Solve the boundary-value problem, if possible.
$$y ^ { tt } + 5 y ^ { t } - 14 y = 0 , y ( 0 ) = 0 , y ( 2 ) = 1$$
A) $y ( x ) = \left( e ^ { 4 } - e ^ { - 14 } \right) ^ { - 1 } \left( e ^ { 2 x } - e ^ { - 7 x } \right)$
B) $y ( x ) = \left( e ^ { 2 } - e ^ { - 14 } \right) \left( e ^ { 2 x } - e ^ { - 7 x } \right)$
C) $y ( x ) = \left( e ^ { 7 } - e ^ { - 14 } \right) ^ { - 1 } \left( e ^ { x } - e ^ { - 7 x } \right)$
D) $y ( x ) = \left( e ^ { 2 } - e ^ { - 14 } \right) ^ { - 1 } \left( e ^ { 2 x } - e ^ { - 7 x } \right)$
E) No solution

Question

Quizplus · Accepted Answer

The characteristic equation of the differential equation is $r^2 + 5r - 14 = 0$, which has roots $r = 2$ and $r = -7$. The general solution is $y(x) = C_1 e^{2x} + C_2 e^{-7x}$. Applying the boundary conditions $y(0) = 0$ and $y(2) = 1$, we solve for $C_1$ and $C_2$ to find the specific solution.

Solve the Boundary-Value Problem, If Possible ytt+5yt−14y=0,y(0)=0,y(2)=1y ^ { tt } + 5 y ^ { t } - 14 y = 0 , y ( 0 ) = 0 , y ( 2 ) = 1ytt+5yt−14y=0,y(0)=0,y(2)=1

Solve the Boundary-Value Problem, If Possible $y ^ { tt } + 5 y ^ { t } - 14 y = 0 , y ( 0 ) = 0 , y ( 2 ) = 1$