Particular solution of an ode: resonance

1. Formula
   Observation

   The dependent variable is written as x in the displayed ODE and in the trial solutions.

x

Unknown function of t in the differential equation.

Function of t; domain not explicitly stated in the clip.

1. Formula
   Observation

   The independent variable appears as t in $e^{-t}$, e^{rt}, and the trial particular solution $A t e^{-t}$.

t

Independent variable (time-like parameter) of the ODE.

Real variable; no explicit interval is shown.

1. Formula
   Observation

   The first derivative is written as $\dot{x}$ in the ODE and later computed explicitly.

$\dot{x}$

First derivative of x with respect to t.

Defined wherever x is differentiable.

1. Formula
   Observation

   The second derivative is written as $\ddot{x}$ in the ODE and later computed explicitly.

$\ddot{x}$

Second derivative of x with respect to t.

Defined wherever x is twice differentiable.

1. Formula
   Observation

   r appears in the trial homogeneous solution $x=e$^{rt} and in the characteristic polynomial $r^2+3r+2=0$.

r

Exponential rate parameter used in the homogeneous trial solution.

Algebraic unknown solved from the characteristic equation.

1. Formula
   Observation

   $c_1$ multiplies $e^{-2t}$ in the displayed homogeneous solution.

$c_1$

Arbitrary constant multiplying the homogeneous mode $e^{-2t}$.

Constant of integration; numerical value not specified.

1. Formula
   Observation

   $c_2$ multiplies $e^{-t}$ in the displayed homogeneous solution.

$c_2$

Arbitrary constant multiplying the homogeneous mode $e^{-t}$.

Constant of integration; numerical value not specified.

1. Formula
   Observation

   A is introduced as the unknown coefficient in the trial particular solution $x=A t e^{-t}$.

1. The final value of A is not reached within this clip.

A

Undetermined coefficient for the resonant particular-solution ansatz.

Constant to be determined by substitution into the ODE.

1. Formula
   Observation

   The variable x appears as the dependent function in the differential equation and trial solution.

x

Dependent variable (function of t) in the ordinary differential equation.

Real-valued function of t.

1. Formula
   Observation

   The variable t appears as the independent variable in $e^{-t}$ and te^{-t}.

t

Independent variable (time).

Real numbers.

1. Formula
   Observation

   The constant A is introduced in the trial particular solution x = Ate^{-t} and solved for later.

A

Undetermined coefficient to be solved for in the particular solution.

Real constant.

1. Formula
   Observation

   The notation $\dot{x}$ appears in the differential equation and its derivative expression.

$\dot{x}$

First derivative of x with respect to t.

Real-valued function of t.

1. Audio
   Observation

   The speaker says they want to find a particular solution of x double dot plus 3 x dot plus 2 x equals 2 times e to the minus t.

2. Formula
   Observation

   The board shows $\ddot{x}+3\dot{x}+2x=2e^{-t}$.

The clip studies the differential equation $\ddot{x}+3\dot{x}+2x=2e^{-t}$ and specifically seeks a particular solution rather than the full general solution. The forcing term is exponential, and the speaker immediately flags that its exponent matches one of the homogeneous modes.

1. Linear second-order ODE with constant coefficients.

2. Right-hand side is $2e^{-t}$.

3. Goal in this clip is only to set up and begin computing a particular solution.

1. Audio
   Observation

   The speaker says, 'for the homogeneous equation if we just try x equals e to the rt, we get the characteristic equation r squared plus 3r plus 2 equals 0'.

2. Formula
   Observation

   The board writes 'hom eq: Try $x = e$^{rt}' and then '$r^2 + 3r + 2 = 0$ => ($r+2$)($r+1$)=0'.

To solve the associated homogeneous equation, the video uses the standard exponential trial $x=e^{rt}$. Substitution produces the characteristic polynomial $r^2+3r+2=0$, which factors as $(r+2)(r+1)=0$. This identifies the two homogeneous exponents $r=-2$ and $r=-1$.

1. Applies to the homogeneous equation associated with constant-coefficient linear ODEs.

2. Distinct real roots are obtained here.

1. Nonhomogeneous second-order linear ODE with exponential forcing

1. Audio
   Observation

   The speaker states the homogeneous solution is some constant times $e^{-2t}$ plus another constant times $e^{-t}$.

