What is the characteristic equation for the second-order case shown here?

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).

The particular solution uses $Ate^{-t}$ because the forcing term $e^{-t}$ is already a solution to the associated homogeneous equation. In the method of undetermined coefficients, if the forcing function overlaps with a homogeneous solution (resonance), the standard guess must be multiplied by $t$ to ensure linear independence.

Conditions: The ODE is linear with constant coefficients.; The forcing term contains $e^{-t}$.; $r=-1$ is a root of the homogeneous characteristic equation.

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 homogeneous roots are obtained by solving the associated homogeneous equation $\ddot{x}+3\dot{x}+2x=0$. By substituting the exponential trial solution $x=e^{rt}$, the differential equation converts into the characteristic polynomial $r^2+3r+2=0$, which factors to $(r+2)(r+1)=0$, yielding the roots $r=-2$ and $r=-1$.

Conditions: The ODE has constant coefficients.; The trial solution is of the form $x=e^{rt}$.; The characteristic equation is a quadratic polynomial.