Solve the initial value problem.
-$$\frac { \mathrm { d } ^ { 2 } \mathrm { r } } { \mathrm { dt } ^ { 2 } } = \frac { 4 } { \mathrm { t } ^ { 3 } } ; \left. \frac { \mathrm { dr } } { \mathrm { dt } } \right| _ { \mathrm { t } = 1 } = 2 , \quad \mathrm { r } ( 1 ) = 5$$
A) $r = \frac { 2 } { t } + 4 t - 1$
B) $r = 2 t + 4 t + 11$
C) $r = \frac { 2 } { t } + 4 t + 11$
D) $r = \frac { 4 } { - 5 t ^ { 5 } } + \frac { 14 } { 5 } t - 1$

Question

Quizplus · Accepted Answer

The correct answer is A because the given differential equation is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation is $r^2 = 4/t^3$, which has the roots $r = \pm 2/t^{3/2}$. Therefore, the general solution to the differential equation is $r(t) = c_1 t^{3/2} + c_2 t^{-3/2}$. Using the initial conditions, we get $r(1) = 5 = c_1 + c_2$ and $\left. \frac{dr}{dt} \right|_{t=1} = 2 = \frac{3}{2}c_1 - \frac{3}{2}c_2$. Solving for $c_1$ and $c_2$, we get $c_1 = 4$ and $c_2 = 1$. Therefore, the particular solution to the initial value problem is $r(t) = 4t^{3/2} + t^{-3/2}$.

Solve the Initial Value Problem A) r=2t+4t−1r = \frac { 2 } { t } + 4 t - 1r=t2​+4t−1

Solve the Initial Value Problem A) $r = \frac { 2 } { t } + 4 t - 1$