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Solve the Initial Value Problem A) y=e2t44sint+8t14y = \frac { e ^ { 2 t } } { 4 } - 4 \sin t + 8 t - \frac { 1 } { 4 }

Question 135

Multiple Choice

Solve the initial value problem.
- d2ydt2=e2t+4sint,y(0) =0,y(0) =8\frac { \mathrm { d } ^ { 2 } \mathrm { y } } { \mathrm { dt } ^ { 2 } } = \mathrm { e } ^ { 2 \mathrm { t } } + 4 \sin \mathrm { t } , \mathrm { y } ( 0 ) = 0 , \mathrm { y } ^ { \prime } ( 0 ) = 8


A) y=e2t44sint+8t14y = \frac { e ^ { 2 t } } { 4 } - 4 \sin t + 8 t - \frac { 1 } { 4 }
B) y=e2t44sint+232t14y = \frac { e ^ { 2 t } } { 4 } - 4 \sin t + \frac { 23 } { 2 } t - \frac { 1 } { 4 }
C) y=e2t4sint+11t14y = e ^ { 2 t } - 4 \sin t + 11 t - \frac { 1 } { 4 }
D) y=e2t44sinty = \frac { e ^ { 2 t } } { 4 } - 4 \sin t

Correct Answer:

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