Answer: Solve {showing details):

Chapter 4, Problem 4.1.91

(choose chapter or problem)

Solve (showing details):

\(y_{1}^{\prime} =4 y_{2}+3 e^{3 t}\)

\(y_{2}^{\prime} =2 y_{2}-15 e^{-3 t}\)

\(y_{1}(0) =2, y_{2}(0)=2\)

Text Transcription:

y_1’  = 4y_2 + 3e^3t

y_2’  = 2y_2 - 15e^-3t

y_1(0) = 2, y_2(0) = 2

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