2.2 Method of Undetermined Coefficients

Finding particular solutions by intelligent guessing

What You'll Learn

By the end of this lesson, you will be able to:

Decompose the general solution as $y = y_c + y_p$
Identify the complementary solution $y_c$ (homogeneous part)
Determine the form of $y_p$ based on the forcing function $g(t)$
Handle critical case: when your guess overlaps with $y_c$
Substitute and match coefficients to solve for unknowns

What is the Method of Undetermined Coefficients?

For second-order linear ODEs with constant coefficients and special forcing functions, we can find the general solution as a sum:

$$y(t) = y_c(t) + y_p(t)$$

where:

$y_c$ = complementary solution (solves the homogeneous equation $ay'' + by' + cy = 0$)
$y_p$ = particular solution (depends on the forcing function $g(t)$)

The key idea: we guess the form of $y_p$ based on what $g(t)$ looks like, then substitute into the full equation to find the unknown coefficients.

Critical Principle

This method only works for constant-coefficient equations with special forcing functions (polynomials, exponentials, sines, cosines, or products/sums of these). For general functions, use Variation of Parameters instead.

“

The method of undetermined coefficients emerged in the 17th century as mathematicians realized that certain forcing functions naturally suggest particular solution forms. Rather than deriving these forms from first principles each time, we simply guess and verify—a principle of elegant mathematical economy.

— A foundational technique in differential equations, streamlined by generations of mathematicians

"O you who have believed, seek help through patience and prayer. Indeed, Allah is with the patient."

Quran 2:153

The Four Cases

The method of undetermined coefficients works by matching the form of $g(t)$ with an appropriate trial form for $y_p$. We study four main cases:

Theorem — Guessing Forms for y_p

Case 1: Polynomial Forcing

If $g(t) = P_n(t)$ (polynomial of degree $n$), try:

y_p = A_n t^n + A_{n-1}t^{n-1} + \cdots + A_1 t + A_0

Exception: If $1$ is a root of the characteristic equation (i.e., $c=0$), multiply by $t$. If $1$ is a repeated root, multiply by $t^2$.

Case 2: Exponential Forcing

If $g(t) = K e^{\alpha t}$, try:

y_p = A e^{\alpha t}

Exception: If $e^{\alpha t}$ appears in $y_c$, multiply by $t$. If $te^{\alpha t}$ also appears in $y_c$, multiply by $t^2$.

Case 3: Trigonometric Forcing

If $g(t) = K_1 \cos(\omega t) + K_2 \sin(\omega t)$, try:

y_p = A\cos(\omega t) + B\sin(\omega t)

Exception: If $\cos(\omega t)$ and $\sin(\omega t)$ appear in $y_c$, multiply by $t$.

Case 4: Superposition (Sum/Product)

If $g(t) = g_1(t) + g_2(t)$, find $y_{p1}$ for $g_1$ and $y_{p2}$ for $g_2$, then:

y_p = y_{p1} + y_{p2}

Most Common Mistake

Always check: does your trial form overlap with $y_c$? If yes, multiply by $t$ (or $t^2$ for repeated roots). Ignoring this leads to linearly dependent solutions and failure of the method.

Quick Reference Table

$g(t)$	First Guess	If Resonance?
Polynomial degree $n$	$A_n t^n + \cdots + A_0$	Multiply by $t$ (or $t^2$)
$e^{\alpha t}$	$Ae^{\alpha t}$	Use $Ate^{\alpha t}$ (or $At^2e^{\alpha t}$)
$\cos(\omega t)$ or $\sin(\omega t)$	$A\cos(\omega t) + B\sin(\omega t)$	Use $t(A\cos + B\sin)$
$g_1(t) + g_2(t)$	$y_{p1} + y_{p2}$	Combine with care

Step-by-Step Solution Procedure

Find the complementary solution $y_c$. Solve the homogeneous equation $ay'' + by' + cy = 0$ using the characteristic equation. This gives $y_c = C_1 y_1 + C_2 y_2$.
Examine $g(t)$. Determine whether it's polynomial, exponential, trig, or a combination. This tells you the form of $y_p$.
Write the trial form. Based on $g(t)$, write $y_p$ with unknown coefficients. Check: does this form (or any of its derivatives) overlap with terms in $y_c$? If yes, multiply by $t$ (or $t^2$).
Compute derivatives. Find $y_p'$ and $y_p''$.
Substitute into the full equation. Replace $y$, $y'$, $y''$ in $ay'' + by' + cy = g(t)$ with $y_p$, $y_p'$, $y_p''$.
Match coefficients. Collect like terms (same powers, same exponentials, same trig functions) and set coefficients equal. Solve the resulting system.
Write the general solution. $y(t) = y_c(t) + y_p(t)$. If initial conditions are given, apply them to find $C_1$ and $C_2$.

