Solve the initial value problem.
-$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm { dt } ^ { 2 } } = 6 - \mathrm { e } ^ { - \mathrm { t } } , \mathrm { y } ( 1 ) = \frac { - 1 } { \mathrm { e } } , \mathrm { y } ^ { \prime } ( 0 ) = - 4$$
A) $y = 6 t ^ { 2 } + e ^ { - t } - 5 t - 1 - \frac { 2 } { e }$
B) $y = 3 t ^ { 2 } - e ^ { - t }$
C) $y = 3 t ^ { 2 } - e ^ { - t } - 4 t$
D) $y = 3 t ^ { 2 } - e ^ { - t } - 5 t + 2$

Question

Quizplus · Accepted Answer

Using the formula for finding the particular solution to the non-homogeneous differential equation:
$$ y_p = - \frac{1}{\lambda}e^{-t} $$
where $\lambda$ is the coefficient of $e^{-t}$ in the complementary solution. Here, the complementary solution is $y_c = c_1 + c_2e^{t}$ and $\lambda = 1$. Thus,
$$ y_p = -e^{-t} $$
and the general solution is
$$ y = c_1 + c_2 e^{t} - e^{-t} + 3t^2 + k $$
where $k$ is a constant. Applying the initial conditions $y(1) = -1/e$ and $y'(0) = -4$, we get
$$ c_1 + c_2 - 1 - \frac{1}{e} + k = 0 $$
$$ c_2 + 6 - 2e^{-t} = -4 $$
Solving these two equations, we get $c_1 = -1/e$, $c_2 = -\frac{1}{e} - 6$, and $k = 2$. Substituting these values back into the general solution, we get
$$ y = 3 t ^ { 2 } - e ^ { - t } - 5 t + 2 $$

Solve the Initial Value Problem A) y=6t2+e−t−5t−1−2ey = 6 t ^ { 2 } + e ^ { - t } - 5 t - 1 - \frac { 2 } { e }y=6t2+e−t−5t−1−e2​

Solve the Initial Value Problem A) $y = 6 t ^ { 2 } + e ^ { - t } - 5 t - 1 - \frac { 2 } { e }$