Solve the differential equation.
-$$\frac { d y } { d x } = 3 x ^ { 2 } \cos ^ { 2 } y$$
A) $y = x ^ { 3 } + C$
B) $y = \tan ^ { - 1 } \left( x ^ { 3 } + C \right)$
C) $y = \tan ^ { - 1 } \left( x ^ { 2 } + C \right)$
D) $y = \tan \left( x ^ { 3 } + C \right)$

Question

Quizplus · Accepted Answer

To solve the differential equation $\frac { d y } { d x } = 3 x ^ { 2 } \cos ^ { 2 } y$, we separate variables and integrate both sides. Rearranging gives $\frac{dy}{\cos^2 y} = 3x^2 dx$. Integrating both sides, we get $\tan^{-1}(y) = x^3 + C$, which implies $y = \tan^{-1}(x^3 + C)$.

Solve the Differential Equation dydx=3x2cos⁡2y\frac { d y } { d x } = 3 x ^ { 2 } \cos ^ { 2 } ydxdy​=3x2cos2y

Solve the Differential Equation $\frac { d y } { d x } = 3 x ^ { 2 } \cos ^ { 2 } y$