Solve the initial value problem.
-$$\frac { d y } { d x } = \frac { 3 } { 1 + x ^ { 2 } } + \frac { 5 } { \sqrt { 1 - x ^ { 2 } } } , \quad y ( 0 ) = 6$$
A) $y = 3 \tan ^ { - 1 } x + 5 \sin ^ { - 1 } x + 6$
B) $y = 3 \tan ^ { - 1 } x + 5 \sec ^ { - 1 } x + 3$
C) $y = 3 \tan ^ { - 1 } x + 5 \sin ^ { - 1 } x + 3$
D) $y = 3 \tan ^ { - 1 } x + 5 \sec ^ { - 1 } x + 6$

Question

Quizplus · Accepted Answer

Integrating the given differential equation $\frac { d y } { d x } = \frac { 3 } { 1 + x ^ { 2 } } + \frac { 5 } { \sqrt { 1 - x ^ { 2 } } }$ gives $y = 3 \tan ^ { - 1 } x + 5 \sin ^ { - 1 } x + C$. Using the initial condition $y(0) = 6$, we find $C = 6$, leading to the solution $y = 3 \tan ^ { - 1 } x + 5 \sin ^ { - 1 } x + 6$.

Solve the Initial Value Problem dydx=31+x2+51−x2,y(0)=6\frac { d y } { d x } = \frac { 3 } { 1 + x ^ { 2 } } + \frac { 5 } { \sqrt { 1 - x ^ { 2 } } } , \quad y ( 0 ) = 6dxdy​=1+x23​+1−x2​5​,y(0)=6

Solve the Initial Value Problem $\frac { d y } { d x } = \frac { 3 } { 1 + x ^ { 2 } } + \frac { 5 } { \sqrt { 1 - x ^ { 2 } } } , \quad y ( 0 ) = 6$