2. Formula
   Observation

   The board displays $x_h(t)=c_1 e^{-2t}+c_2 e^{-t}$.

3. Diagram
   Observation

   An arrow is drawn from the forcing term $2e^{-t}$ toward the $e^{-t}$ term in the homogeneous solution.

From the roots $r=-2$ and $r=-1$, the homogeneous solution is written as $x_h(t)=c_1e^{-2t}+c_2e^{-t}$. The visual arrow emphasizes that the forcing term $2e^{-t}$ shares the same exponential factor as one homogeneous mode, so the usual non-resonant guess would fail.

1. Roots of the characteristic equation are distinct and real.

2. This is the complementary solution of the homogeneous ODE.

1. Characteristic-equation method for the homogeneous ODE

1. Audio
   Observation

   The speaker says that to find the particular solution, they must try x equal to some unknown coefficient and multiply the right-hand side by t, giving $A t e^{-t}$.

2. Formula
   Observation

   The board writes 'Part sol: Try $x = A t e^{-t}$'.

Because $e^{-t}$ already solves the homogeneous equation, the video modifies the usual undetermined-coefficients guess by multiplying by $t$. The chosen ansatz is $x_p(t)=Ate^{-t}$, where $A$ is the unknown coefficient to be determined after substitution into the ODE.

1. Use when the forcing term overlaps a homogeneous solution mode.

2. Here the overlap is with $e^{-t}$, so one factor of $t$ is inserted.

1. Homogeneous solution and overlap with the forcing term

1. Audio
   Observation

   The speaker verbally applies the product rule: derivative of the first times the second plus the first times derivative of the second.

2. Formula
   Observation

   The board shows $\dot{x}$=Ae^{-t}(1-t).

Differentiating $x=Ate^{-t}$ with the product rule gives $\dot{x}=Ae^{-t}-Ate^{-t}=Ae^{-t}(1-t)$. This expression is then used to compute the second derivative.

1. Differentiate with respect to t.

2. Product rule applied to $t$ and $e^{-t}$.

1. Resonance-adjusted trial for the particular solution

1. Audio
   Observation

   The speaker again uses the product rule and says the result is minus one plus t plus the first times derivative of the second, giving minus one.

2. Formula
   Observation

   The board shows $\ddot{x}$=Ae^{-t}(-1+t-1).

3. Diagram
   Observation

   A brace groups the terms inside the parentheses as the speaker comments on the combination.

1. The unsimplified form is shown, but the clip ends before the simplified second derivative or substituted ODE result is written.

Differentiating $\dot{x}=Ae^{-t}(1-t)$ once more yields $\ddot{x}=Ae^{-t}(-1+t-1)$. The bracketed terms combine to $t-2$, so the simplified form is $\ddot{x}=Ae^{-t}(t-2)$, although the simplification is not fully completed on screen within this clip.

1. Continue from the previously computed first derivative.

2. Product rule applied again to $e^{-t}$ and $(1-t)$.

1. First derivative of the resonant trial solution

1. Formula
   Observation

   The board shows 'Part sol. Try x = Ate^{-t}' along with its first and second derivatives.

For a non-homogeneous term proportional to $e^{-t}$, when -1 is a root of the characteristic equation, the standard method of undetermined coefficients multiplies the usual guess by t. The video uses this resonant form x = Ate^{-t}.

1. The forcing term contains $e^{-t}$.

2. -1 is a simple root of the homogeneous characteristic equation.

1. Audio
   Observation

   The speaker says, 'So we substitute into the differential equation, so x double dot plus 3x dot plus 2x.'

2. Formula
   Observation

   The left-hand side operator $\ddot{x} + 3\dot{x} + 2x$ is rewritten below the trial solution block.

The candidate particular solution and its derivatives are substituted into the left-hand side of the original differential equation to determine the unknown coefficient A.

1. x, $\dot{x}$, and $\ddot{x}$ are evaluated using the trial form x = Ate^{-t}.

1. Trial Particular Solution Form

1. Audio
   Observation

   The speaker explains collecting terms proportional to t and terms not proportional to t after canceling $e^{-t}$.