Why This Works

The principle is simple: if a function of a certain form (e.g., $e^{\alpha t}$) is a solution to $ay'' + by' + cy = 0$, then $te^{\alpha t}$ might be needed for $y_p$. By building in these multiplicities from the start, we avoid contradictions and find a unique solution.

Engineering Perspective: The System as a Gain

Key Insight: Filtering and Amplification

A linear constant-coefficient system $ay'' + by' + cy = g(t)$ acts like a filter or amplifier on the input signal $g(t)$. The particular solution $y_p$ represents how the system processes the input. The system either:

Amplifies the input (gain > 1)
Attenuates the input (gain < 1)
Passes through the input unchanged (gain = 1)

Examples of Gain Interpretation

Input $g(t)$	Particular Solution $y_p$	Gain Interpretation
$e^{\alpha t}$	$Ae^{\alpha t}$	Scalar gain $A$; system multiplies the exponential
$\sin(\omega t)$	$A\sin(\omega t) + B\cos(\omega t)$	Amplitude change plus phase shift
$t^n$ (polynomial)	Polynomial of degree $n$	Polynomial structure preserved; coefficients scaled
$e^{rt}$ (where $r$ is a root)	$Ate^{rt}$ or $At^2e^{rt}$	Resonance! Response grows linearly or quadratically

Resonance: The Special "Overlap" Case

When $g(t)$ is itself a solution of the homogeneous equation (that is, when the forcing frequency matches a natural frequency of the system), we have resonance. The system can't simply amplify the input because that would already be a homogeneous solution! Instead, the response grows: we multiply by $t$ or $t^2$ to create a genuinely new solution.

This is exactly the concept of resonance in mechanical vibrations (undamped springs driven at their natural frequency) and electrical circuits (RLC circuits driven at their resonant frequency). The system is being driven at its own natural frequency, and the energy builds up unboundedly—hence the $t$ factor.

Complementary vs. Forced Response

Solution Components

The general solution decomposes into two essential parts:

$y_c(t)$ — the complementary (or homogeneous) solution: represents the natural dynamics of the system. It decays (if damped) or oscillates (if undamped), depending only on the coefficients $a, b, c$.
$y_p(t)$ — the particular solution: represents the forced response to the input. It shows how the system directly processes the external signal $g(t)$.

The general solution $y(t) = y_c(t) + y_p(t)$ combines both: the forced response (driven by $g(t)$) plus the system's own natural dynamics.

This perspective helps you understand not just how to find $y_p$, but what it means physically: every particular solution is the system's way of responding to a given input.

Worked Examples

Click each example to see the full solution. Watch for the critical overlap checking!

Step 1: Find $y_c$. Characteristic equation: $r^2 - 5r + 6 = 0$. Factor: $(r-2)(r-3) = 0$, so $r = 2, 3$.

y_c = C_1 e^{2t} + C_2 e^{3t}

Step 2: Examine $g(t)$. We have $g(t) = 2t$, a polynomial of degree 1.

Step 3: Write trial form. For polynomial degree 1, try $y_p = At + B$. Check overlap: are $1$ and $t$ in $y_c$? No (only exponentials). So this form is fine.

Step 4: Derivatives. $y_p' = A$, $y_p'' = 0$.

Step 5: Substitute.

$$0 - 5A + 6(At + B) = 2t$$

$$6At + (6B - 5A) = 2t$$

Step 6: Match coefficients.

Coefficient of $t$: $6A = 2 \Rightarrow A = \frac{1}{3}$
Constant term: $6B - 5A = 0 \Rightarrow 6B = \frac{5}{3} \Rightarrow B = \frac{5}{18}$

Step 7: General solution.

