Solve the initial value problem.
-$$\theta ^ { 2 } \frac { d y } { d \theta } - 3 \theta y = \theta ^ { 5 } \sec \theta \tan \theta ; \theta 0 , y ( \pi ) = 0$$
A) $\mathrm { y } = \frac { \theta ^ { 3 } } { \cos \theta } + \theta ^ { 3 } , \theta 0$
B) $y = - \frac { \theta ^ { 3 } } { \cos \theta } - \theta ^ { 3 } , \theta 0$
C) $y = - \frac { \theta ^ { 2 } } { \cos \theta } - \theta ^ { 2 } , \theta 0$
D) $y = \frac { \theta ^ { 2 } } { \cos \theta } + \theta ^ { 2 } , \theta 0$

Question

Solve the initial value problem.
-$$\theta ^ { 2 } \frac { d y } { d \theta } - 3 \theta y = \theta ^ { 5 } \sec \theta \tan \theta ; \theta > 0 , y ( \pi ) = 0$$
A) $\mathrm { y } = \frac { \theta ^ { 3 } } { \cos \theta } + \theta ^ { 3 } , \theta > 0$
B) $y = - \frac { \theta ^ { 3 } } { \cos \theta } - \theta ^ { 3 } , \theta > 0$
C) $y = - \frac { \theta ^ { 2 } } { \cos \theta } - \theta ^ { 2 } , \theta > 0$
D) $y = \frac { \theta ^ { 2 } } { \cos \theta } + \theta ^ { 2 } , \theta > 0$

Quizplus · Accepted Answer

The Answer of Solve the initial value problem.
-$$\theta ^ { 2 } \frac { d y } { d \theta } - 3 \theta y = \theta ^ { 5 } \sec \theta \tan \theta ; \theta > 0 , y ( \pi ) = 0$$
A) $\mathrm { y } = \frac { \theta ^ { 3 } } { \cos \theta } + \theta ^ { 3 } , \theta > 0$
B) $y = - \frac { \theta ^ { 3 } } { \cos \theta } - \theta ^ { 3 } , \theta > 0$
C) $y = - \frac { \theta ^ { 2 } } { \cos \theta } - \theta ^ { 2 } , \theta > 0$
D) $y = \frac { \theta ^ { 2 } } { \cos \theta } + \theta ^ { 2 } , \theta > 0$

Solve the Initial Value Problem θ2dydθ−3θy=θ5sec⁡θtan⁡θ;θ>0,y(π)=0\theta ^ { 2 } \frac { d y } { d \theta } - 3 \theta y = \theta ^ { 5 } \sec \theta \tan \theta ; \theta > 0 , y ( \pi ) = 0θ2dθdy​−3θy=θ5secθtanθ;θ>0,y(π)=0

Solve the Initial Value Problem $\theta ^ { 2 } \frac { d y } { d \theta } - 3 \theta y = \theta ^ { 5 } \sec \theta \tan \theta ; \theta > 0 , y ( \pi ) = 0$