2. Formula
   Observation

   The board shows the grouped expression ($A - 3A + 2A$)$t + (-2A + 3A)$.

After substituting the trial solution and factoring out $e^{-t}$, the resulting expression is organized into coefficients of t and constant terms. This allows matching against the right-hand side of the differential equation.

1. The common factor $e^{-t}$ has been canceled from both sides.

1. Substituting Trial Solution into ODE

1. Formula
   Observation

   $\ddot{x} + 3\dot{x} + 2x = 2$

The video displays a second-order linear ordinary differential equation with constant coefficients and a constant non-homogeneous term on the right side.

1. The equation is linear.

2. The coefficients are constants (1, 3, 2).

3. The right-hand side is a constant (2).

1. Formula
   Observation

   Substitution of terms involving A into the ODE to solve for A.

1. The exact assumed form of the particular solution before substitution is not explicitly written out in the visible frame.

A technique for finding a particular solution to a non-homogeneous linear ODE by assuming a solution form with unknown coefficients, substituting it into the ODE, and solving for those coefficients.

1. Applies to linear ODEs with constant coefficients.

2. The non-homogeneous term must be of a specific type (e.g., polynomial, exponential, sine/cosine).

1. Second-order linear non-homogeneous ODE

1. Audio
   Observation

   The speaker says the complication is that $e^{-t}$ is a solution of the homogeneous equation.

2. Formula
   Observation

   The homogeneous solution contains the term $c_2 e^{-t}$.

3. Diagram
   Observation

   An arrow visually links the forcing term $2e^{-t}$ to the $e^{-t}$ part of $x_h(t)$.

For the ODE $\ddot{x}+3\dot{x}+2x=2e^{-t}$, the forcing factor $e^{-t}$ belongs to the homogeneous solution space because $r=-1$ is a root of the characteristic equation.

1. Consider the associated homogeneous equation $\ddot{x}+3\dot{x}+2x=0$.

2. Use the exponential trial $x=e^{rt}$.

3. The characteristic equation is $r^2+3r+2=0$.

For this specific ODE and its homogeneous solution.

1. Audio
   Observation

   The speaker says they need to multiply now the right hand side by t, giving $A t e^{-t}$.

2. Formula
   Observation

   The board writes Part sol: Try $x = A t e^{-t}$.

1. The clip demonstrates the rule in this example but does not state a general theorem with full hypotheses.

When the forcing term has the same exponential factor as a homogeneous solution, the trial particular solution is modified by multiplying by $t$; here the chosen form is $x_p(t)=Ate^{-t}$.

1. The ODE is linear with constant coefficients.

2. The nonhomogeneous term contains a factor already present in the homogeneous solution.

Stated for the worked example in the clip; the broader general rule is implied but not formally proved here.

1. Audio
   Observation

   The speaker says trying $x=e$^{rt} gives the characteristic equation $r^2+3r+2=0$, which factors to ($r+2$)($r+1$)=0.

2. Formula
   Observation

   The board shows hom eq: Try $x=e$^{rt}, then $r^2+3r+2=0$ => ($r+2$)($r+1$)=0.

1. Expression

   $$\ddot{x}+3\dot{x}+2x=0$$
   Explanation

   Start from the homogeneous version of the given ODE.

   Justification

   Associated homogeneous equation of the displayed nonhomogeneous ODE.

   Shown in the video
2. Expression

   $$x=e^{rt}$$
   Explanation

   Try an exponential solution.

   Justification

   Standard method for constant-coefficient linear ODEs, as stated in the video.

   Shown in the video
3. Expression

   $$r^2+3r+2=0$$
   Explanation

   Substitution yields the characteristic polynomial.

   Justification

   Audio explicitly says this is the characteristic equation obtained from the trial.

   Shown in the video
4. Expression

   $$(r+2)(r+1)=0$$
   Explanation

   Factor the quadratic.

   Justification

   Algebraic factorization shown on the board.

   Shown in the video
5. Expression

   $$r=-2,\ r=-1$$
   Explanation

   Read off the two roots.

   Justification

   Zero-product property applied to the factored equation.