\boxed{y(t) = C_1 e^{2t} + C_2 e^{3t} + \frac{t}{3} + \frac{5}{18}}

Interpretation

The first two terms represent the natural (homogeneous) response, decaying to zero. The last two terms form the particular solution—the steady-state response to the polynomial forcing.

Step 1: Find $y_c$. Characteristic equation: $r^2 - 7r + 12 = 0$. Factor: $(r-3)(r-4) = 0$, so $r = 3, 4$.

y_c = C_1 e^{3t} + C_2 e^{4t}

Step 2: Examine $g(t)$. We have $g(t) = 4e^{2t}$, an exponential with $\alpha = 2$.

Step 3: Write trial form. For exponential, try $y_p = Ae^{2t}$. Check: is $e^{2t}$ in $y_c$? No (only $e^{3t}$ and $e^{4t}$). So this form is fine.

Step 4: Derivatives. $y_p' = 2Ae^{2t}$, $y_p'' = 4Ae^{2t}$.

Step 5: Substitute.

4Ae^{2t} - 7(2Ae^{2t}) + 12(Ae^{2t}) = 4e^{2t}

Ae^{2t}(4 - 14 + 12) = 4e^{2t}

2Ae^{2t} = 4e^{2t}

Step 6: Solve for $A$. $A = 2$.

Step 7: General solution.

\boxed{y(t) = C_1 e^{3t} + C_2 e^{4t} + 2e^{2t}}

Step 1: Find $y_c$. Same as Example 2: $y_c = C_1 e^{3t} + C_2 e^{4t}$.

Step 2: Examine $g(t)$. We have $g(t) = 5e^{4t}$, an exponential with $\alpha = 4$.

Step 3: Write trial form — WITH RESONANCE CHECK. Naively, try $y_p = Ae^{4t}$. But wait! Is $e^{4t}$ in $y_c$? Yes! It appears as $C_2 e^{4t}$. This is resonance. We must multiply by $t$:

y_p = Ate^{4t}

Step 4: Derivatives.

y_p' = Ae^{4t} + 4Ate^{4t} = Ae^{4t}(1 + 4t)

y_p'' = 4Ae^{4t}(1 + 4t) + 4Ae^{4t} = Ae^{4t}(4 + 16t + 4) = Ae^{4t}(8 + 16t)

Step 5: Substitute.

Ae^{4t}(8 + 16t) - 7Ae^{4t}(1 + 4t) + 12Ate^{4t} = 5e^{4t}

Ae^{4t}[8 + 16t - 7 - 28t + 12t] = 5e^{4t}

Ae^{4t}(1 + 0 \cdot t) = 5e^{4t}

Step 6: Solve for $A$. The constant term gives $A = 5$. (The coefficient of $t$ is automatically zero—this confirms our choice was correct!)

Step 7: General solution.

\boxed{y(t) = C_1 e^{3t} + C_2 e^{4t} + 5te^{4t}}

Critical Lesson

Without the $t$ factor, we would have gotten $0 = 5$, a contradiction! The resonance check saved us. This is why Step 3 must always include overlap verification.

Step 1: Find $y_c$. Characteristic equation: $r^2 - 8r + 16 = 0$. Factor: $(r-4)^2 = 0$, so $r = 4$ (double root).

y_c = (C_1 + C_2 t)e^{4t}

Step 2: Examine $g(t)$. We have $g(t) = 2e^{4t}$.

Step 3: Resonance check. Both $e^{4t}$ and $te^{4t}$ appear in $y_c$! This is double resonance. We must multiply by $t^2$:

y_p = At^2 e^{4t}

Step 4: Derivatives.

y_p' = 2Ate^{4t} + 4At^2e^{4t} = Ae^{4t}(2t + 4t^2)

y_p'' = A[2e^{4t} + 4te^{4t} + 8te^{4t} + 16t^2e^{4t}] = Ae^{4t}(2 + 12t + 16t^2)

Step 5: Substitute.

Ae^{4t}(2 + 12t + 16t^2) - 8Ae^{4t}(2t + 4t^2) + 16At^2e^{4t} = 2e^{4t}

Ae^{4t}(2 + 12t + 16t^2 - 16t - 32t^2 + 16t^2) = 2e^{4t}

Ae^{4t}(2 - 4t + 0 \cdot t^2) = 2e^{4t}

Step 6: Solve for $A$. Constant term: $2A = 2 \Rightarrow A = 1$. (Coefficient of $t$ is automatically zero.)

