Calculus and Complex Numbers for Differential Equations, differential equations lesson

Does the function $y = e^{-t}\cos 2t$ solve $y'' + 2y' + 5y = 0$ ? Answering takes two derivatives, each needing the product rule and the chain rule, and seeing where such a function comes from takes complex numbers and Euler’s formula. This lesson collects the calculus used to build and check solutions, with the conditions each tool needs, and settles that question at the end. Throughout, the independent variable is $t$ and $y' = dy/dt$ .

Key terms

Product and chain rules

Definition If $u$ and $v$ are differentiable at $t$ , then so is their product, and

(uv)' = u'v + uv'.

If $g$ is differentiable at $t$ and $f$ is differentiable at $g(t)$ , then the composition $f(g(t))$ is differentiable at $t$ , and

\frac{d}{dt} f\bigl(g(t)\bigr) = f'\bigl(g(t)\bigr)\,g'(t).

Both rules need differentiability at the point in question. A formula that differentiates cleanly on each side of a corner, a jump or an excluded point does not by itself give a derivative at that point.

In plain words When two factors both change, each one contributes its own change while the other is held as it is. When one function sits inside another, the rate of the outer layer is multiplied by the rate of the inner layer.

Example For $te^{-2t}$ the first factor has derivative $1$ , and the second has derivative $-2e^{-2t}$ , where the $-2$ is the chain rule factor from the inner function $-2t$ . So

\begin{aligned} \frac{d}{dt}\left(te^{-2t}\right) &= e^{-2t} - 2te^{-2t} \\[0.4em] &= (1 - 2t)e^{-2t}. \end{aligned}

Treating the factor $t$ as a constant would lose the term $e^{-2t}$ . The chain rule also applies when the inner function is the unknown itself. For a differentiable function $y(t)$ ,

\frac{d}{dt}\bigl[y(t)^2\bigr] = 2y(t)\,y'(t),

and the factor $y'(t)$ is exactly what implicit differentiation produces.

Antiderivatives and constants of integration

Definition An antiderivative of $f$ on an interval $I$ is a function $F$ with $F'(t) = f(t)$ for every $t$ in $I$ . If $F$ is one antiderivative, then every antiderivative on $I$ is $F + C$ for some constant $C$ , because a function with zero derivative on an interval is constant. The indefinite integral $\int f(t)\,dt = F(t) + C$ names this whole family. On a set made of several separate intervals, the constant can be different on each piece.

Two rules for finding antiderivatives come from the derivative rules. Substitution reverses the chain rule, and integration by parts reverses the product rule. For $u$ and $v$ continuously differentiable on $I$ ,

\int uv'\,dt = uv - \int u'v\,dt,

often written $\int u\,dv = uv - \int v\,du$ .

In plain words Differentiating erases constants, so going back from a rate to a function brings a free constant with it. The constant stores the starting information that the derivative threw away, and an initial value fixes it.

Example To integrate $te^{-2t}$ by parts, take $u = t$ and $v' = e^{-2t}$ , so that $u' = 1$ and $v = -\tfrac{1}{2}e^{-2t}$ . Then

\begin{aligned} &\int te^{-2t}\,dt \\[0.4em] &\quad = -\tfrac{1}{2}te^{-2t} + \tfrac{1}{2}\int e^{-2t}\,dt \\[0.4em] &\quad = -\tfrac{1}{4}(2t + 1)e^{-2t} + C. \end{aligned}

Differentiating the result with the product rule gives back $te^{-2t}$ , which confirms it on all of $\mathbb{R}$ .

Fundamental theorem of calculus

Definition Let $f$ be continuous on an open interval $I$ and let $t_0$ be a point of $I$ . The first part of the theorem says that the function

A(t) = \int_{t_0}^{t} f(s)\,ds

is differentiable on $I$ with $A'(t) = f(t)$ . The second part says that $\int_a^b f(s)\,ds = F(b) - F(a)$ for every antiderivative $F$ of $f$ on $I$ and all $a$ and $b$ in $I$ . Consequently the initial-value problem $y' = f(t)$ , $y(t_0) = y_0$ has exactly one solution on $I$ ,

y(t) = y_0 + \int_{t_0}^{t} f(s)\,ds,

because any two solutions differ by a constant, and that constant is $0$ at $t_0$ . The letter $s$ is a dummy variable, kept different from the upper limit $t$ . Continuity of $f$ is the hypothesis. If $f$ has a jump at a point, the accumulated function $A$ is still continuous there but is not differentiable at that point.

In plain words The value now is the starting value plus everything the rate has added since the start. The lower limit anchors the function at the initial time, and the upper limit lets it move.

