Solve $\frac { d y } { d x } = \sqrt { 36 - y ^ { 2 } }$ if $y = 3$ when $x = \frac { \pi } { 6 }$ .
A) $y = 6 \sin x$
B) $y = 6 \cos x$
C) $y = 36 \sin x$
D) $y = 36 \cos x$

Question

Quizplus · Accepted Answer

The differential equation $\frac{dy}{dx} = \sqrt{36 - y^2}$ suggests a trigonometric substitution. Using $y = 6 \sin x$, we have $\frac{dy}{dx} = 6 \cos x$, which matches $\sqrt{36 - y^2} = \sqrt{36 - 36 \sin^2 x} = 6 \cos x$. The initial condition $y = 3$ when $x = \frac{\pi}{6}$ satisfies $3 = 6 \sin(\frac{\pi}{6})$.

Solve dydx=36−y2\frac { d y } { d x } = \sqrt { 36 - y ^ { 2 } }dxdy​=36−y2​

Solve $\frac { d y } { d x } = \sqrt { 36 - y ^ { 2 } }$