Differential-Equations · Unit 9 · Video 3 · Interactive Practice

Complex Roots, Real Solutions: The Underdamped Spring y'' + 4y' + 5y = 0

IKey Formulas

Formula	Name	What it says
$r^2 + Ar + B = 0 \;\Rightarrow\; r = a \pm bi$	Complex conjugate roots	Small damping: $A^2 < 4B$
$e^{(a+bi)t} = e^{at}\cos bt + i\,e^{at}\sin bt$	Real and imaginary parts	Euler's formula on $e^{ibt}$
$y = e^{at}(c_1\cos bt + c_2\sin bt)$	General real solution	$a$ sets the envelope, $b$ the frequency
$y = C e^{at}\cos(bt - \phi)$	Amplitude–phase form	$(C, \phi)$ = polar coordinates of $(c_1, c_2)$

Key Insight: If $u + iv$ solves a linear equation with real coefficients, then $u$ and $v$ are each real solutions — constant coefficients are not needed.

IIOne Root, Two Real Solutions

The real part $a$ of the root goes into the exponential; the imaginary part $b$ becomes the frequency.

IIIAmplitude and Phase Lag

For $y = e^{-2t}(c_1\cos t + c_2\sin t)$ , the polar coordinates $(C, \phi)$ of $(c_1, c_2)$ give the envelope $\pm Ce^{-2t}$ and the phase lag.

IVMagnify the Tail

Drawn to scale the tail of $y = \sqrt{5}\,e^{-2t}\cos(t - \phi)$ looks flat, yet it keeps crossing zero at $t = \tfrac{\pi}{2} + \phi + n\pi$ .

Vertical scale ×1

💡 In theory the oscillation never stops: every interval of length $\pi$ holds another zero crossing, each swing smaller than the last by the factor $e^{-2\pi} \approx 0.0019$ .

VQuiz Questions

Problem 1 · Read Off the Real Solutions

Given: $y'' + 6y' + 13y = 0$ — find the general real solution.

✅ Correct! The roots are

-3 \pm 2i

: the real part

-3

goes into the exponential, the imaginary part

2

is the frequency.

❌ Real and imaginary parts swapped. For

r = a \pm bi

the exponential carries

a

(here

-3

) and the cosine and sine carry

b

(here

2

❌ Check the sign of the real part.

r = \frac{-6 \pm \sqrt{36 - 52}}{2}

, so the real part is

-3

, not

3

— a damped spring decays.

❌ Not quite. The roots

-3 \pm 2i

are complex;

-3

and

2

are not two real roots. Split

e^{(-3+2i)t}

into its real and imaginary parts.

Show solution

The characteristic equation is $r^2 + 6r + 13 = 0$ :

r = \frac{-6 \pm \sqrt{36 - 52}}{2} = \frac{-6 \pm \sqrt{-16}}{2} = \frac{-6 \pm 4i}{2} = -3 \pm 2i.

So $a = -3$ , $b = 2$ . The complex solution splits as

e^{(-3+2i)t} = e^{-3t}\cos 2t + i\,e^{-3t}\sin 2t,

and its real and imaginary parts are two real solutions. The general solution is

y = e^{-3t}(c_1\cos 2t + c_2\sin 2t).

Problem 2 · When Does the Theorem Apply?

Given: $u + iv$ is a complex solution, with $u$ and $v$ real functions of $t$ . For which equation can you conclude that $u$ and $v$ are each real solutions?

✅ Correct! The equation is linear and its coefficients

t^2

and

e^t

are real, so the real and imaginary parts separate.

❌ The coefficient $i$ is not real.

i(u' + iv') = -v' + iu'

, so

v'

lands in the real part and the real part no longer involves

u

alone.

❌ This equation is not linear. The product

y\,y'

mixes

u

and

v

(u+iv)(u'+iv')

has real part

uu' - vv'

❌ Constant coefficients are not needed. The proof only uses that each coefficient is a real number at every

t

, so real coefficients that vary with

t

work too.

Show solution

Substitute $y = u + iv$ into $y'' + p(t)y' + q(t)y = 0$ with real $p$ , $q$ and collect the terms with and without $i$ :

(u'' + p\,u' + q\,u) + i\,(v'' + p\,v' + q\,v) = 0.

Because $u, v, p, q$ are real, the first bracket is the real part and the second the imaginary part. A complex number is zero only when both parts are zero, so both brackets vanish: $u$ and $v$ are solutions. Here $p = t^2$ , $q = e^t$ — real, so (b) works.

