Solve the initial value problem.
-$\frac { \mathrm { dv } } { \mathrm { dt } } = \frac { 1 } { 4 } \csc t \cot t , v \left( \frac { 3 \pi } { 2 } \right) = \frac { 3 } { 4 }$
A) $v = - \frac { \csc t } { 4 } + \frac { 1 } { 2 }$
B) $v = \csc t - \frac { 1 } { 4 }$
C) $v = - \frac { \sec t } { 4 } + \frac { 3 } { 4 }$
D) $v = - \frac { \operatorname { csct } } { 4 } + 1$

Question

Quizplus · Accepted Answer

First, we can write the differential equation as:
$$ \frac { \mathrm { dv } } { \mathrm { dt } } = \frac { 1 } { 4 } \csc t \cot t = \frac { 1 } { 4 } \frac { \cos t } { \sin ^ { 2 } t }  $$

Then we can integrate both sides with respect to $t$:
$$ \int \mathrm { dv } = \int \frac { 1 } { 4 } \frac { \cos t } { \sin ^ { 2 } t } \mathrm { d } t $$
Using substitution $u  = \sin t$, $\mathrm{d} u = \cos t \mathrm{d} t$ into the integral on the right hand side gives:
$$\int \frac{\cos t}{\sin^2 t} \mathrm{d}t = \int \frac{1}{u^2}\mathrm{d}u = -\frac{1}{u} = -\csc t $$
So, we have:
$$ v = - \frac { 1 } { 4 } \csc t + C $$
To find the value of $C$, we use the initial condition $v \left( \frac { 3 \pi } { 2 } \right) = \frac { 3 } { 4 }$:
$$ \frac { 3 } { 4 } = - \frac { 1 } { 4 } \csc \left( \frac { 3 \pi } { 2 } \right) + C = \frac { 1 } { 4 } + C $$
Therefore, $C = \frac { 3 } { 4 } - \frac { 1 } { 4 } = \frac { 1 } { 2 }$, which leads to the final solution:
$$ v = - \frac { 1 } { 4 } \csc t + \frac { 1 } { 2 } $$
Hence, the answer is A.

Solve the Initial Value Problem dvdt=14csc⁡tcot⁡t,v(3π2)=34\frac { \mathrm { dv } } { \mathrm { dt } } = \frac { 1 } { 4 } \csc t \cot t , v \left( \frac { 3 \pi } { 2 } \right) = \frac { 3 } { 4 }dtdv​=41​csctcott,v(23π​)=43​

Solve the Initial Value Problem $\frac { \mathrm { dv } } { \mathrm { dt } } = \frac { 1 } { 4 } \csc t \cot t , v \left( \frac { 3 \pi } { 2 } \right) = \frac { 3 } { 4 }$