Differential-Equations · Unit 8 · Video 1 · Interactive Practice

Cartesian Method: Rationalise, Then Take the Real Part

IKey Formulas

Formula	Name	Role
$\tilde{y} = \dfrac{1}{1 + i(\omega/k)}\,e^{i\omega t},\quad y = \mathrm{Re}(\tilde{y})$	Complex solution	Solves $\tilde{y}' + k\tilde{y} = k e^{i\omega t}$ ; its real part solves $y' + ky = k\cos\omega t$
$\dfrac{1}{1 + i(\omega/k)} = \dfrac{1 - i(\omega/k)}{1 + (\omega/k)^2}$	Rationalise	Multiply top and bottom by the conjugate $1 - i(\omega/k)$
$e^{i\omega t} = \cos\omega t + i\sin\omega t$	Euler's definition	Puts the exponential in Cartesian form
$y = \dfrac{\cos\omega t + (\omega/k)\sin\omega t}{1 + (\omega/k)^2}$	Cartesian answer	A cosine plus a multiple of a sine, same frequency $\omega$

Key Insight: Of the four terms in $\left(1 - i\frac{\omega}{k}\right)(\cos\omega t + i\sin\omega t)$ , only two are real — and in the last one the minus from the conjugate and the minus from $i^2 = -1$ cancel, giving $+\frac{\omega}{k}\sin\omega t$ .

IIVisualization 1 — Multiply by the Conjugate

Multiplying $\alpha = 1 + i(\omega/k)$ by its conjugate $\bar{\alpha}$ lands on the real axis at $1 + (\omega/k)^2$ .

ω/k 0.5

IIIVisualization 2 — Two of the Four Terms Are Real

In $\left(1 - i\frac{\omega}{k}\right)(\cos\omega t + i\sin\omega t)$ , the terms with no factor of $i$ or two factors of $i$ are real.

ω/k 0.5 ωt 0.79

IVVisualization 3 — Same Curve, Two Formulas

The Cartesian sum and the polar single cosine trace one curve; only the polar form displays its amplitude and lag.

ω/k 0.5 ωt 1.20

💡 Reading the amplitude and the lag straight off $\cos\omega t + (\omega/k)\sin\omega t$ takes the sinusoidal identity, which the next video states.

VQuiz Questions

Problem 1 · Rationalise the Coefficient

Given: $\omega/k = 2$ , so the coefficient of $e^{i\omega t}$ is $\dfrac{1}{1 + 2i}$ — write it in Cartesian form.

✅ Correct! The denominator

(1 + 2i)(1 - 2i) = 1 + 4 = 5

is real, and the numerator is the conjugate

1 - 2i

❌ The reciprocal of a sum is not the sum of the reciprocals.

\frac{1}{1 + 2i} \neq 1 + \frac{1}{2i}

; multiply top and bottom by the conjugate instead.

❌ Check the numerator. Multiplying by

\frac{1 - 2i}{1 - 2i}

puts the conjugate

1 - 2i

on top, not

1 + 2i

❌ Check the denominator.

(1 + 2i)(1 - 2i) = 1 - (2i)^2 = 1 - 4i^2

, and

i^2 = -1

makes it

1 + 4 = 5

, not

1 - 4

❌ Not quite. Multiply top and bottom by the conjugate

1 - 2i

Show solution

Multiply top and bottom by the conjugate of the denominator:

\frac{1}{1 + 2i}\cdot\frac{1 - 2i}{1 - 2i} = \frac{1 - 2i}{(1 + 2i)(1 - 2i)}.

The cross terms $-2i + 2i$ cancel, and $i^2 = -1$ :

(1 + 2i)(1 - 2i) = 1 - 4i^2 = 1 + 4 = 5.

\frac{1}{1 + 2i} = \frac{1 - 2i}{5} = \frac{1}{5} - \frac{2}{5}i.

This is the general formula $\frac{1 - i(\omega/k)}{1 + (\omega/k)^2}$ with $\omega/k = 2$ .

Problem 2 · Keep the Real Terms (Watch the Sign)

Given: $\omega/k = 3$ — find $\mathrm{Re}\big[(1 - 3i)(\cos\omega t + i\sin\omega t)\big]$ .

✅ Correct! The real terms are

1\cdot\cos\omega t

and

(-3i)(i\sin\omega t) = +3\sin\omega t

❌ Watch the sign.

(-3i)(i\sin\omega t) = -3i^2\sin\omega t = +3\sin\omega t

: the minus from the conjugate and the minus from

i^2

cancel.

❌ That is the imaginary part.

