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

Question 90

Multiple Choice

Solve the initial value problem.
- θdydθ+y=cosθ;θ>0,y(π) =1\theta \frac { \mathrm { dy } } { \mathrm { d } \theta } + \mathrm { y } = \cos \theta ; \theta > 0 , \mathrm { y } ( \pi ) = 1


A) y=sinθ+πθθ2,θ>0y = \frac { - \sin \theta + \pi \theta } { \theta ^ { 2 } } , \theta > 0
B) y=sinθ+πθ,θ>0\mathrm { y } = \frac { \sin \theta + \pi } { \theta } , \theta > 0
C) y=sinθ+πθθ2,θ>0\mathrm { y } = \frac { \sin \theta + \pi \theta } { \theta ^ { 2 } } , \theta > 0
D) y=sinθ+πθ,θ>0y = \frac { - \sin \theta + \pi } { \theta } , \theta > 0

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