MATH 2410 — Differential Equations

Lecture Notes. This is NOT meant to be a substitute for coming to lecture or consulting the textbook directly, but rather a helpful resource to review for exams/ review what was discussed in lecture.

Section 7.1

Introduction to the Laplace Transform

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What Is a Transform?

A transform is an operation that converts one type of function into a different kind of function. We have already seen several examples of this idea: differentiation transforms \(x^2\) into \(2x\) and \(\ln(x)\) into \(1/x\); integration transforms \(e^{kx}\) into \(\tfrac1ke^{kx}\).

Remark All of the transforms we will study have a linearity property.

Integral transforms. Suppose we have a function \(f(x,y)\) and hold \(y\) constant. Then

\[ \int_a^b f(x,y)\,dx = F(b,y)-F(a,y), \qquad \text{where } F_x(x,y)=f(x,y), \]

turns a function of two variables into a function of one variable \(y\).

Example \[ \int_1^3 2xy^2\,dx = x^2y^2\Big|_{x=1}^{x=3} = 9y^2 - y^2 = 8y^2. \]

We now generalize this idea in two ways: (1) integrate \(f(t)\) against some fixed "kernel" function \(K(s,t)\), and (2) integrate over the unbounded interval \([0,\infty)\):

\[ \int_0^\infty K(s,t)f(t)\,dt = \lim_{b\to\infty}\int_0^b K(s,t)f(t)\,dt, \]

where \(s\) is a real variable. Choosing \(K(s,t)=e^{-st}\) gives the Laplace transform.

The Laplace Transform

Definition Let \(f\) be defined on \([0,\infty)\). The Laplace transform of \(f\) is \[ \mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st}f(t)\,dt, \] defined for all values of \(s\) where the improper integral converges.

Theorem (Linearity) For constants \(c_1,c_2\) and functions \(f(t),g(t)\) defined on \([0,\infty)\), \[ \mathcal{L}\{c_1f(t)+c_2g(t)\} = c_1\mathcal{L}\{f(t)\}+c_2\mathcal{L}\{g(t)\}. \]

This follows immediately from linearity of the integral:

\[ \mathcal{L}\{c_1f+c_2g\} = \int_0^\infty e^{-st}(c_1f+c_2g)\,dt = c_1\int_0^\infty e^{-st}f(t)\,dt + c_2\int_0^\infty e^{-st}g(t)\,dt = c_1\mathcal{L}\{f(t)\}+c_2\mathcal{L}\{g(t)\}. \]

Example Find \(\mathcal{L}\{k\}\) where \(k\) is a constant.

By linearity, \(\mathcal{L}\{k\}=k\,\mathcal{L}\{1\}\), so it suffices to find

\[ \mathcal{L}\{1\} = \int_0^\infty e^{-st}\,dt = \lim_{b\to\infty}\int_0^b e^{-st}\,dt = \lim_{b\to\infty}\left(-\frac1se^{-st}\right)\Big|_0^b = \lim_{b\to\infty}\left(\frac1s - \frac{e^{-sb}}{s}\right). \]

This limit is \(\tfrac1s\) if \(s>0\) (and diverges if \(s\le 0\)). So

\[ \boxed{\mathcal{L}\{k\} = \frac{k}{s}, \qquad s>0.} \]

Example Find \(\mathcal{L}\{mt+k\}\), where \(m,k\) are constants.

By linearity, \(\mathcal{L}\{mt+k\} = m\,\mathcal{L}\{t\}+\mathcal{L}\{k\}\), so we need

\[ \mathcal{L}\{t\} = \int_0^\infty te^{-st}\,dt = \lim_{b\to\infty}\int_0^b te^{-st}\,dt. \]

Using integration by parts (\(u=t\), \(dv=e^{-st}dt\), so \(du=dt\), \(v=-\tfrac1se^{-st}\)),

\[ \int_0^b te^{-st}\,dt = \left(-\frac ts e^{-st} - \frac{1}{s^2}e^{-st}\right)\Big|_0^b = -\frac bs e^{-sb} - \frac{1}{s^2}e^{-sb} + \frac{1}{s^2}. \]

As \(b\to\infty\), the first two terms vanish for \(s>0\), leaving \(\mathcal{L}\{t\} = \tfrac{1}{s^2}\) for \(s>0\). Thus

\[ \boxed{\mathcal{L}\{mt+k\} = \frac{m}{s^2}+\frac{k}{s}, \qquad s>0.} \]

Example Find \(\mathcal{L}\{\sin t\}\). \[ \mathcal{L}\{\sin t\} = \int_0^\infty e^{-st}\sin(t)\,dt. \]

Integrating by parts twice (first with \(u=e^{-st}\), \(dv=\sin(t)dt\), then again on the resulting integral),

\[ \int_0^\infty e^{-st}\sin(t)\,dt = \Big[-e^{-st}\cos(t) - se^{-st}\sin(t)\Big]_0^\infty - s^2\int_0^\infty e^{-st}\sin(t)\,dt. \]

