i\frac{\partial a(t)}{\partial t}\frac{\partial y}{\partial z} = 0 $$ Here $$ a = \frac{J_{\frac{1}{4}}(t^{2})}{\sqrt{t}} \approx -\frac{3}{4}\frac{cos\left(t^{2} - \frac{3\pi}{8}\right)}{t^{\frac{3}{2}}} \sqrt{\frac{2}{\pi }} = a_{0}\frac{cos\left(t^{2} - \frac{3\pi}{8}\right)}{t^{\frac{3}{2}}} $$ For large values of time $t$ I may also simplify the derivative $\frac{\partial a}{\partial t} = a'$ to the form $$ \tag 2 a{'} \approx - 2a_{0}\frac{sin\left(t^{2} - \frac{3\pi}{8}\right)}{\sqrt{t}} $$ I need to analyze Eq. $(1)$ on instabilities. In Fourier space it reads $$ y_{k}{''} + (k^{2} + Ak a{'})y_{k} = 0, \quad A = const, \quad y_{k} \equiv y(k, t) $$ It is Mathieu-like equation: by using $(2)$ it takes the form $$ \tag 3 y_{k}{''} + \left[k^{2} - \tilde{A}k \frac{sin\left( t^{2} - \frac{3 \pi}{8}\right)}{\sqrt{t}}\right]y_{k} = 0 $$ Or how to build approximate solution? Especially I look for exponentially growing solutions.
An edit. For large argument $t$ is is possible to transform $(3)$ to Mathieu-like equation with time-dependent parameters. Due to adiabaticity of these parameters it is possible to get exponent of the unstable solution and the bands of parameters which correspond to unstable solution.
By $y_k$ do you mean $y(k,t)$? What you have is a PDE with only time derivatives, and so the general solution is given by $$y(k,t)=f_1(k)\cos(k^2+Aka')t+f_2(k)\sin(k^2+Aka')t,$$ for general functions $f_1,f_2$, which would be determined if you had an initial condition for $y$ and $\frac{\partial y}{\partial t}$.