Differential-Equations · Unit 7 · Video 3 · Interactive Practice

Solving y' + ky = k cos ωt by Complexification

IKey Formulas

Formula	Name	Role
$y' + ky = k\cos\omega t$	Real equation, cosine input	The problem: find the response $y$
$\tilde{y}' + k\tilde{y} = k\,e^{i\omega t}$	Complexified equation	$\cos\omega t = \operatorname{Re}(e^{i\omega t})$ , $\sin\omega t = \operatorname{Im}(e^{i\omega t})$
$\tilde{y} = y_1 + i\,y_2$	Real and imaginary parts	$y_1$ is the response to $\cos\omega t$ , $y_2$ the response to $\sin\omega t$
$\tilde{y} = \dfrac{k}{k + i\omega}\,e^{i\omega t} = \dfrac{e^{i\omega t}}{1 + i(\omega/k)}$	Complex solution	Integrating factor $e^{kt}$ , no $+\,c$

Key Insight: Integrating $e^{(k + i\omega)t}$ is one division by $k + i\omega$ , where the cosine would need integration by parts twice, and the result depends on $k$ and $\omega$ only through $\omega/k$ .

IIAngular Frequency and Frequency

$\omega$ counts oscillations in a $t$ -interval of length $2\pi$ ; the frequency $\omega/2\pi$ counts them per unit of $t$ .

Angular frequency ω 2

IIIReal Part for Cosine, Imaginary Part for Sine

For $y' + y = \cos t$ ( $k = \omega = 1$ ), the real part of $\tilde{y}$ answers $\cos t$ and the imaginary part answers $\sin t$ .

t 2.00 Part

IVOnly the Ratio ω/k Matters

The coefficient $\dfrac{k}{k + i\omega} = \dfrac{1}{1 + i(\omega/k)}$ depends on $k$ and $\omega$ only through the ratio $\omega/k$ .

k 2 ω 3

VQuiz Questions

Problem 1 · Angular Frequency vs Frequency

Given: the input $q_e(t) = \cos 3t$ . Find its frequency in the everyday sense: the number of complete oscillations in a $t$ -interval of length $1$ .

✅ Correct!

\cos 3t

makes

3

complete oscillations per interval of length

2\pi

, so

\tfrac{3}{2\pi} \approx 0.477

per unit of

t

❌ Close, but…

3

is the angular frequency

\omega

: it counts oscillations in an interval of length

2\pi

, not of length

1

. Divide by

2\pi

❌ Not quite.

\tfrac{2\pi}{3}

is the period, the length of one oscillation; the frequency is its reciprocal, the number of oscillations per unit length.

❌ Not quite. The frequency is

\omega/2\pi

, so the

2\pi

belongs in the denominator.

Show solution

For $\cos\omega t$ the angular frequency is $\omega = 3$ : three complete oscillations in a $t$ -interval of length $2\pi$ . One unit of $t$ is the fraction $\tfrac{1}{2\pi}$ of that interval, so it holds

\frac{\omega}{2\pi} = \frac{3}{2\pi} \approx 0.477

oscillations. Equivalently, the period is $\tfrac{2\pi}{3}$ and the frequency is its reciprocal $\tfrac{3}{2\pi}$ .

Problem 2 · A Sine Input

Given: $y' + 2y = 2\sin 3t$ . Which complexification produces the response $y$ ?

✅ Correct!

\sin 3t = \operatorname{Im}(e^{3it})

, and the imaginary parts of the complex equation are exactly

y_2' + 2y_2 = 2\sin 3t

❌ Close, but… The real part of

\tilde{y}

solves

y_1' + 2y_1 = 2\cos 3t

, the cosine input. The sine is the imaginary part of

e^{3it}

❌ Check the sign.

\operatorname{Im}(e^{-3it}) = -\sin 3t

, so this gives the response to

-2\sin 3t

, the negative of

y

❌ Not quite. The right side lost its factor

k = 2

\operatorname{Im}

of this

\tilde{y}

solves

y' + 2y = \sin 3t

, so it is only half the response.

❌ Not quite. Replace

2\sin 3t

2e^{3it}

, whose imaginary part it is.