   Derived from the video

The homogeneous exponents are $r=-2$ and $r=-1$, leading to the complementary solution built from $e^{-2t}$ and $e^{-t}$.

1. Audio
   Observation

   The speaker verbally computes the first and second derivatives using the product rule.

2. Formula
   Observation

   The board writes x=Ate^{-t}, then $\dot{x}$=Ae^{-t}(1-t), then $\ddot{x}$=Ae^{-t}(-1+t-1).

1. The clip stops before substituting these derivatives back into the ODE and solving for A.

1. Expression

   $$x=Ate^{-t}$$
   Explanation

   Begin with the chosen particular-solution ansatz.

   Justification

   Selected because $e^{-t}$ overlaps the homogeneous solution.

   Shown in the video
2. Expression

   $$\dot{x}=Ae^{-t}-Ate^{-t}$$
   Explanation

   Apply the product rule to $t$ and $e^{-t}$.

   Justification

   Speaker explicitly describes derivative of the first times the second plus the first times derivative of the second.

   Shown in the video
3. Expression

   $$\dot{x}=Ae^{-t}(1-t)$$
   Explanation

   Factor out $Ae^{-t}$.

   Justification

   Algebraic simplification shown on the board.

   Shown in the video
4. Expression

   $$\ddot{x}=Ae^{-t}(-1)+(Ae^{-t})(-1)+A(-e^{-t})(1-t)$$
   Explanation

   Differentiate $\dot{x}=Ae^{-t}(1-t)$ again using the product rule.

   Justification

   This is the standard expansion corresponding to the spoken derivation; the board presents the collected result rather than every intermediate line.

   Derived from the video
5. Expression

   $$\ddot{x}=Ae^{-t}(-1+t-1)$$
   Explanation

   Collect terms into the displayed bracketed form.

   Justification

   Matches the formula written on the board.

   Shown in the video
6. Expression

   $$\ddot{x}=Ae^{-t}(t-2)$$
   Explanation

   Simplify the bracket.

   Justification

   Arithmetic simplification of $-1+t-1=t-2$; this final simplification is not fully written before the clip ends.

   Derived from the video

The trial particular solution has derivatives $\dot{x}=Ae^{-t}(1-t)$ and $\ddot{x}=Ae^{-t}(t-2)$, setting up the next step of substitution into the ODE.

1. Formula
   Observation

   The board displays the step-by-step substitution and simplification leading to $A = 2$.

2. Audio
   Observation

   The speaker narrates the cancellation of the t-proportional terms and the solving of the remaining equation.

1. Expression

   $$\ddot{x} + 3\dot{x} + 2x = 2e^{-t}$$
   Explanation

   Start with the original non-homogeneous differential equation.

   Justification

   Given problem statement visible on the board.

   Shown in the video
2. Expression

   $$x = Ate^{-t}, \quad \dot{x} = Ae^{-t}(1-t), \quad \ddot{x} = Ae^{-t}(-2+t)$$
   Explanation

   Use the trial particular solution and its computed first and second derivatives.

   Justification

   Method of undetermined coefficients for a resonant forcing term.

   Shown in the video
3. Expression

   $$Ae^{-t}(-2+t) + 3Ae^{-t}(1-t) + 2Ate^{-t} = 2e^{-t}$$
   Explanation

   Substitute x, $\dot{x}$, and $\ddot{x}$ into the left-hand side of the ODE.

   Justification

   Direct algebraic substitution.

   Shown in the video
4. Expression

   $$e^{-t} [ A(-2+t) + 3A(1-t) + 2At ] = 2e^{-t}$$
   Explanation

   Factor out the common exponential term $e^{-t}$.

   Justification

   Distributive property of multiplication over addition.

   Derived from the video
5. Expression

   $$A(-2+t) + 3A(1-t) + 2At = 2$$
   Explanation

   Divide both sides by $e^{-t}$ (which is never zero) to simplify the equation.

   Justification

   Algebraic cancellation of a non-zero common factor.

   Shown in the video
6. Expression

   $$(A - 3A + 2A)t + (-2A + 3A) = 2$$
   Explanation

   Expand and collect terms by powers of t.

   Justification

   Polynomial grouping.

