Solve the initial value problem.
-$$\theta \frac { \mathrm { dy } } { \mathrm { d } \theta } + \mathrm { y } = \cos \theta ; \theta 0 , \mathrm { y } ( \pi ) = 1$$
A) $y = \frac { - \sin \theta + \pi \theta } { \theta ^ { 2 } } , \theta 0$
B) $\mathrm { y } = \frac { \sin \theta + \pi } { \theta } , \theta 0$
C) $\mathrm { y } = \frac { \sin \theta + \pi \theta } { \theta ^ { 2 } } , \theta 0$
D) $y = \frac { - \sin \theta + \pi } { \theta } , \theta 0$

Question

Solve the initial value problem.
-$$\theta \frac { \mathrm { dy } } { \mathrm { d } \theta } + \mathrm { y } = \cos \theta ; \theta > 0 , \mathrm { y } ( \pi ) = 1$$
A) $y = \frac { - \sin \theta + \pi \theta } { \theta ^ { 2 } } , \theta > 0$
B) $\mathrm { y } = \frac { \sin \theta + \pi } { \theta } , \theta > 0$
C) $\mathrm { y } = \frac { \sin \theta + \pi \theta } { \theta ^ { 2 } } , \theta > 0$
D) $y = \frac { - \sin \theta + \pi } { \theta } , \theta > 0$

Quizplus · Accepted Answer

The Answer of Solve the initial value problem.
-$$\theta \frac { \mathrm { dy } } { \mathrm { d } \theta } + \mathrm { y } = \cos \theta ; \theta > 0 , \mathrm { y } ( \pi ) = 1$$
A) $y = \frac { - \sin \theta + \pi \theta } { \theta ^ { 2 } } , \theta > 0$
B) $\mathrm { y } = \frac { \sin \theta + \pi } { \theta } , \theta > 0$
C) $\mathrm { y } = \frac { \sin \theta + \pi \theta } { \theta ^ { 2 } } , \theta > 0$
D) $y = \frac { - \sin \theta + \pi } { \theta } , \theta > 0$

Solve the Initial Value Problem θdydθ+y=cos⁡θ;θ>0,y(π)=1\theta \frac { \mathrm { dy } } { \mathrm { d } \theta } + \mathrm { y } = \cos \theta ; \theta > 0 , \mathrm { y } ( \pi ) = 1θdθdy​+y=cosθ;θ>0,y(π)=1

Solve the Initial Value Problem $\theta \frac { \mathrm { dy } } { \mathrm { d } \theta } + \mathrm { y } = \cos \theta ; \theta > 0 , \mathrm { y } ( \pi ) = 1$