Solve the boundary-value problem, if possible. $$y ^ {tt } + 5 y ^ {t } - 24 y = 0 , y ( 0 ) = 0 , y ( 2 ) = 1$$
A) $y ( x ) = \left( e ^ { 6 } - e ^ { - 16 } \right) ^ { - 1 } \left( e ^ { 3 x } - e ^ { - 8 x } \right)$
B) $y ( x ) = \left( e ^ { 8 } - e ^ { - 16 } \right) ^ { - 1 } \left( e ^ { x } - e ^ { - 8 x } \right)$
C) $y ( x ) = \left( e ^ { 3 } - e ^ { - 16 } \right) \left( e ^ { 3 x } - e ^ { - 8 x } \right)$
D) $y ( x ) = \left( e ^ { 3 } - e ^ { - 16 } \right) ^ { - 1 } \left( e ^ { 3 x } - e ^ { - 8 x } \right)$
E) No solution

Question

Quizplus · Accepted Answer

The characteristic equation for the differential equation is $r^2 + 5r - 24 = 0$, which factors to $(r - 3)(r + 8) = 0$. The general solution is $y(x) = C_1 e^{3x} + C_2 e^{-8x}$. Applying the boundary conditions $y(0) = 0$ and $y(2) = 1$, we solve for $C_1$ and $C_2$, leading to the solution in choice A.

Solve the Boundary-Value Problem, If Possible ytt+5yt−24y=0,y(0)=0,y(2)=1y ^ {tt } + 5 y ^ {t } - 24 y = 0 , y ( 0 ) = 0 , y ( 2 ) = 1ytt+5yt−24y=0,y(0)=0,y(2)=1

Solve the Boundary-Value Problem, If Possible $y ^ {tt } + 5 y ^ {t } - 24 y = 0 , y ( 0 ) = 0 , y ( 2 ) = 1$