For the 2nd order linear PDE below, please give method(s) to solve it, working, a solution, and what conditions the solution can exist?
$$\sin(t)\frac{\partial^2y}{\partial t^2}+\cos(t)\frac{\partial y}{\partial t}+\cos(t)\frac{\partial ^2y}{\partial x^2}=0$$
I tried a solution like $y=y(x,t)=Ae^{\alpha t}sin(Ct)+Be^{\beta t}sin(Dt)+Fe^{\zeta x}sin(Ex)+Ge^{\gamma x}sin(Hx)$ via method of undetermined coeffs. However I found this made coeffs hard to determine.
Can it be solved via transformations?
$$\sin(t)\frac{\partial^2y}{\partial t^2}+\cos(t)\frac{\partial y}{\partial t}+\cos(t)\frac{\partial ^2y}{\partial x^2}=0$$
You tried a solution like $y=y(x,t)=Ae^{\alpha t}sin(Ct)+Be^{\beta t}sin(Dt)+Fe^{\zeta x}sin(Ex)+Ge^{\gamma x}sin(Hx)$ via method of undetermined coefficients.
This can succeed only if each of the terms is solution of the PDE. Putting each one into the PDE shows that no one of these terms satisfies the PDE. So, the method of undetermined coefficients cannot be successful on this basis.
The PDE has an infinity of solutions. Some are easy to obtain. For example :
Case of solutions including only the variable $x \quad\to\quad \frac{d^2y}{dx^2}=0$ : $$y(x,t)=c_1t+c_2 \quad\text{ is solution of the PDE}$$
Case of solutions including only the variable $t \quad\to\quad \sin(t)\frac{d^2y}{dt^2}+\cos(t)\frac{dy}{dt}=0$ : $$y(x,t)=c_1\ln\left(\tan\left(\frac{t}{2} \right) \right)+c_2 \quad\text{ is solution of the PDE}$$
Case of solutions on the form $y(x,t)=f(x)g(t) \quad\to\quad \begin{cases} f''=C\:f \quad\text{any constant } C \\ \tan(t)g''+g'=-C\:g \end{cases}$
(As far as I know, there is no closed form for $g(t)$ : Numerical solving required in this case)
General case :
The PDE has an infinity of solutions. Without giving the boundary conditions, the question is too wide. You have to add the boundary conditions in the wording of the question if you expect a more specific answer.