Why does the complex-root case use Euler's formula?

The repeated roots case occurs when the discriminant $b^2 - 4ac$ is zero, causing the two characteristic roots $r_1$ and $r_2$ to be identical ($r_1 = r_2$). The video identifies this as a distinct scenario where the standard linear combination of two exponentials $c_1 e^{rt} + c_2 e^{rt}$ collapses to a single term, requiring a more involved solution method (typically involving a $t e^{rt}$ term) which is deferred to a later video.

Conditions: The discriminant $b^2 - 4ac = 0$.; $r_1 = r_2$.

The $\pm$ symbol indicates that the quadratic formula produces two distinct roots for $r$, denoted as $r_1$ and $r_2$. These two roots correspond to two independent exponential solutions, $e^{r_1 t}$ and $e^{r_2 t}$, which form the basis of the general solution.

Conditions: The characteristic equation is quadratic.; The discriminant $b^2 - 4ac$ is positive (yielding two distinct real roots in the context of this specific clip's introduction).

When the characteristic roots $r_1$ and $r_2$ are real and distinct, each root generates a valid exponential solution $e^{r_1 t}$ and $e^{r_2 t}$. Because the differential equation is linear and homogeneous, the superposition principle applies, meaning any linear combination of these solutions is also a solution.

Conditions: The differential equation is linear, homogeneous, and has constant coefficients.; The characteristic roots $r_1$ and $r_2$ are real and distinct.

When the discriminant $b^2 - 4ac$ is negative, the square root term in the quadratic formula becomes imaginary. This results in complex characteristic roots $r_1$ and $r_2$.

Conditions: The coefficients $a, b, c$ are real.; The discriminant $b^2 - 4ac < 0$.

The characteristic equation is the quadratic algebraic equation $0 = ar^2 + br + c$. It is obtained by substituting the trial solution $y = e^{rt}$ into the second-order linear homogeneous differential equation $ay'' + by' + cy = 0$, factoring out the nonzero exponential term $e^{rt}$, and setting the remaining polynomial in $r$ to zero.

Conditions: The differential equation is $ay'' + by' + cy = 0$.; The trial solution is $y = e^{rt}$.; $a, b, c$ are constants.