Proving $g(y)f^{''}(x)+f(x)g^{''}(y)+ (\alpha ^2 + \beta ^2)f(x)g(y)= 0 \implies f^{''}(x)+\alpha ^2f(x) = 0 , g^{''}(y)+ \beta ^2 g(y)=0$

Question

I was reading a proof in physics, suddenly I'm stuck at proving this passage:

$g(y)f^{''}(x)+f(x)g^{''}(y)+ (\alpha ^2 + \beta ^2)f(x)g(y)= 0 \implies f^{''}(x)+\alpha ^2f(x) = 0 , g^{''}(y)+ \beta ^2 g(y)=0$

How can I prove this implication?

Answer 1

Hint $$g(y)[f''(x)+a^2f(x)]+f(x)[g''(y)+\beta^2g(y)]=0$$ so we have $$f''(x)+a^2f(x)=0,g''(y)+\beta^2g(y)=0$$