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Solve $$y'' - 4y' + 5y = 0 $$

Where $y(0) = 0 \ , \ y'(0) = 2$.

So I solve this as a second degree polynomial (no idea why)

$$\frac{4 \pm \sqrt{16-20}}{2} = 2 \pm 2i$$

So the CASE III solution as my book calls it is: $$Ae^{kt} \cos(wt) + B e^{kt} \sin(wt)$$

Where $k = Re$ and $w = Im$.

So anyhow, $$y(0) = A \cos(0) + B \sin (0) \Rightarrow A = 0$$

$$y'(0) = -A \sin(0) + B \cos (0) = 2 \Rightarrow B = 2$$

So the solution is thus $$y(t) = 2 \cos(t)$$

Am I even doing this right? I have no idea what I am doing and it seems that differential equations are just taught this way. Plug this and that into these magical formulas.

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We have:

$$m^2-4m+5 = 0 \implies m_1 = 2+i, m_2 = 2 - i$$

This means we have a solution:

$$y(x) = e^{2t}(c_1 \cos t + c_2 \sin t)$$

Now, substitute in the ICs and find:

$$c_1 = 0, c_2 = 2$$

This makes the final solution:

$$y(t) = 2 e^{2t} \sin t$$

Do you see the slight issue when you found the $\sqrt{-4}$ term in your answer? You also found the correct constants, but used the wrong trig term in your solution.

As an alternative to prove this to yourself, let $y(t) = e^{\lambda t}$. Substitute that back into the ODE and see that it gives you the same result.

Also, here are some notes on Complex Roots if you want to understand the theory a bit more.

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2
  • $\begingroup$ So I'm doing it right? $\endgroup$
    – Paze
    Apr 12 '14 at 1:33
  • $\begingroup$ Why do you write $sin 2t$?? $\endgroup$
    – Paze
    Apr 13 '14 at 15:27

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