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Let $A, B, C\geq 0$ be constants. Is there an explicit formula to a nontrivial solution to the homogeneous linear ODE $$y''(t) -(A+B\,\sin t)\,y'(t) -C\, y(t)=0$$ for $t\in(0,2\pi)$ with periodic boundary condition $y(0)=y(2\pi)$? This is not clear to me even for the case $A=0,B=C=1$.

p.s. This equation does not seem to be taken cared of in any "Handbook of differential equation". Please correct me if this statement is wrong. Thanks!

Let $A, B, C\geq 0$ be constants. Is there an explicit formula to a nontrivial solution to the homogeneous linear ODE $$y''(t) -(A+B\,\sin t)\,y'(t) -C\, y(t)=0$$ for $t\in(0,2\pi)$ with periodic boundary condition $y(0)=y(2\pi)$?

p.s. This equation does not seem to be taken cared of in any "Handbook of differential equation". Please correct me if this statement is wrong. Thanks!

Let $A, B, C\geq 0$ be constants. Is there an explicit formula to a nontrivial solution to the homogeneous linear ODE $$y''(t) -(A+B\,\sin t)\,y'(t) -C\, y(t)=0$$ for $t\in(0,2\pi)$ with periodic boundary condition $y(0)=y(2\pi)$? This is not clear to me even for the case $A=0,B=C=1$. Thanks!

Let $A, B, C\geq 0$ be constants. Is there an explicit formula to a nontrivial solution to the homogeneous linear ODE $$y''(t) -(A+B\,\sin t)\,y'(t) -C\, y(t)=0$$ for $t\in(0,2\pi)$ with periodic boundary condition $y(0)=y(2\pi)$? This is not clear to me even for the case $A=0,B=C=1$.

p.s. This equation does not seem to be taken cared of in any "Handbook of differential equation". Please correct me if this statement is wrong. Thanks!

Let $A, B, C\geq 0$ be constants. Is there an explicit solution to the homogeneous linear ODE $$y''(t) -(A+B\,\sin t)\,y'(t) -C\, y(t)=0$$ for $t\in(0,2\pi)$ with periodic boundary condition $y(0)=y(2\pi)$? This is not clear to me even for the case $A=0,B=C=1$. Thanks!

Let $A, B, C\geq 0$ be constants. Is there an explicit formula to a nontrivial solution to the homogeneous linear ODE $$y''(t) -(A+B\,\sin t)\,y'(t) -C\, y(t)=0$$ for $t\in(0,2\pi)$ with periodic boundary condition $y(0)=y(2\pi)$? This is not clear to me even for the case $A=0,B=C=1$. Thanks!