(a): with $p = i$ , the term $i(u' + iv') = -v' + iu'$ moves $v'$ into the real part; the argument fails.
(c): $y\,y'$ is not linear, so the equation does not split into one equation for $u$ and one for $v$ .

Problem 3 · Fit the Initial Conditions

Given: $y'' + 6y' + 13y = 0$ with $y(0) = 2$ and $y'(0) = 0$ — find the solution.

✅ Correct!

y(0) = c_1 = 2

and

y'(0) = -3c_1 + 2c_2 = 0

give

c_2 = 3

❌ Missing the chain-rule factor. The derivative of

\sin 2t

2\cos 2t

, so

y'(0) = -3c_1 + 2c_2

, not

-3c_1 + c_2

❌ Check the sign.

y'(0) = -3c_1 + 2c_2 = -6 + 2c_2 = 0

gives

c_2 = +3

❌ The exponential has a derivative too. The product rule gives the term

-3e^{-3t}(c_1\cos 2t + c_2\sin 2t)

, which contributes

-3c_1

y'(0)

❌ Not quite. Use

y(0) = c_1

and

y'(0) = ac_1 + bc_2

with

a = -3

b = 2

Show solution

From Problem 1, $y = e^{-3t}(c_1\cos 2t + c_2\sin 2t)$ . The product rule gives

y' = -3e^{-3t}(c_1\cos 2t + c_2\sin 2t) + e^{-3t}(-2c_1\sin 2t + 2c_2\cos 2t).

At $t = 0$ : $e^0 = 1$ , $\cos 0 = 1$ , $\sin 0 = 0$ , so

y(0) = c_1 = 2, \qquad y'(0) = -3c_1 + 2c_2 = -6 + 2c_2 = 0 \;\Rightarrow\; c_2 = 3.

y = e^{-3t}(2\cos 2t + 3\sin 2t).

Problem 4 · Amplitude and First Crossing

Given: $y = e^{-2t}(\cos t + \sqrt{3}\sin t)$ , a solution of $y'' + 4y' + 5y = 0$ — find its amplitude $Ce^{at}$ and the first time $t > 0$ at which $y = 0$ .

What is the amplitude?

What is the first zero crossing?

✅ Correct!

y = 2e^{-2t}\cos\!\left(t - \tfrac{\pi}{3}\right)

, which first crosses zero at

t = \tfrac{\pi}{2} + \tfrac{\pi}{3} = \tfrac{5\pi}{6}

❌ Check the amplitude.

C

is the hypotenuse of the triangle with legs

1

and

\sqrt{3}

C = \sqrt{1 + 3}

❌ The legs don't add.

C

is the hypotenuse,

\sqrt{1^2 + (\sqrt{3})^2} = 2

, not

1 + \sqrt{3}

❌ The amplitude changes with time.

C = 2

multiplies the decaying factor

e^{-2t}

; the amplitude is

Ce^{at}

❌ Check the crossing.

\cos(t - \phi) = 0

first when

t - \phi = \tfrac{\pi}{2}

, with

\tan\phi = \sqrt{3}

❌ The phase is a lag. The curve is a cosine shifted right by

\phi

, so the crossing is at

\tfrac{\pi}{2} + \phi

, not

\tfrac{\pi}{2} - \phi

❌ That is the crossing of an ordinary cosine. This one is shifted right by

\phi = \tfrac{\pi}{3}

Show solution

Polar coordinates of $(c_1, c_2) = (1, \sqrt{3})$ :

C = \sqrt{1^2 + (\sqrt{3})^2} = 2, \qquad \tan\phi = \sqrt{3},\ \text{first quadrant} \;\Rightarrow\; \phi = \tfrac{\pi}{3}.

y = 2e^{-2t}\cos\!\left(t - \tfrac{\pi}{3}\right), \quad \text{amplitude } 2e^{-2t}.

The factor $2e^{-2t}$ is never zero, so $y = 0$ exactly when $\cos(t - \tfrac{\pi}{3}) = 0$ . For $t > 0$ the first such value is $t - \tfrac{\pi}{3} = \tfrac{\pi}{2}$ :

t = \tfrac{\pi}{2} + \tfrac{\pi}{3} = \tfrac{5\pi}{6}.

Check: $\cos\tfrac{5\pi}{6} + \sqrt{3}\sin\tfrac{5\pi}{6} = -\tfrac{\sqrt{3}}{2} + \tfrac{\sqrt{3}}{2} = 0$ .

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