$\text { Use the differential equation } \frac { d y } { d x } = \frac { 6 } { \sqrt { 64 - x ^ { 2 } } } \text { and the initial condition } y ( 0 ) = \pi \text { to }$ find y.
A) $y = 6 \arccos ( 64 x ) + \pi$
B) $y = 6 \arcsin \left( \frac { 1 } { 8 } x \right) + \pi$
C) $y = 6 \arccos \left( \frac { 1 } { 8 } x \right) + \pi$
D)$y = 6 \arccos \left( \frac { 1 } { 64 } x \right) + \pi$
E) $y = 6 \arcsin ( 8 x ) + \pi$

Question

$\text { Use the differential equation } \frac { d y } { d x } = \frac { 6 } { \sqrt { 64 - x ^ { 2 } } } \text { and the initial condition } y ( 0 ) = \pi \text { to }$ find y. 
A) $y = 6 \arccos ( 64 x ) + \pi$
B) $y = 6 \arcsin \left( \frac { 1 } { 8 } x \right) + \pi$
C) $y = 6 \arccos \left( \frac { 1 } { 8 } x \right) + \pi$
D)$y = 6 \arccos \left( \frac { 1 } { 64 } x \right) + \pi$
E) $y = 6 \arcsin ( 8 x ) + \pi$

Quizplus · Accepted Answer

To solve the differential equation, integrate both sides with respect to $x$. The integral of $\frac{6}{\sqrt{64-x^2}}$ with respect to $x$ is $6 \arcsin\left(\frac{x}{8}\right) + C$, where $C$ is the constant of integration. Using the initial condition $y(0) = \pi$, we find $C = \pi$. Therefore, the solution is $y = 6 \arcsin\left(\frac{x}{8}\right) + \pi$.

Find Y A) y=6arccos⁡(64x)+πy = 6 \arccos ( 64 x ) + \piy=6arccos(64x)+π

Find Y
A) $y = 6 \arccos ( 64 x ) + \pi$