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The differential equation I am trying to solve is $$ \dfrac{d^2y}{dt^2} + 2\dfrac{dy}{dt} + 1y = \cos(3t) $$ I know how to start off. I have done the $s^2 + 2s + 1 = 0$ to get $s = -1$

I understand how to do the solution for the right hand side, but with repeated eigenvalues how do i find the homogenous solution for the left hand side?

Thank you for the help

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This is a standard situation. When the indicial equation has a repeated root $\lambda$, you have $e^{\lambda t}$ and $te^{\lambda t}$ as solutions.

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  • $\begingroup$ I know that when dealing with matrices you have $e^{\lambda t}V_o + te^{\lambda t}V_1$ Do I not need to worry about the V's $\endgroup$
    – Bob Dale
    Apr 24, 2013 at 4:18
  • $\begingroup$ This is a scalar ODE. Perhaps you saw in your class how to solve it by converting to a first-order system, where the vectors would have emerged. $\endgroup$ Apr 24, 2013 at 11:01
  • $\begingroup$ What is often passed over in introductory ODE courses is that the "power of $t$ rule" comes from the variation of parameters method (which gets skipped over frequently). If you know the one solution from the characteristic equation, $e^{-t}$, you could construct a second solution of the form $u(t) \cdot e^{-t}$. Inserting this into the homogeneous DE (and applying the Product Rule) tells us that $u'' = 0 \Rightarrow u = Ct$, making our second solution $Cte^{-t}$. $\endgroup$ Apr 25, 2013 at 17:43

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