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Kaydian

  • 2 years ago

3d2y/dt2-dy/dt +2y=e-3t Differential Equation

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  1. malevolence19
    • 2 years ago
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    \[\frac{d^2 y}{dt^2}-\frac{1}{3} \frac{dy}{dt}+\frac{2 y}{3}=\frac{e^{-3t}}{3}\] Solve the homogenous part (=0). Then try a solution of the form: \[y_p(t)=Ae^{-3t}\] And differentiate and plug in. Be careful that your homogenous solution doesn't have a root of 3 or else you need something like: \[y_p(t)=Ate^{-3t}\]

  2. sirm3d
    • 2 years ago
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    the characteristic root of the homogeneous DE are complex numbers, different from the characteristic root of the RHS. writing the LHS as \(f(D)y\) you can determine right away the particular solution \[y_p=\frac{e^{-3t}}{f(-3)}\]

  3. sirm3d
    • 2 years ago
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    minor correction:\[\Huge y_p=\frac{\frac{e^{-3t}}{3}}{f(-3)}\]

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