Using Laplace transform methods, the solution of $y ^ { \prime } + y = 2 \sin t , y ( 0 ) = 1$ is (Hint: the previous problem might be useful.)
A) $y = 2 e ^ { - t } + \sin t + \cos t$
B) $y = e ^ { t } + e ^ { - t } - \sin t - \cos t$
C) $y = 2 e ^ { - t } - \sin t - \cos t$
D) $y = 2 e ^ { - t } + \sin t - \cos t$
E) $y = e ^ { t } + e ^ { - t } + \sin t - \cos t$

Question

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The Answer of Using Laplace transform methods, the solution of $y ^ { \prime } + y = 2 \sin t , y ( 0 ) = 1$ is (Hint: the previous problem might be useful.)
A) $y = 2 e ^ { - t } + \sin t + \cos t$
B) $y = e ^ { t } + e ^ { - t } - \sin t - \cos t$
C) $y = 2 e ^ { - t } - \sin t - \cos t$
D) $y = 2 e ^ { - t } + \sin t - \cos t$
E) $y = e ^ { t } + e ^ { - t } + \sin t - \cos t$

Using Laplace Transform Methods, the Solution Of y′+y=2sin⁡t,y(0)=1y ^ { \prime } + y = 2 \sin t , y ( 0 ) = 1y′+y=2sint,y(0)=1

Using Laplace Transform Methods, the Solution Of $y ^ { \prime } + y = 2 \sin t , y ( 0 ) = 1$