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Solve the Initial Value Problem dydt=etsec2(πet),y(ln5)=1π\frac { d y } { d t } = e ^ { - t } \sec ^ { 2 } \left( \pi e ^ { - t } \right) , y ( - \ln 5 ) = \frac { 1 } { \pi }

Question 129

Multiple Choice

Solve the initial value problem.
- dydt=etsec2(πet) ,y(ln5) =1π\frac { d y } { d t } = e ^ { - t } \sec ^ { 2 } \left( \pi e ^ { - t } \right) , y ( - \ln 5 ) = \frac { 1 } { \pi }


A) y=tan(πet) +6πy = \frac { \tan \left( \pi \mathrm { e } ^ { - \mathrm { t } } \right) + 6 } { \pi }
B) y=etcot(πet) +0πy = \frac { - e ^ { - t } \cot \left( \pi e ^ { - t } \right) + 0 } { \pi }
C) y=cot(πet) +1y = \cot \left( \pi e ^ { - t } \right) + 1
D) y=tan(πet) +1πy = \frac { - \tan \left( \pi e ^ { - t } \right) + 1 } { \pi }

Correct Answer:

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