Question

Solve BVP: y''-4y=0. y(0)=1; y(1)=2

Answer 1

\[y''-4y=0\]
yields the characteristic equation
\[r^2-4=0~\Rightarrow~r=\pm 2\]
giving you the solution
\[y=C_1e^{2t}+C_2e^{-2t}\]
Given the boundary condition \(y(0)=1\), you have
\[1=C_1e^0+C_2e^0\\
C_1+C_2=1\]
and given \(y(1)=2\),
\[C_1e^{2}+C_2e^{-2}=2\]
Solve for your constants.