What would you like to understand?

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 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.

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).

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).

The characteristic equation $ar^2 + br + c = 0$ is solved by applying the quadratic formula. This yields the roots $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, which provide the values for the parameter $r$ in the exponential trial solution.

Conditions: The characteristic equation is quadratic in $r$.; $a \neq 0$.

The characteristic equation $ar^2 + br + c = 0$ is solved by applying the quadratic formula. This yields the roots $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, which provide the values for the parameter $r$ in the exponential trial solution.

Conditions: The characteristic equation is quadratic in $r$.; $a \neq 0$.