Going over past exams (without solutions) I came across this question:

$\frac{d^2y}{dt^2}+2 \frac{dy}{dt}+5y=0$.

I work out using guess $y=\exp\lambda t$


By quadratic formula: $\lambda =-1 \pm2i$

Therefore $y(t)=\exp(t(-1+2i))$

$= \exp(t)\left( \exp(-1+2i) \right)$

$=\exp(t)\left( \exp(-1(\cos(2)+i\sin(2)\right)))$


But when I substitute the found $y$ into the original equation I don't identically get 0, and I'm not sure where I went wrong.

Thanks for your help.

  • $\begingroup$ You've forgotten to multiply t times the 2i. $\endgroup$ Nov 16, 2017 at 5:41

1 Answer 1


It should be $$\lambda^2+5\lambda+5=0\implies\lambda=\dfrac{-5\pm\sqrt5}2$$

$$\displaystyle y=Ae^{\dfrac{(-5+\sqrt5)t}2}+Be^{\dfrac{(-5-\sqrt5)t}2}$$


After the change in the question,

If $y=e^{-1+2i}$

$$\dfrac{\frac{d^2y}{dt^2}+2 \frac{dy}{dt}+5y}{e^{t(-1+2i})}=(-1+2i)^2+2(-1+2i)+5=-3-4i-2+4i+5=0$$

Similarly for $y=e^{-1-2i}$

  • $\begingroup$ Hi there, thanks for pointing out my typo, I put the question in wrong, it should actually by f''(y)+2f'(y)+5y=0 $\endgroup$
    – Tyler Bond
    Nov 16, 2017 at 6:03
  • $\begingroup$ @TylerBond, Updated the answer $\endgroup$ Nov 16, 2017 at 6:15

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