Differential-Equations · Unit 9 · Video 1 · Interactive Practice

Second-Order Linear ODEs and the Spring–Mass–Dashpot

IKey Formulas

Formula	Name	Role
$y'' + Ay' + By = 0$	Standard form	Second order, linear, homogeneous; $A$ , $B$ constants
$y = c_1 y_1 + c_2 y_2$	General solution	$y_1$ , $y_2$ are solutions; two constants for a second-order equation
$mx'' = -kx - cx' \;\Rightarrow\; mx'' + cx' + kx = 0$	Spring–mass–dashpot	Newton's law with Hooke's spring force and the dashpot force
$A = \dfrac{c}{m},\quad B = \dfrac{k}{m}$	Coefficients	Divide by $m$ : $A$ is the damping, $B$ the spring

Key Insight: Solving $y'' + Ay' + By = 0$ comes down to finding two solutions $y_1$ , $y_2$ ; the two initial conditions $y(0)$ , $y'(0)$ then fix $c_1$ and $c_2$ .

IIVisualization 1 — Two Conditions Fix Two Constants

For $y'' = 0$ , every solution is a line $y = c_1 t + c_2$ : the starting value fixes $c_2$ , the starting slope fixes $c_1$ .

IIIVisualization 2 — Newton's Law on the Cart

The spring force $-kx$ depends only on position, the dashpot force $-cx'$ only on velocity; their sum is $mx''$ .

Position x 1 Velocity x′ 0

💡 With $x' = 0$ the dashpot exerts no force, however far the spring is stretched.

IVVisualization 3 — Only $c/m$ and $k/m$ Matter

Dividing $mx'' + cx' + kx = 0$ by $m$ leaves $A = c/m$ (damping) and $B = k/m$ (spring) to govern the motion.

Mass m 2 Damping c 2 Spring k 8

💡 Challenge: find a second triple $(m, c, k)$ whose curve lands exactly on the dashed one, which has $A = 1$ , $B = 4$ .

VQuiz Questions

Problem 1 · Put the Model in Standard Form

Given: $3x'' + 6x' + 12x = 0$ — find the coefficients $A$ and $B$ of the standard form $x'' + Ax' + Bx = 0$ .

✅ Correct! Dividing every term by

3

gives

x'' + 2x' + 4x = 0

❌ The leading coefficient is not yet $1$ . Standard form starts with

x''

alone, so every term must be divided by

3

❌ The ratios are upside down.

A = \frac{c}{m} = \frac{6}{3}

, not

\frac{3}{6}

; you divide by the coefficient of

x''

❌ Only one term was divided. Dividing the equation by

3

divides every term, including

12x

❌ Not quite. Divide every term by the coefficient of

x''

Show solution

Divide each term by the leading coefficient $3$ (the role played by $m$ ):

\frac{3}{3}x'' + \frac{6}{3}x' + \frac{12}{3}x = 0 \;\Longrightarrow\; x'' + 2x' + 4x = 0.

Comparing with $x'' + Ax' + Bx = 0$ : $A = 2$ and $B = 4$ .

Problem 2 · Signs of the Two Forces

Given: $m = 2$ , $c = 2$ , $k = 8$ . At one instant the cart is compressed toward the wall at $x = -0.5$ and moving away from the wall with $x' = 1$ — find its acceleration $x''$ .

✅ Correct! The compressed spring pushes with

+4

, the dashpot resists the motion with

-2

, so

2x'' = 2

❌ Check the spring force's sign.

-kx = -8(-0.5) = +4

: a compressed spring pushes the mass away from the wall.

❌ That is $mx''$ , the net force. Divide by

m = 2

to get the acceleration.

❌ Check the dashpot force's sign.

-cx' = -2(1) = -2

: the dashpot always opposes the velocity.

❌ Not quite. Use

mx'' = -kx - cx'

with the given signs.

Show solution

Spring (Hooke's law): $F_{\text{spring}} = -kx = -8(-0.5) = 4$ , pointing away from the wall.