Step 7: General solution.

\boxed{y(t) = (C_1 + C_2 t)e^{4t} + t^2 e^{4t} = (C_1 + C_2 t + t^2)e^{4t}}

Step 1: Find $y_c$. Characteristic equation: $r^2 - 2r + 1 = 0$. Factor: $(r-1)^2 = 0$, so $r = 1$ (double root).

y_c = (C_1 + C_2 t)e^{t}

Step 2: Examine $g(t)$. We have trig: $g(t) = 5\cos(2t) + 10\sin(2t)$ with $\omega = 2$.

Step 3: Resonance check. Do $\cos(2t)$ or $\sin(2t)$ appear in $y_c = (C_1 + C_2 t)e^{t}$? No. So try:

y_p = A\cos(2t) + B\sin(2t)

Step 4: Derivatives.

y_p' = -2A\sin(2t) + 2B\cos(2t)

y_p'' = -4A\cos(2t) - 4B\sin(2t)

Step 5: Substitute.

-4A\cos(2t) - 4B\sin(2t) - 2[-2A\sin(2t) + 2B\cos(2t)] + [A\cos(2t) + B\sin(2t)] = 5\cos(2t) + 10\sin(2t)

(-4A - 4B + A)\cos(2t) + (-4B + 4A + B)\sin(2t) = 5\cos(2t) + 10\sin(2t)

(-3A - 4B)\cos(2t) + (4A - 3B)\sin(2t) = 5\cos(2t) + 10\sin(2t)

Step 6: Match coefficients.

$\cos(2t)$: $-3A - 4B = 5$
$\sin(2t)$: $4A - 3B = 10$

Solving: From the second equation, $4A - 3B = 10$. Multiply by 3: $12A - 9B = 30$. Multiply the first by 4: $-12A - 16B = 20$. Add: $-25B = 50$, so $B = -2$. Then $4A + 6 = 10$, so $A = 1$.

Step 7: General solution.

\boxed{y(t) = (C_1 + C_2 t)e^{t} + \cos(2t) - 2\sin(2t)}

Step 1: Find $y_c$. Characteristic equation: $r^2 + 4 = 0$. Roots: $r = \pm 2i$, with $\omega_0 = 2$.

y_c = C_1\cos(2t) + C_2\sin(2t)

Step 2: Examine $g(t)$. We have $g(t) = 8\cos(2t)$ with $\omega = 2$.

Step 3: Resonance check. Both $\cos(2t)$ and $\sin(2t)$ appear in $y_c$! This is trig resonance (natural frequency matches forcing frequency). Multiply by $t$:

y_p = t[A\cos(2t) + B\sin(2t)]

Step 4: Derivatives (product rule!).

y_p' = A\cos(2t) + B\sin(2t) + t[-2A\sin(2t) + 2B\cos(2t)]

y_p' = (A + 2Bt)\cos(2t) + (B - 2At)\sin(2t)

And $y_p'' = 4B\cos(2t) - 4A\sin(2t) + \text{(terms with } t \text{)}$. [Detailed expansion: apply product rule twice.]

Step 5: Substitute into $y'' + 4y = 8\cos(2t)$. After expansion and matching coefficients:

4B\cos(2t) - 4A\sin(2t) = 8\cos(2t)

Step 6: Solve. $4B = 8 \Rightarrow B = 2$, and $-4A = 0 \Rightarrow A = 0$.

Step 7: General solution.

\boxed{y(t) = C_1\cos(2t) + C_2\sin(2t) + 2t\sin(2t)}

Physical Insight

The term $2t\sin(2t)$ grows without bound! This is resonance — when the forcing frequency matches the natural frequency, energy accumulates and amplitude grows linearly with time. This is why bridges must avoid matching the natural frequency of pedestrians' footsteps.

Step 1: Find $y_c$. Characteristic equation: $r^2 - 4r = r(r - 4) = 0$, so $r = 0, 4$.

y_c = C_1 + C_2 e^{4t}

Step 2: Decompose $g(t)$. We have $g(t) = te^{t} + \cos(2t)$. Split:

For $te^{t}$: try $y_{p1}$
For $\cos(2t)$: try $y_{p2}$

For $te^{t}$: This is a polynomial times exponential with $\alpha = 1$. Try $y_{p1} = (At + B)e^{t}$. Check: is $e^{t}$ in $y_c$? No. Proceed.

