@Differential Equations

I don’t understand boundary-value problems. Could someone help me out?

For example, take
$$y''+2y'+y=0,$$$$y(1)=2,$$$$y'(0)=2,$$whose solution (after applying one of the above boundary values) is
$$y(t)=y(0)e^{-t}+2te^{-t}.$$I don't know what y(0) is. However, can I solve for it by setting
$$2=y(0)e^{-1}+2e^{-1},$$$$y(0)=\frac{2}{e^{-1}}-2,$$$$y(t)=2e^{-t+1}-2e^{-t}+2te^{-t}?$$I am so confused!

Question

@Differential Equations

I don’t understand boundary-value problems. Could someone help me out?

For example, take
$$y''+2y'+y=0,$$$$y(1)=2,$$$$y'(0)=2,$$whose solution (after applying one of the above boundary values) is
$$y(t)=y(0)e^{-t}+2te^{-t}.$$I don't know what y(0) is. However, can I solve for it by setting
$$2=y(0)e^{-1}+2e^{-1},$$$$y(0)=\frac{2}{e^{-1}}-2,$$$$y(t)=2e^{-t+1}-2e^{-t}+2te^{-t}?$$I am so confused!

anonymous · Answer

I think I made a mistake in the last three lines.
$$y(t)=y(0)e^{-t}+2te^{-t},$$$$y(0)=\frac{2}{e^{-1}}-2t,$$$$y(t)=2e^{-t+1}.$$I still think it's not correct, though. :/