Complex Roots
Differential Equations · Section 3.3 · generated practice set
Negative discriminant, Euler's formula, and the split between the decay rate and the frequency.
Practice this set → Fresh numbers on every attempt. No account needed.
Try one now
Solve $$y'' - 6y' + 13y = 0, \qquad y(0) = -2,\ y'(0) = 5,$$ writing it as $y = e^{\alpha t}\left(c_1\cos\beta t + c_2\sin\beta t\right)$ with $\beta > 0$.
Solution
- Characteristic equation $r^2 - 6r + 13 = 0$; discriminant $= -16 < 0$.
- $r = \dfrac{6 \pm \sqrt{-16}}{2} = 3 \pm 2i$, so $\alpha = 3$, $\beta = 2$.
- $y = e^{3t}\left(c_1\cos 2t + c_2\sin 2t\right)$; $y(0) = c_1 = -2$.
- $y' (0)= \alpha c_1 + \beta c_2 = 5$, so $c_2 = \dfrac{5 - (3)(-2)}{2} = 5.5$.
- The sign of $\alpha$ alone decides growth vs decay; $\beta$ only sets the frequency.
Next problem, fresh numbers → No account needed. Your accuracy is tracked on this device.
Method
When $b^2 - 4ac < 0$ the roots are $r = \alpha \pm i\beta$ and Euler's formula turns them into real solutions: $$y = e^{\alpha t}\left(c_1\cos\beta t + c_2\sin\beta t\right).$$ The real part $\alpha$ controls growth or decay; $\beta$ sets the frequency.
Definitions and theorems
Worked example
Solve $y'' + 4y' + 13y = 0$, $y(0) = 2$, $y'(0) = -1$.
- $r^2 + 4r + 13 = 0 \Rightarrow r = \frac{-4 \pm \sqrt{-36}}{2} = -2 \pm 3i$.
- $y = e^{-2t}(c_1\cos 3t + c_2\sin 3t)$.
- $y(0) = c_1 = 2$.
- $y'(0) = -2c_1 + 3c_2 = -1 \Rightarrow c_2 = 1$.
- $y = e^{-2t}(2\cos 3t + \sin 3t)$, a decaying oscillation.
Common mistakes
- Dropping $e^{\alpha t}$. It multiplies the entire bracket.
- Product rule at $y'(0)$. $y' = \alpha e^{\alpha t}(\ldots) + e^{\alpha t}(\ldots)'$ — both terms contribute.
Practice it
Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.