   Shown in the video
7. Expression

   $$0 \cdot t + A = 2$$
   Explanation

   Simplify the coefficients: $A - 3A + 2A = 0$ and -2A + $3A = A$.

   Justification

   Arithmetic simplification.

   Shown in the video
8. Expression

   $$A = 2$$
   Explanation

   Solve the resulting linear equation for the undetermined coefficient.

   Justification

   Basic algebra.

   Shown in the video

The undetermined coefficient in the trial particular solution is $A = 2$.

1. Formula
   Observation

   = ($A - 3A + 2A$)$t + (-2A + 3A) = 2 \Rightarrow A = 2$

1. The initial guess for the particular solution is not shown, but the algebraic steps imply a specific form was substituted.

1. Expression

   $$(A - 3A + 2A)t + (-2A + 3A) = 2$$
   Explanation

   After substituting the assumed particular solution and its derivatives into the ODE, the terms are grouped by powers of t.

   Justification

   Algebraic simplification and grouping like terms.

   Shown in the video
2. Expression

   $$0 \cdot t + A = 2$$
   Explanation

   The coefficient of t simplifies to zero ($A - 3A + 2A = 0$), and the constant term simplifies to $A (-2A + 3A = A)$.

   Justification

   Arithmetic calculation.

   Derived from the video
3. Expression

   $$A = 2$$
   Explanation

   Equating the remaining constant term to the right-hand side of the ODE yields the value of A.

   Justification

   Equality of polynomials/functions.

   Shown in the video

The undetermined coefficient A is equal to 2.

1. Audio
   Observation

   The entire clip is a worked example of finding a particular solution for the displayed ODE.

2. Formula
   Observation

   The board carries the example from the ODE through the homogeneous solution, resonant ansatz, and derivative computations.

1. The example is unfinished within this clip because the coefficient A is not solved before the video segment ends.

Find a particular solution of $\ddot{x}+3\dot{x}+2x=2e^{-t}$ when the forcing exponential overlaps a homogeneous solution.

1. The ODE is $\ddot{x}+3\dot{x}+2x=2e^{-t}$.

2. The associated homogeneous equation has characteristic polynomial $r^2+3r+2=0$.

3. The homogeneous solution is $x_h(t)=c_1e^{-2t}+c_2e^{-t}$.

Set up the correct resonant trial for a particular solution and compute the derivatives needed for substitution.

1. Expression

   $$r^2+3r+2=0\Rightarrow (r+2)(r+1)=0$$
   Explanation

   Solve the homogeneous characteristic equation.

   Justification

   Shown directly on the board after trying $x=e^{rt}$.

   Shown in the video
2. Expression

   $$x_h(t)=c_1e^{-2t}+c_2e^{-t}$$
   Explanation

   Write the homogeneous solution from the two roots.

   Justification

   Standard consequence of distinct real roots; displayed on the board.

   Shown in the video
3. Expression

   $$x_p(t)=Ate^{-t}$$
   Explanation

   Because $e^{-t}$ is already a homogeneous mode, multiply the naive guess by $t$.

   Justification

   Explicitly stated by the speaker as the needed modification for resonance.

   Shown in the video
4. Expression

   $$\dot{x}_p=Ae^{-t}(1-t)$$
   Explanation

   Differentiate the trial solution once.

   Justification

   Product rule applied in the audio and written on the board.

   Shown in the video
5. Expression

   $$\ddot{x}_p=Ae^{-t}(-1+t-1)$$
   Explanation

   Differentiate again.

   Justification

   Product rule applied a second time; the unsimplified form is shown on the board.

   Shown in the video

The clip establishes the correct resonant ansatz $x_p(t)=Ate^{-t}$ and computes $\dot{x}_p=Ae^{-t}(1-t)$ and $\ddot{x}_p=Ae^{-t}(-1+t-1)=Ae^{-t}(t-2)$, but it does not reach the final value of $A$.

Verification would require substituting $x_p$, $\dot{x}_p$, and $\ddot{x}_p$ into the original ODE and matching coefficients of $e^{-t}$; that step is not shown in this segment.

1. Formula
   Observation

   The entire segment is dedicated to solving for the particular solution of $\ddot{x} + 3\dot{x} + 2x = 2e^{-t}$.

