I have $Y=Y(y)$ and $$Y''-(n\pi)^2Y=0 \qquad \text{with} \qquad Y(0)=0 \,\,\,\, Y(1)=0$$
My book writes $Y = A_n \sinh{n\pi y} + B_n\cosh{n\pi y}$ and then says that this gives $Y \equiv 0$
Why does it give the trivial solution?
My trial $$Y''-(n\pi)^2Y=0 $$ gives exponential solution $Y = Ae^{n\pi y}+Be^{-n\pi y}$ using $Y(0)=0$ we have $A=-B$ and therefore $Y= A(e^{n\pi y}-e^{-n\pi y})$ or if you want $Y=2A\sinh{n\pi y}$
Applying the other doesn't give $A=0$ though.. $Y(1)=0$ gives $0 = 2A\sinh{n\pi}$
Can someone explain to me how we get the trivial solution?