Example For $y' = 2t$ with $y(1) = 4$ ,

\begin{aligned} y(t) &= 4 + \int_1^t 2s\,ds \\[0.4em] &= 4 + (t^2 - 1) \\[0.4em] &= t^2 + 3. \end{aligned}

The same formula works when no elementary antiderivative exists. The problem $y' = e^{-t^2}$ , $y(0) = 1$ has the solution $y(t) = 1 + \int_0^t e^{-s^2}\,ds$ on $\mathbb{R}$ , and the first part of the theorem confirms it directly, since the integrand is continuous.

Exponential and natural logarithm

Definition For every real constant $a$ , the function $e^{at}$ is differentiable on $\mathbb{R}$ , with

\frac{d}{dt}e^{at} = ae^{at}.

The natural logarithm $\ln t$ is defined for $t > 0$ and is the inverse of the exponential, so $e^{\ln t} = t$ for $t > 0$ and $\ln(e^{t}) = t$ for every real $t$ . Its derivative is $1/t$ for $t > 0$ . Since $\ln\lvert t \rvert$ also has derivative $1/t$ for $t < 0$ , on any interval that does not contain $0$

\int \frac{dt}{t} = \ln\lvert t \rvert + C.

The formula never applies across $0$ , where $1/t$ is undefined, and it is used on one of the intervals $(-\infty, 0)$ and $(0, \infty)$ at a time.

In plain words An exponential is a function whose rate is proportional to its own value, which is exactly what an equation like $y' = ay$ asks for. The logarithm undoes the exponential, and the absolute value lets you integrate $1/t$ on the negative side too, but never across zero.

Example Solve $y' = ay$ for a real constant $a$ . On an interval where $y$ is never zero, dividing by $y$ gives $y'/y = a$ , and the chain rule shows that the left side is the derivative of $\ln\lvert y \rvert$ . Integrating gives $\ln\lvert y \rvert = at + c$ , so $y = \pm e^{c}e^{at}$ , which is $Ce^{at}$ with $C \ne 0$ . The division has lost the zero function, which also solves the equation. The complete answer is $y = Ce^{at}$ for every real $C$ , and the product rule confirms that nothing else is missing. If $y' = ay$ on an interval, then

\begin{aligned} \frac{d}{dt}\bigl(e^{-at}y\bigr) &= e^{-at}(y' - ay) \\[0.4em] &= 0, \end{aligned}

so $e^{-at}y$ is a constant $C$ on that interval.

Complex numbers

Definition A complex number has the form $z = a + ib$ with real numbers $a$ and $b$ , where the imaginary unit $i$ satisfies $i^2 = -1$ . The real part is $\operatorname{Re} z = a$ and the imaginary part is $\operatorname{Im} z = b$ . Complex numbers are added part by part and multiplied by expanding brackets and replacing $i^2$ by $-1$ , so $(a + ib)(c + id) = (ac - bd) + i(ad + bc)$ . The conjugate of $z$ is $\bar{z} = a - ib$ and its modulus is $\lvert z \rvert = \sqrt{a^2 + b^2}$ , with $z\bar{z} = \lvert z \rvert^2$ . Two complex numbers are equal exactly when their real parts are equal and their imaginary parts are equal, so one complex equation carries two real equations.

In plain words A complex number is a point of the plane, with the real part along one axis and the imaginary part along the other. The one new rule, $i^2 = -1$ , is all you need to compute with it.

Example The quadratic $r^2 + 2r + 5 = 0$ has discriminant $4 - 20 = -16$ , which is negative, so it has no real roots. The two complex numbers whose square is $-16$ are $\pm 4i$ , and the quadratic formula gives

r = \frac{-2 \pm 4i}{2} = -1 \pm 2i.

To check the root $-1 + 2i$ , square it to get $1 - 4i + 4i^2 = -3 - 4i$ . Adding $2r = -2 + 4i$ and $5$ gives the real part $-3 - 2 + 5 = 0$ and the imaginary part $-4 + 4 = 0$ , so $r^2 + 2r + 5 = 0$ . The two roots are conjugates of each other, as the roots of a real quadratic with negative discriminant always are.

Euler’s formula and the complex exponential

Definition Euler’s formula says that for every real $\theta$

e^{i\theta} = \cos\theta + i\sin\theta.

One way to see it uses the exponential series $e^{z} = \sum_{n=0}^{\infty} z^n/n!$ , which converges for every complex $z$ . With $z = i\theta$ the powers of $i$ repeat in the cycle $i, -1, -i, 1$ , the even-power terms add up to the cosine series, and the odd-power terms add up to $i$ times the sine series. The regrouping is allowed because the series converges absolutely. For a complex constant $r = a + ib$ with real $a$ and $b$ and a real variable $t$ , the complex exponential is

e^{rt} = e^{at}(\cos bt + i\sin bt).

