Question

If y"=2y' and if y=y'=e when x=0, find y when x=1. (Answer: (e/2)(e^2+1).)

Answer 1

y"=2y'--> y"-2y'=0
characteristic equation r^2 -2r =0---> r = 0 and r = 2
general solution: \[y = C_1+C_2e^{2x}\]

Answer 2

y(0) = e --> C1 + C2 = e
\[y'(x) = 2C_2e^{2x}\]
and y'(0) =e --> \(C_2 = \dfrac{e}{2}\) and \(C_1=e/2\)

Answer 3

Replace all, \(y = \dfrac{e}{2}+\dfrac{e}{2}e^{2x}\)
y(1) = \(\dfrac{e}{2}+\dfrac{e^3}{2}\)