Solve the boundary-value problem, if possible. $$y ^ { tt } + 14 y ^ { t } + 49 y = 0 , y ( 0 ) = 0 , y ( 1 ) = 3$$
A) $y ( x ) = 7 x e ^ { ( - 3 x + 3 ) }$
B) $y ( x ) = 3 e ^ { ( - 7 x - 7 ) }$
C) $y ( x ) = 3 x e ^ { ( - 7 x - 7 ) }$
D) $y ( x ) = 3 x e ^ { ( - 7 x + 7 ) }$
E) No solution

Question

Quizplus · Accepted Answer

The characteristic equation for $y'' + 14y' + 49y = 0$ is $r^2+14r+49 = 0$, which has a double root of $-7$. Therefore, the general solution is of the form $y(x) = e^{-7x}(c_1+c_2x)$. Using the boundary conditions, we can solve for $c_1$ and $c_2$:

$y(0) = 0 \implies c_1 = 0$

$y(1) = 3 \implies 3 = c_2e^{-7} \implies c_2 = 3e^7$

Thus, the solution is $y(x) = 3xe^{-7x+7}$. Therefore, the answer is D.

Solve the Boundary-Value Problem, If Possible ytt+14yt+49y=0,y(0)=0,y(1)=3y ^ { tt } + 14 y ^ { t } + 49 y = 0 , y ( 0 ) = 0 , y ( 1 ) = 3ytt+14yt+49y=0,y(0)=0,y(1)=3

Solve the Boundary-Value Problem, If Possible $y ^ { tt } + 14 y ^ { t } + 49 y = 0 , y ( 0 ) = 0 , y ( 1 ) = 3$