Question

1. Solve the given boundary-value problem.
y'' − 16y' + 64y = 0, y(0) = 1, y(1) = 0

y(x) = _____

Answer 1

$...$$y^{\prime \prime}-16 y^{\prime}+64 y=0 \\[1 em] r^{2}-16 r+64=0 \longrightarrow(r-8)^{2}=0 \\[1 em] \therefore r_{1}=r_{2}=8 \\[1 em] \therefore y=y_{h}=\left(c_{1} x+c_{2}\right) e^{8 x} \\[1 em] y(1)=0 \quad \rightarrow 0=\left(c_{1}+c_{2}\right) e^{8} \\[1 em] \therefore c_{1}+c_{2}=0 \\[1 em] y(0)=1 \longrightarrow 1=c_{2} \\[1 em] \therefore c_{1}=-1 \quad, c_{2}=1 \\[1 em] \therefore y=(1-x) e^{8 x}$