Show solution

By Euler, $2e^{3it} = 2\cos 3t + 2i\sin 3t$ . Substituting $\tilde{y} = y_1 + i\,y_2$ into $\tilde{y}' + 2\tilde{y} = 2e^{3it}$ and matching imaginary parts ( $k = 2$ is real) gives

y_2' + 2y_2 = 2\sin 3t,

so the response is $y = \operatorname{Im}\tilde{y}$ . Solving with the integrating factor $e^{2t}$ :

\tilde{y} = \frac{2}{2 + 3i}\,e^{3it} = \frac{4 - 6i}{13}\,(\cos 3t + i\sin 3t),

y = \operatorname{Im}\tilde{y} = \frac{4\sin 3t - 6\cos 3t}{13}.

Check: $y' + 2y = \dfrac{12\cos 3t + 18\sin 3t}{13} + \dfrac{8\sin 3t - 12\cos 3t}{13} = 2\sin 3t$ ✓

Problem 3 · Taking the Real Part

Given: $y' + y = \cos 2t$ , so $k = 1$ , $\omega = 2$ and $\tilde{y} = \dfrac{e^{2it}}{1 + 2i}$ . Find $y_1 = \operatorname{Re}\tilde{y}$ .

✅ Correct! Multiplying

\tfrac{1 - 2i}{5}

\cos 2t + i\sin 2t

puts

\cos 2t + 2\sin 2t

in the real part.

❌ Close, but…

\operatorname{Re}(c\,e^{2it}) \neq \operatorname{Re}(c)\cos 2t

: the imaginary part

-\tfrac{2}{5}i

of the coefficient times

i\sin 2t

is real too. Multiply out first.

❌ Check the sign.

\tfrac{1}{1 + 2i} = \tfrac{1 - 2i}{5}

, and

(-2i)(i\sin 2t) = +2\sin 2t

❌ Not quite. Multiplying top and bottom by

1 - 2i

puts

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

in the denominator, not

\sqrt{5}

❌ Not quite. Rationalize

\tfrac{1}{1 + 2i}

, multiply by

\cos 2t + i\sin 2t

, then keep the real terms.

Show solution

Step 1: rationalize the coefficient.

\frac{1}{1 + 2i} = \frac{1 - 2i}{(1 + 2i)(1 - 2i)} = \frac{1 - 2i}{5}

Step 2: multiply out with $e^{2it} = \cos 2t + i\sin 2t$ .

\tilde{y} = \frac{1}{5}\big(\cos 2t + 2\sin 2t\big) + \frac{i}{5}\big(\sin 2t - 2\cos 2t\big)

Step 3: keep the real part: $y_1 = \tfrac{1}{5}(\cos 2t + 2\sin 2t)$ .

Check: $y_1' = \tfrac{1}{5}(-2\sin 2t + 4\cos 2t)$ , so $y_1' + y_1 = \tfrac{1}{5}(5\cos 2t) = \cos 2t$ ✓

Problem 4 · Same Ratio, Same Coefficient

Given: $y' + 2y = 2\cos 3t$ , whose complex solution is $\tilde{y} = \dfrac{2}{2 + 3i}\,e^{3it}$ . Which equation's complex solution has the same coefficient in front of its $e^{i\omega t}$ ?

✅ Correct! Doubling both

k

and

\omega

keeps

\omega/k = \tfrac{6}{4} = \tfrac{3}{2}

, so

\tfrac{4}{4 + 6i} = \tfrac{2}{2 + 3i}

❌ Close, but… Swapping

k

and

\omega

gives

\omega/k = \tfrac{2}{3}

, not

\tfrac{3}{2}

: the coefficient becomes

\tfrac{3}{3 + 2i}

❌ Not quite. Doubling

\omega

alone doubles the ratio to

\omega/k = 3

, giving

\tfrac{2}{2 + 6i} = \tfrac{1}{1 + 3i}

❌ Not quite. Doubling

k

alone halves the ratio to

\omega/k = \tfrac{3}{4}

, giving

\tfrac{4}{4 + 3i}

Show solution

Dividing top and bottom by $k$ , the coefficient of $y' + ky = k\cos\omega t$ is

\frac{k}{k + i\omega} = \frac{1}{1 + i(\omega/k)},

so two equations share it exactly when they share $\omega/k$ . Here $\omega/k = \tfrac{3}{2}$ . The options have ratios (a) $\tfrac{2}{3}$ , (b) $\tfrac{6}{2} = 3$ , (c) $\tfrac{3}{4}$ and (d) $\tfrac{6}{4} = \tfrac{3}{2}$ .

Only (d) matches: $\dfrac{4}{4 + 6i} = \dfrac{2}{2 + 3i} = \dfrac{4 - 6i}{13}$ .

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