Solve $$y''+9y=0$$ where $$y(0)+y'(0)=0$$ and $$y(π)-y'(π)=0$$Plot the solution on domain $0<x<π$
so, $$y=D\cos3x+E\sin3x$$ and $$y'=-3D\sin3x+3E\cos3x$$ with the first given equation, I substitute these two and have $$D=3E$$ with the second given equation, I substitute again and have $$3E=D$$ finally, my general solution is $$y=E(3\cos3x+\sin3x).$$ But how can we find $E$ as we need to plot the solution?
I suppose a small mistake in the problem since $$y(x)=c_1 \cos (3 x)+c_2 \sin (3 x)$$ $$y'(x)=-3 c_1 \sin (3 x)+3 c_2 \cos (3 x)$$ Now the conditions $$y(0)+y'(0)=c_1+3 c_2=0$$ $$y'(\pi)-y'(\pi)=-c_1+3 c_2=0$$ The only possible solutions are $c_1=c_2=0$ which make $y(x)=0$ as you showed it already.
I suppose that one of the conditions is not $0$. It could be $y'(\pi)-y'(\pi)=1$.