Find a particular solution to the differential equation $\ddot{x} + 3\dot{x} + 2x = 2e^{-t}$.

1. The differential equation is $\ddot{x} + 3\dot{x} + 2x = 2e^{-t}$.

2. The homogeneous solution contains the term $e^{-t}$, indicating resonance with the forcing function.

Determine the specific form and coefficient of the particular solution $X_p(t)$.

1. Expression

   $$x = Ate^{-t}$$
   Explanation

   Propose a trial solution multiplied by t due to resonance.

   Justification

   Standard rule for undetermined coefficients when the forcing exponent matches a homogeneous root.

   Shown in the video
2. Expression

   $$\dot{x} = Ae^{-t}(1-t), \quad \ddot{x} = Ae^{-t}(-2+t)$$
   Explanation

   Compute the necessary derivatives of the trial solution.

   Justification

   Product rule and chain rule of differentiation.

   Shown in the video
3. Expression

   $$(A - 3A + 2A)t + (-2A + 3A) = 2$$
   Explanation

   Substitute into the ODE, cancel $e^{-t}$, and group like terms.

   Justification

   Algebraic manipulation and equating coefficients.

   Shown in the video
4. Expression

   $$A = 2$$
   Explanation

   Solve for the unknown coefficient A.

   Justification

   Simplifying the grouped equation yields $A = 2$.

   Shown in the video
5. Expression

   $$X_p(t) = 2te^{-t}$$
   Explanation

   Write the final particular solution by substituting A back into the trial form.

   Justification

   Definition of the particular solution based on the solved coefficient.

   Shown in the video

$X_p(t) = 2$te^{-t}

Substituting $X_p(t) = 2$te^{-t} and its derivatives back into the original ODE satisfies the equation.

1. Diagram
   Observation

   An arrow is drawn from the forcing term $2e^{-t}$ in the top equation toward the $e^{-t}$ term in the homogeneous solution.

1. Top equation $\ddot{x}+3\dot{x}+2x=2e^{-t}$

2. Homogeneous solution $x_h(t)=c_1e^{-2t}+c_2e^{-t}$

3. Arrow annotation

1. A new arrow annotation appears connecting the right-hand-side exponential to the matching homogeneous exponential.

1. The algebraic formulas themselves remain unchanged while the arrow is added.

The arrow marks the resonance issue: the forcing contains the same exponential factor as one homogeneous solution, motivating the later multiplication by $t$ in the particular-solution trial.

1. Diagram
   Observation

   A brace is drawn under the parenthetical expression in $\ddot{x}=Ae^{-t}(-1+t-1)$ as the speaker comments on combining the terms.

1. The simplified result is discussed but not fully rewritten before the clip ends.

1. Formula $\ddot{x}=Ae^{-t}(-1+t-1)$

2. Brace annotation under the parentheses

1. A brace is added beneath the grouped terms to emphasize their combination.

1. The prefactor $Ae^{-t}$ remains outside the grouped expression.

The visual grouping highlights that the bracketed terms should be combined, leading to the simplified factor $t-2$.

1. Animation
   Observation

   The digital pen writes the substituted equation line by line, grouping terms with t and without t.

1. Handwritten mathematical expressions on a black background.

1. New lines of algebraic expressions appear sequentially as the speaker talks.

2. Terms are grouped and simplified visually on the board.

1. The original differential equation and trial solution remain visible at the top of the screen.

The visual progression mirrors the algebraic steps of substituting the trial solution into the ODE and collecting like terms to solve for A.

1. Animation
   Observation

   The speaker writes '$X_p(t) = 2$te^{-t}' at the bottom of the board.

1. The final equation $X_p(t) = 2$te^{-t}.

1. The value $A=2$ is substituted into the general trial form to produce the specific particular solution.

1. The structure of the trial solution x = Ate^{-t} remains the same, only the constant A is fixed.

This visual event concludes the derivation by presenting the explicit formula for the particular solution.

1. Diagram
   Observation

   Handwritten mathematical equations on a black background.