Dashpot: $F_{\text{dashpot}} = -cx' = -2(1) = -2$ , opposing the motion.

Newton's law:

mx'' = -kx - cx' = 4 - 2 = 2 \;\Longrightarrow\; x'' = \frac{2}{2} = 1.

Equivalently, in standard form $x'' + x' + 4x = 0$ : $x'' = -x' - 4x = -1 + 2 = 1$ .

Problem 3 · Fit the Constants Together

Given: $y_1 = e^{t}$ and $y_2 = e^{-t}$ are solutions of $y'' - y = 0$ , so $y = c_1 e^{t} + c_2 e^{-t}$ . The initial conditions are $y(0) = 5$ , $y'(0) = 1$ — find $c_1$ and $c_2$ .

What is $c_1$ ?

What is $c_2$ ?

✅ Correct!

y = 3e^{t} + 2e^{-t}

has

y(0) = 3 + 2 = 5

and

y'(0) = 3 - 2 = 1

❌ Here no condition fixes a constant on its own. At

t = 0

both exponentials equal

1

, so

y(0) = c_1 + c_2

and

y'(0) = c_1 - c_2

: two equations, solved together.

❌ Check $c_1$ . Add the equations

c_1 + c_2 = 5

and

c_1 - c_2 = 1

❌ Check $c_2$ . Subtract

c_1 - c_2 = 1

from

c_1 + c_2 = 5

; the derivative of

e^{-t}

carries a minus sign.

Show solution

Differentiate the general solution: $y' = c_1 e^{t} - c_2 e^{-t}$ .

Set $t = 0$ , where $e^{0} = 1$ :

y(0) = c_1 + c_2 = 5, \qquad y'(0) = c_1 - c_2 = 1.

Adding: $2c_1 = 6$ , so $c_1 = 3$ . Subtracting: $2c_2 = 4$ , so $c_2 = 2$ .

y = 3e^{t} + 2e^{-t}.

Check: $y(0) = 3 + 2 = 5$ ✓ and $y'(0) = 3 - 2 = 1$ ✓.

Problem 4 · Two Carts, One Motion?

Given: cart $P$ has $m = 1$ , $c = 3$ , $k = 5$ ; cart $Q$ has $m = 2$ , $c = 6$ , $k = 10$ . Both start from the same $x(0)$ and $x'(0)$ — compare their motions.

✅ Correct! Both reduce to

x'' + 3x' + 5x = 0

, and the same initial conditions fix the same

c_1

c_2

❌ The heavier mass comes with a stiffer spring. What matters is

B = k/m = 10/2 = 5

, the same as for

P

❌ Damping counts relative to the mass.

A = c/m = 6/2 = 3

, the same as for

P

❌ The amplitude is set by the initial conditions, which are the same for both carts; the equations reduce to the same standard form.

❌ Not quite. Divide each equation by its mass and compare.

Show solution

Cart $P$ : $x'' + 3x' + 5x = 0$ directly, since $m = 1$ .

Cart $Q$ : $2x'' + 6x' + 10x = 0$ ; dividing by $2$ gives $x'' + 3x' + 5x = 0$ .

Both have $A = c/m = 3$ and $B = k/m = 5$ , so they satisfy the same equation. Its general solution $c_1 y_1 + c_2 y_2$ is the same for both, and the same $x(0)$ , $x'(0)$ fix the same $c_1$ , $c_2$ . The motions are identical: doubling $m$ , $c$ and $k$ together changes nothing.

Solved: 0 / 4

Second-Order Linear ODEs and the Spring–Mass–Dashpot

IKey Formulas

IIVisualization 1 — Two Conditions Fix Two Constants

IIIVisualization 2 — Newton's Law on the Cart

IVVisualization 3 — Only c/mc/mc/m and k/mk/mk/m Matter

VQuiz Questions

IVVisualization 3 — Only $c/m$ and $k/m$ Matter