\sin\omega t

and

-3\cos\omega t

come from the two cross terms, each carrying a single factor of

i

❌ One real term is missing. The last term has two factors of

i

, and

i \cdot i = -1

is real.

❌ Not quite. Keep the terms with zero or two factors of

i

Show solution

The product has four terms:

1\cdot\cos\omega t,\quad 1\cdot i\sin\omega t,\quad -3i\cos\omega t,\quad -3i\cdot i\sin\omega t.

The middle two each carry one factor of $i$ , so they are imaginary and drop out of the real part. The last one is $-3i^2\sin\omega t = +3\sin\omega t$ .

\mathrm{Re}\big[(1 - 3i)(\cos\omega t + i\sin\omega t)\big] = \cos\omega t + 3\sin\omega t.

Problem 3 · Solve a Specific Equation

Given: $y' + 4y = 4\cos 2t$ — find the sinusoidal solution by the Cartesian method.

✅ Correct! With

\omega/k = \frac{1}{2}

y = \frac{\cos 2t + \frac{1}{2}\sin 2t}{5/4}

❌ Watch the sign. The real part of

\left(1 - \tfrac{1}{2}i\right)(\cos 2t + i\sin 2t)

\cos 2t + \tfrac{1}{2}\sin 2t

, with a plus.

❌ The real factor in front is missing. Rationalising leaves

\frac{1}{1 + (\omega/k)^2} = \frac{4}{5}

in front of everything.

❌ Check $\omega/k$ . Here

\omega = 2

and

k = 4

, so

\omega/k = \frac{1}{2}

, not

2

❌ Not quite. Identify

k

and

\omega

, then use

y = \frac{\cos\omega t + (\omega/k)\sin\omega t}{1 + (\omega/k)^2}

Show solution

Matching $y' + ky = k\cos\omega t$ gives $k = 4$ , $\omega = 2$ , so $\omega/k = \frac{1}{2}$ .

Rationalise:

\frac{1}{1 + \tfrac{1}{2}i} = \frac{1 - \tfrac{1}{2}i}{1 + \tfrac{1}{4}} = \frac{4}{5}\left(1 - \tfrac{1}{2}i\right).

Real part:

y = \frac{4}{5}\,\mathrm{Re}\Big[\left(1 - \tfrac{1}{2}i\right)(\cos 2t + i\sin 2t)\Big] = \frac{4}{5}\left(\cos 2t + \tfrac{1}{2}\sin 2t\right) = \frac{4}{5}\cos 2t + \frac{2}{5}\sin 2t.

Check: $y' = -\frac{8}{5}\sin 2t + \frac{4}{5}\cos 2t$ and $4y = \frac{16}{5}\cos 2t + \frac{8}{5}\sin 2t$ , so $y' + 4y = \frac{20}{5}\cos 2t = 4\cos 2t$ ✓

Problem 4 · A Sine Input (Transfer)

Given: $y' + ky = k\sin\omega t$ . Since $\sin\omega t = \mathrm{Im}(e^{i\omega t})$ , the same $\tilde{y}$ works with $y = \mathrm{Im}(\tilde{y})$ — find $y$ .

✅ Correct! The imaginary terms are

1\cdot i\sin\omega t

and

-i\frac{\omega}{k}\cos\omega t

; here the minus survives because no

i^2

appears.

❌ Watch the sign. The term

-i\frac{\omega}{k}\cos\omega t

has only one factor of

i

, so nothing cancels its minus.

❌ That is the real part. It solves the cosine equation; a sine input needs the imaginary part of

\tilde{y}

❌ Check the denominator. Rationalising gives

1 + (\omega/k)^2

; the square root belongs to the polar amplitude, which multiplies a single cosine.

❌ Not quite. Keep the terms with exactly one factor of

i

Show solution

As before,

\tilde{y} = \frac{1}{1 + (\omega/k)^2}\left(1 - i\frac{\omega}{k}\right)(\cos\omega t + i\sin\omega t).

The front factor is real, so it stays in front. Of the four terms, the imaginary ones are the two cross terms:

1\cdot i\sin\omega t \quad\text{and}\quad -i\frac{\omega}{k}\cos\omega t.

Taking their coefficients of $i$ :

y = \mathrm{Im}(\tilde{y}) = \frac{\sin\omega t - (\omega/k)\cos\omega t}{1 + (\omega/k)^2}.

Check: write $r = \omega/k$ , so $kr = \omega$ and $r\omega = kr^2$ . Then

y' + ky = \frac{(\omega\cos\omega t + r\omega\sin\omega t) + (k\sin\omega t - \omega\cos\omega t)}{1 + r^2} = \frac{k(1 + r^2)\sin\omega t}{1 + r^2} = k\sin\omega t\ \checkmark

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