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Dugee

  • 2 years ago

Does anyone know how to solve this? Solve the differential equation by variation of parameters, subject to the initial conditions y(0) = 1, y'(0) = 0. y'' + 2y' − 8y = 5e^(−3x) − e^(−x) I get y = c1e^(-4x)+c2e^(2x)-e^(-3x)+e^(-x)/9 but idk how to carry on with the initial condition part.

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  1. sirm3d
    • 2 years ago
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    \[\Large {y=c_1e^{-4x}+c_2e^{2x}-e^{-3x}+e^{-x}/9\\1=c_1+c_2-1+1/9}\] \[\Large {y'=-4c_1e^{-4x}+2c_2e^{2x}+3e^{-3x}-e^{-x}/9\\0=-4c_1+2c_2+3-1/9}\] solve the unknowns \(\large c_1,\;c_2\)

  2. Dugee
    • 2 years ago
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    Thanks. I was able to solve the unknowns. \[c_{1} = \frac{ 121 }{ 54 } , c_{2} = -\frac{ 7 }{54 }\] Is that what you got as well?

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