The derivative of a complex-valued function of $t$ is taken part by part, and with it $\tfrac{d}{dt}e^{rt} = re^{rt}$ , the same rule as for a real exponent. If a homogeneous linear equation has real coefficients and a complex-valued function $u + iv$ solves it, then its real part $u$ and its imaginary part $v$ are real solutions. Substituting $u + iv$ gives the result for $u$ plus $i$ times the result for $v$ , and a complex number is zero only when both parts are zero.

In plain words The number $e^{i\theta}$ is the point at angle $\theta$ on the unit circle. For $b \ne 0$ the point $e^{ibt}$ turns around that circle as $t$ grows, and the real factor $e^{at}$ scales its distance from the origin. So the real part $e^{at}\cos bt$ is an oscillation whose amplitude decays when $a < 0$ , grows when $a > 0$ and stays fixed when $a = 0$ .

Example The root $r = -1 + 2i$ found above makes $z = e^{rt}$ a solution of $y'' + 2y' + 5y = 0$ , since $z' = rz$ and $z'' = r^2z$ give $z'' + 2z' + 5z = (r^2 + 2r + 5)z = 0$ . The coefficients are real, so the real part $y = e^{-t}\cos 2t$ is a real solution. A direct check with the product and chain rules agrees,

\begin{aligned} y' &= e^{-t}(-\cos 2t - 2\sin 2t), \\[0.4em] y'' &= e^{-t}(-3\cos 2t + 4\sin 2t). \end{aligned}

In $y'' + 2y' + 5y$ the cosine coefficients are $-3$ , $-2$ and $5$ , which add up to $0$ , and the sine coefficients are $4$ and $-4$ , which cancel. So the function solves the equation for every real $t$ , with $y(0) = 1$ and $y'(0) = -1$ .

Check yourself

Differentiate $y = t^2e^{-3t}$ and name the rule behind each step.

Show the answer

The product rule with $u = t^2$ and $v = e^{-3t}$ gives $u'v + uv'$ . Here $u' = 2t$ , and the chain rule gives $v' = -3e^{-3t}$ , since the inner function $-3t$ has derivative $-3$ . So

\begin{aligned} y' &= 2te^{-3t} - 3t^2e^{-3t} \\[0.4em] &= t(2 - 3t)e^{-3t}. \end{aligned}

A classmate computes the integral below and gets $0$ . What is wrong?

\int_{-1}^{1} \frac{dt}{t} = \ln\lvert 1 \rvert - \ln\lvert -1 \rvert = 0

Show the answer

The second part of the fundamental theorem needs the integrand to be continuous on the whole interval $[-1, 1]$ , and $1/t$ is not even defined at $t = 0$ . The antiderivative $\ln\lvert t \rvert$ is valid only on intervals that do not contain $0$ , so the computation jumps across the point where the formula breaks down.

The integral is improper at $0$ , and it diverges. For $0 < \varepsilon < 1$ ,

\int_{\varepsilon}^{1} \frac{dt}{t} = -\ln\varepsilon,

which grows without bound as $\varepsilon \to 0^+$ , and the piece over $[-1, -\varepsilon]$ equals $\ln\varepsilon$ , which goes to $-\infty$ . So the integral has no value, and in particular it is not $0$ . Cancelling the two pieces symmetrically defines a principal value, which is a different notion.

Find the roots of $r^2 + 4r + 13 = 0$ , use Euler’s formula to write two real solutions of $y'' + 4y' + 13y = 0$ , and check one of them directly.

Show the answer

The discriminant is $16 - 52 = -36$ , so the roots are $r = (-4 \pm 6i)/2 = -2 \pm 3i$ . For $r = -2 + 3i$ , the function $e^{rt} = e^{-2t}(\cos 3t + i\sin 3t)$ solves the equation by the same argument as in the last key term. The coefficients are real, so its real and imaginary parts

e^{-2t}\cos 3t, \qquad e^{-2t}\sin 3t

are real solutions. To check $y = e^{-2t}\sin 3t$ , write $c = \cos 3t$ and $s = \sin 3t$ . The product and chain rules give

\begin{aligned} y' &= e^{-2t}(3c - 2s), \\[0.4em] y'' &= e^{-2t}(-12c - 5s). \end{aligned}

In $y'' + 4y' + 13y$ the cosine coefficients are $-12$ and $12$ , and the sine coefficients are $-5$ , $-8$ and $13$ , so both groups add up to $0$ and the function solves the equation for every real $t$ .

Calculus and Complex Numbers for Differential Equations

Key terms

Product and chain rules

Antiderivatives and constants of integration

Fundamental theorem of calculus

Exponential and natural logarithm

Complex numbers

Euler’s formula and the complex exponential

Check yourself

How this lesson connects

Builds on

Used later in