1. Differential equation

2. Algebraic substitution steps

3. Final particular solution

1. The text remains static throughout the 4-second clip.

The visual presents the final algebraic steps of solving for a particular solution using the method of undetermined coefficients.

1. Audio
   Observation

   The speaker says the complication is that $e^{-t}$ is a solution of the homogeneous equation, and therefore the trial particular solution must be multiplied by $t$.

2. Diagram
   Observation

   The arrow from $2e^{-t}$ to the $e^{-t}$ term in $x_h(t)$ visually marks the overlap.

One might try a particular solution of the form $Ae^{-t}$ directly because the right-hand side is proportional to $e^{-t}$.

The video shows that this fails in the resonant case because $e^{-t}$ already belongs to the homogeneous solution, so the correct trial is $Ate^{-t}$.

1. Formula
   Observation

   The video explicitly uses x = Ate^{-t} instead of x = Ae^{-t} for the particular solution.

Students might incorrectly guess the particular solution as x = Ae^{-t} when the forcing term $e^{-t}$ is already part of the homogeneous solution.

When there is resonance (the forcing exponent matches a root of the characteristic equation), the trial solution must be multiplied by t to ensure linear independence from the homogeneous solutions.

1. Formula
   Observation

   The board moves from the nonhomogeneous ODE to 'hom eq: Try $x=e$^{rt}'.

The particular-solution discussion depends on first understanding the associated homogeneous equation and its characteristic roots.

1. Formula
   Observation

   The factorization $(r+2)(r+1)=0$ is followed by the displayed solution $x_h(t)=c_1e^{-2t}+c_2e^{-t}$.

The homogeneous solution is derived directly from the roots found by the characteristic-equation method.

1. Diagram
   Observation

   An arrow connects the forcing term $2e^{-t}$ to the $e^{-t}$ component of the homogeneous solution.

2. Audio
   Observation

   The speaker explains that this overlap forces multiplication by $t$ in the particular-solution trial.

Recognizing that $e^{-t}$ is a homogeneous mode is what motivates the resonant ansatz $Ate^{-t}$.

1. Formula
   Observation

   After writing $x=Ate^{-t}$, the board successively computes $\dot{x}$ and $\ddot{x}$.

The chosen trial solution is then differentiated so it can be substituted into the differential equation.

1. Formula
   Observation

   The second derivative is computed from the previously written first derivative $Ae^{-t}(1-t)$.

The second-derivative calculation depends directly on the already obtained first derivative.

1. Audio
   Observation

   The speaker transitions from stating the trial solution to saying 'So we substitute into the differential equation'.

The chosen trial solution form is directly applied in the substitution step to evaluate the left-hand side of the ODE.

1. Audio
   Observation

   The speaker says, 'So we collect the terms,' immediately after writing the substituted expression.

After substitution, the resulting complex expression must be collected and simplified to isolate the undetermined coefficient.

1. Formula
   Observation

   The method is applied directly to the displayed ODE.

The Method of Undetermined Coefficients is the technique being used to solve the specific second-order linear non-homogeneous ODE shown.

1. Audio
   Observation

   The speaker explicitly says the complication is that $e^{-t}$ solves the homogeneous equation and therefore the right-hand side must be multiplied by $t$.

2. Diagram
   Observation

   The arrow annotation links the forcing term to the homogeneous mode.

1. Homogeneous solution and overlap with the forcing term
2. Resonance-adjusted trial for the particular solution
3. Do not use the unmodified guess for resonant forcing

1. Formula
   Observation

   The board shows $r^2+3r+2=0\Rightarrow(r+2)(r+1)=0$ and then $x_h(t)=c_1e^{-2t}+c_2e^{-t}$.

1. Characteristic-equation method for the homogeneous ODE
2. Homogeneous solution and overlap with the forcing term

1. Formula
   Observation

   The board displays $\dot{x}=Ae^{-t}(1-t)$ and $\ddot{x}=Ae^{-t}(-1+t-1)$.

2. Audio
   Observation

   The speaker verbally applies the product rule twice.

1. The final simplification of the second derivative is not fully written before the clip ends.

1. First derivative of the resonant trial solution
2. Second derivative of the resonant trial solution

1. Formula
   Observation

   The clip ends after computing derivatives but before substituting them into the ODE.