Differential-Equations · Unit 12 · Video 1 · Interactive Practice

Inhomogeneous ODEs: Forced Springs and RLC Circuits

IKey Formulas

Formula	Name	What it says
$y'' + p(x)\,y' + q(x)\,y = f(x)$	Inhomogeneous equation	$f$ is the input (signal, driving or forcing term); $y$ is the response
$y'' + p(x)\,y' + q(x)\,y = 0,\qquad y_c = c_1y_1 + c_2y_2$	Associated homogeneous (reduced) equation	Its general solution is the complementary solution $y_c$
$m\,x'' + b\,x' + k\,x = f(t)$	Forced spring–mass–dashpot	$f(t)$ is an external force; $f = 0$ is the passive system
$L\,i' + R\,i + \dfrac{q}{C} = E(t),\quad q' = i\ \Rightarrow\ L\,i'' + R\,i' + \dfrac{1}{C}\,i = E'(t)$	RLC circuit (Kirchhoff's voltage law)	The input for the current is $E'(t)$ , so a constant battery gives the homogeneous case

Key Insight: The left side describes the system and the right side the push from outside. The passive behaviour ( $f = 0$ ) is always part of the forced solution, which is why the homogeneous equation is solved first.

IIVisualization 1 — Passive versus Forced Spring

With $x'' + 0.5x' + 2x = f(t)$ and the same start $x(0) = 1$ , $x'(0) = 0$ , what does the input change?

Input f(t) t 16

IIIVisualization 2 — What Drives the Current in an RLC Circuit?

In $i'' + i' + 4i = E'(t)$ the current responds to the derivative of the source voltage, not the voltage itself.

Source E(t)

IVQuiz Questions

Problem 1 · The Complementary Solution

Given: $y'' + 4y = x$ — find the complementary solution $y_c$ .

✅ Correct!

y_c

is the general solution of the reduced equation

y'' + 4y = 0

, whose characteristic roots are

\pm 2i

❌ Not quite. The term

\tfrac{x}{4}

answers the full equation, not the reduced one:

y_c

solves

y'' + 4y = 0

, with the input replaced by

0

❌ Check the sign.

r^2 + 4 = 0

gives

r = \pm 2i

, not

\pm 2

; the exponentials

e^{\pm 2x}

solve

y'' - 4y = 0

❌ Not quite.

(\cos 4x)'' + 4\cos 4x = -12\cos 4x \neq 0

. The roots of

r^2 + 4 = 0

are

\pm 2i

, so the frequency is

2

Show solution

Replace the input $f(x) = x$ by $0$ to get the associated homogeneous equation $y'' + 4y = 0$ .

r^2 + 4 = 0 \ \Rightarrow\ r = \pm 2i \ \Rightarrow\ y_c = c_1\cos 2x + c_2\sin 2x.

The input plays no part in $y_c$ ; it only enters through a particular solution of the full equation.

Problem 2 · A Battery in the Circuit

Given: an RLC circuit driven by a battery of constant voltage $E(t) = 12$ — which equation does the current $i(t)$ satisfy?

✅ Correct! The current equation has input

E'(t)

, and the derivative of a constant is

0

: a battery gives the passive circuit.

❌ Not quite.

12

is the right side of the equation for the charge,

L\,q'' + R\,q' + q/C = E(t)

. Differentiating to get the current equation turns

E

into

E'

❌ Not quite. Dividing by

L

would also divide the left side; and the input for

i

E'(t)

, which is

0

here.

❌ Not quite. Going from

q

i

differentiates, it does not integrate: the input becomes

E'(t) = 0

Show solution

Kirchhoff's voltage law gives $L\,i' + R\,i + q/C = E(t)$ . Differentiate each term and use $q' = i$ :

L\,i'' + R\,i' + \frac{1}{C}\,i = E'(t).

With $E(t) = 12$ constant, $E'(t) = 0$ , so $L\,i'' + R\,i' + \tfrac{1}{C}\,i = 0$ — homogeneous, even though a battery is connected.

Problem 3 · Standard Form of a Forced Spring

Given: $2x'' + 6x' + 10x = 4\cos 3t$ — find the coefficients $p$ , $q$ and the input when the equation is written as $x'' + p\,x' + q\,x = f(t)$ .

✅ Correct! Dividing every term by

m = 2

gives

x'' + 3x' + 5x = 2\cos 3t

❌ Almost. The input must be divided by

m = 2

as well:

f(t)/m = 2\cos 3t

❌ Not quite. The standard form starts with a plain

x''

, so divide every term by the leading coefficient

2

first.

❌ Check the order.

p

multiplies

x'

and

q

multiplies

x

p = b/m = 3

and

q = k/m = 5

Show solution

Here $m = 2$ , $b = 6$ , $k = 10$ . Divide the whole equation by $m$ :

x'' + \frac{6}{2}\,x' + \frac{10}{2}\,x = \frac{4\cos 3t}{2} \ \Rightarrow\ x'' + 3x' + 5x = 2\cos 3t.

So $p = b/m = 3$ , $q = k/m = 5$ and $f(t) = 2\cos 3t$ .

Problem 4 · An AC Circuit in Standard Form

Given: an RLC circuit with $L = 0.5$ , $R = 4$ , $C = 0.02$ driven by $E(t) = 5\sin 10t$ — find the equation for the current $i(t)$ written with a plain $i''$ .

✅ Correct!

E'(t) = 50\cos 10t

, and dividing

0.5\,i'' + 4i' + 50i = 50\cos 10t

L = 0.5

doubles every term.

❌ Not quite. That uses

E(t)/L

. The current equation has input

E'(t)/L

, and

E'(t) = 50\cos 10t

❌ Check the last coefficient. After dividing by

L

it is

\frac{1}{LC} = \frac{1}{0.5 \cdot 0.02} = 100

;

50

1/C

before dividing.

❌ Almost. The right side must be divided by

L = 0.5

too:

50\cos 10t / 0.5 = 100\cos 10t

Show solution

The current satisfies $L\,i'' + R\,i' + \tfrac{1}{C}\,i = E'(t)$ with $\tfrac{1}{C} = 50$ and $E'(t) = 50\cos 10t$ :

0.5\,i'' + 4i' + 50i = 50\cos 10t.

Divide by $L = 0.5$ : $p = R/L = 8$ , $q = \tfrac{1}{LC} = 100$ , input $E'(t)/L = 100\cos 10t$ :

i'' + 8i' + 100i = 100\cos 10t.

Solved: 0 / 4