Compute derivatives and substitute into $y'' - 4y' = te^{t}$. After algebra: $A = -\frac{1}{3}$, $B = -\frac{2}{9}$.

For $\cos(2t)$: Try $y_{p2} = C\cos(2t) + D\sin(2t)$. Check: are trig in $y_c$? No. Proceed.

Compute and substitute into $y'' - 4y' = \cos(2t)$:

-4C\cos(2t) - 4D\sin(2t) - 4[-2C\sin(2t) + 2D\cos(2t)] = \cos(2t)

(-4C - 8D)\cos(2t) + (-4D + 8C)\sin(2t) = \cos(2t)

Matching: $-4C - 8D = 1$ and $8C - 4D = 0$. Solve: $D = 2C$ from the second, so $-4C - 16C = 1 \Rightarrow C = -\frac{1}{20}$, $D = -\frac{1}{10}$.

General solution (superposition):

\boxed{y(t) = C_1 + C_2 e^{4t} - \frac{1}{3}te^t - \frac{2}{9}e^t - \frac{1}{20}\cos(2t) - \frac{1}{10}\sin(2t)}

Superposition Principle

When $g(t) = g_1 + g_2$, we can solve for $y_{p1}$ and $y_{p2}$ separately, then combine. This reduces a complex problem to simpler pieces.

Step 1: Find $y_c$. Characteristic equation: $r^2 + 3r + 2 = (r+1)(r+2) = 0$, so $r = -1, -2$.

y_c = C_1 e^{-x} + C_2 e^{-2x}

Step 2: Examine $g(x)$. Polynomial times trig. The highest degree is 1 (from $20x$), with trig $\cos(x)$ and $\sin(x)$ at $\omega = 1$.

Step 3: Trial form. For $(ax + b)[\cos(\omega x) + \text{const} \sin(\omega x)]$, try:

y_p = (A_0 + A_1 x)\cos(x) + (B_0 + B_1 x)\sin(x)

Check: are any of $\cos(x), \sin(x), x\cos(x), x\sin(x)$ in $y_c$? No (only exponentials). Proceed.

Step 4 & 5: Derivatives and substitution. [Lengthy algebra omitted.] After computing $y_p'$ and $y_p''$ and substituting into the full equation, matching coefficients of $\cos(x)$ and $\sin(x)$ (and their $x$ multiples):

Constant in $\cos(x)$ coeff: $A_0 = 1$
Linear in $\cos(x)$ coeff: $A_1 = 2$
Constant in $\sin(x)$ coeff: $B_0 = -1$
Linear in $\sin(x)$ coeff: $B_1 = 6$

Step 7: General solution.

\boxed{y(x) = C_1 e^{-x} + C_2 e^{-2x} + (1+2x)\cos(x) + (-1+6x)\sin(x)}

Complex Guesses

When $g(t)$ is a product (polynomial $\times$ trig, or polynomial $\times$ exponential), combine the forms: include all coefficients for all powers of the polynomial, applied to each trig/exp term.

📝 Exam-Style Practice Problems

These problems are similar in style and difficulty to past exam questions. Click each problem to reveal the step-by-step solution.

Practice 1: Find the general solution of $y'' - 4y' + 3y = 5e^{2t}$ ▶

Solution

Step 1: Find $y_c$ (homogeneous solution). Characteristic equation: $r^2 - 4r + 3 = 0$.

(r-1)(r-3) = 0 \Rightarrow r = 1, 3

y_c = c_1 e^{t} + c_2 e^{3t}

Step 2: Check for resonance. Is $e^{2t}$ in $y_c$? No. So we do not need to multiply by $t$.

Step 3: Guess the particular solution. Try $y_p = Ae^{2t}$.

Step 4: Compute derivatives. $y_p' = 2Ae^{2t}$, $y_p'' = 4Ae^{2t}$.

Step 5: Substitute into the equation.

4Ae^{2t} - 4(2Ae^{2t}) + 3(Ae^{2t}) = 5e^{2t}

Ae^{2t}(4 - 8 + 3) = 5e^{2t}

-Ae^{2t} = 5e^{2t} \Rightarrow A = -5

Step 6: General solution.

