Solve the differential equation with the given initial condition.
-$y ^ { \prime } = \sin t \cos ^ { 3 } t , y \left( \frac { \pi } { 3 } \right) = 0$
A) y = $\frac { 1 } { 64 }$ - $\frac { \cos ^ { 4 } t } { 4 }$
B) y = $\frac { \cos ^ { 4 } t } { 4 }$
C) y = 16 + $\frac { \cos ^ { 4 } t } { 4 }$
D) y = 64 - $\cos 4$ t

Question

Solve the differential equation with the given initial condition. 
-$y ^ { \prime } = \sin t \cos ^ { 3 } t , y \left( \frac { \pi } { 3 } \right) = 0$
A) y = $\frac { 1 } { 64 }$ - $\frac { \cos ^ { 4 } t } { 4 }$
B) y = $\frac { \cos ^ { 4 } t } { 4 }$
C) y = 16 + $\frac { \cos ^ { 4 } t } { 4 }$
D) y = 64 - $\cos 4$ t

Quizplus · Accepted Answer

To solve the differential equation $y' = \sin t \cos^3 t$, we integrate the right-hand side with respect to $t$. The integral of $\sin t \cos^3 t$ can be found by using a substitution method, where we let $u = \cos t$, thus $du = -\sin t dt$. Integrating, we get $-\int u^3 du = -\frac{u^4}{4} + C = -\frac{\cos^4 t}{4} + C$. To find $C$, we use the initial condition $y\left(\frac{\pi}{3}\right) = 0$, which gives us $C = \frac{1}{64}$. Therefore, the solution is $y = \frac{1}{64} - \frac{\cos^4 t}{4}$.

Solve the Differential Equation with the Given Initial Condition y′=sin⁡tcos⁡3t,y(π3)=0y ^ { \prime } = \sin t \cos ^ { 3 } t , y \left( \frac { \pi } { 3 } \right) = 0y′=sintcos3t,y(3π​)=0

Solve the Differential Equation with the Given Initial Condition $y ^ { \prime } = \sin t \cos ^ { 3 } t , y \left( \frac { \pi } { 3 } \right) = 0$