\boxed{y = c_1 e^{t} + c_2 e^{3t} - 5e^{2t}}

Practice 2: Find the general solution of $y'' + y = 3\cos(t)$ ▶

Solution

Step 1: Find $y_c$. Characteristic equation: $r^2 + 1 = 0 \Rightarrow r = \pm i$.

y_c = c_1\cos(t) + c_2\sin(t)

Step 2: Check for resonance. The forcing is $\cos(t)$. Are $\cos(t)$ and $\sin(t)$ in $y_c$? Yes! This is resonance. Multiply by $t$.

Step 3: Guess the particular solution. Try $y_p = t(A\cos(t) + B\sin(t))$.

Step 4: Compute derivatives (using product rule).

y_p' = (A\cos(t) + B\sin(t)) + t(-A\sin(t) + B\cos(t))

y_p'' = -A\sin(t) + B\cos(t) + (-A\sin(t) + B\cos(t)) + t(-A\cos(t) - B\sin(t))

y_p'' = -2A\sin(t) + 2B\cos(t) + t(-A\cos(t) - B\sin(t))

Step 5: Substitute into $y'' + y = 3\cos(t)$.

-2A\sin(t) + 2B\cos(t) + t(-A\cos(t) - B\sin(t)) + t(A\cos(t) + B\sin(t)) = 3\cos(t)

-2A\sin(t) + 2B\cos(t) = 3\cos(t)

Step 6: Match coefficients. $-2A = 0 \Rightarrow A = 0$, and $2B = 3 \Rightarrow B = \frac{3}{2}$.

Step 7: General solution.

\boxed{y = c_1\cos(t) + c_2\sin(t) + \frac{3}{2}t\sin(t)}

Interactive Visualizer

Explore how changing the equation affects $y_c$, $y_p$, and the general solution. Watch the resonance case!

Preset Equation:

How to Read the Plot

Blue curve: $y_c(t)$ — complementary solution (homogeneous response). Notice how it decays or grows depending on eigenvalues.

Green curve: $y_p(t)$ — particular solution (steady-state or forced response). Depends entirely on $g(t)$.

Red curve: $y(t) = y_c + y_p$ — the complete general solution with $C_1 = C_2 = 0$ (you can adjust with initial conditions).

Practice Quiz

Test your understanding with 5 multiple-choice problems. Choose the correct general form for $y_p$ or identify the solution.

Problem 1

For the equation $y'' + 2y' + y = t^2 + 1$, what is the complementary solution $y_c$?

Problem 2

For $y'' - 3y' + 2y = 5e^{2t}$, should we include a factor of $t$ in $y_p$ (i.e., resonance)?

Problem 3

If the characteristic equation is $(r-2)^2(r+1) = 0$ and $g(t) = 3e^{2t}$, which is the correct trial form for $y_p$?

Problem 4

For $y'' + y = 2\cos(t)$, the characteristic equation gives $r = \pm i$. What is the correct trial form for $y_p$?

Problem 5

For $y'' - 4y' + 3y = t + e^{3t}$, how should we split the problem using superposition?

\(g(t)\)	First Guess	If Resonance?
Polynomial degree \(n\)	\(A_n t^n + \cdots + A_0\)	Multiply by \(t\) (or \(t^2\))
\(e^{\alpha t}\)	\(Ae^{\alpha t}\)	Use \(Ate^{\alpha t}\) (or \(At^2e^{\alpha t}\))
\(\cos(\omega t)\) or \(\sin(\omega t)\)	\(A\cos(\omega t) + B\sin(\omega t)\)	Use \(t(A\cos + B\sin)\)
\(g_1(t) + g_2(t)\)	\(y_{p1} + y_{p2}\)	Combine with care

Input \(g(t)\)	Particular Solution \(y_p\)	Gain Interpretation
\(e^{\alpha t}\)	\(Ae^{\alpha t}\)	Scalar gain \(A\); system multiplies the exponential
\(\sin(\omega t)\)	\(A\sin(\omega t) + B\cos(\omega t)\)	Amplitude change plus phase shift
\(t^n\) (polynomial)	Polynomial of degree \(n\)	Polynomial structure preserved; coefficients scaled
\(e^{rt}\) (where \(r\) is a root)	\(Ate^{rt}\) or \(At^2e^{rt}\)	Resonance! Response grows linearly or quadratically