Moving the last term to the left-hand side,

\[ (1+s^2)\,\mathcal{L}\{\sin t\} = \lim_{b\to\infty}\Big(-e^{-sb}\cos(b)-se^{-sb}\sin(b)\Big) - (-1-0) = 1 \qquad (s>0), \]

since the boundary terms vanish as \(b\to\infty\) for \(s>0\). Thus

\[ \boxed{\mathcal{L}\{\sin t\} = \frac{1}{1+s^2}, \qquad s>0.} \]

\(f(t)\)	\(\mathcal{L}\{f(t)\}\)
\(1\)	\(\dfrac1s\)
\(t^n\)	\(\dfrac{n!}{s^{n+1}}\)
\(e^{at}\)	\(\dfrac{1}{s-a}\)
\(\sin(kt)\)	\(\dfrac{k}{s^2+k^2}\)
\(\cos(kt)\)	\(\dfrac{s}{s^2+k^2}\)

Common Laplace transforms (all valid for \(s\) sufficiently large).

Section 7.2

Inverse Transforms and Solving IVPs

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Inverse Laplace Transforms

Definition If \(\mathcal{L}\{f(t)\} = F(s)\), then \(f(t)\) is called the inverse Laplace transform of \(F(s)\), written \(f(t) = \mathcal{L}^{-1}\{F(s)\}\).

Remark Just like the Laplace transform itself, the inverse transform has the linearity property: \[ \mathcal{L}^{-1}\{c_1F(s)+c_2G(s)\} = c_1\mathcal{L}^{-1}\{F(s)\} + c_2\mathcal{L}^{-1}\{G(s)\}. \]

Reading the transform table from Section 7.1 in reverse gives us a small table of inverse transforms:

\[ \mathcal{L}^{-1}\left\{\frac1s\right\}=1, \qquad \mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\}=t, \qquad \mathcal{L}^{-1}\left\{\frac{1}{s+3}\right\}=e^{-3t}. \]

Example Find \(\mathcal{L}^{-1}\left\{\dfrac{-2s+6}{s^2+4}\right\}\).

By linearity,

\[ \mathcal{L}^{-1}\left\{\frac{-2s+6}{s^2+4}\right\} = -2\,\mathcal{L}^{-1}\left\{\frac{s}{s^2+4}\right\} + 6\,\mathcal{L}^{-1}\left\{\frac{1}{s^2+4}\right\}. \]

Using \(\mathcal{L}\{\cos(2t)\}=\dfrac{s}{s^2+4}\) and \(\mathcal{L}\{\sin(2t)\}=\dfrac{2}{s^2+4}\),

\[ \boxed{\mathcal{L}^{-1}\left\{\frac{-2s+6}{s^2+4}\right\} = -2\cos(2t)+6\sin(2t).} \]

When the denominator doesn't match a known form directly, we turn to partial fractions.

Example Find \(\mathcal{L}^{-1}\left\{\dfrac{6s+2}{s^2-1}\right\}\).

Since the inverse transform works with sums and differences but not with products or quotients, we first factor and decompose:

\[ \frac{6s+2}{(s-1)(s+1)} = \frac{A}{s-1}+\frac{B}{s+1}. \]

Then \(6s+2 = A(s+1)+B(s-1)\). Setting \(s=1\): \(8=2A \Rightarrow A=4\). Setting \(s=-1\): \(-4=-2B \Rightarrow B=2\). So

\[ \mathcal{L}^{-1}\left\{\frac{6s+2}{(s-1)(s+1)}\right\} = \mathcal{L}^{-1}\left\{\frac{4}{s-1}\right\} + \mathcal{L}^{-1}\left\{\frac{2}{s+1}\right\} = \boxed{4e^{t}+2e^{-t}.} \]

Solving Initial Value Problems with the Laplace Transform

Theorem Suppose \(f,f',\dots,f^{(n-1)}\) are continuous on \([0,\infty)\) and \(f^{(n)}\) is piecewise continuous on \([0,\infty)\). Then for \(s\) sufficiently large, \[ \mathcal{L}\{f^{(n)}(t)\} = s^nF(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \cdots - f^{(n-1)}(0), \] where \(F(s)=\mathcal{L}\{f(t)\}\).

Method of Solving

Apply the Laplace transform to both sides of the differential equation.
Use the theorem above to rewrite each derivative term, and solve the resulting algebraic equation for \(F(s)=\mathcal{L}\{y(t)\}\).
Apply the inverse Laplace transform (using partial fractions if needed) to recover \(y(t) = \mathcal{L}^{-1}\{F(s)\}\).

Example Solve the initial value problem \(y'+3y=16e^t\), \(y(0)=1\).

Taking the Laplace transform of both sides,

\[ \mathcal{L}\{y'\}+3\mathcal{L}\{y\} = \big(sF(s)-y(0)\big)+3F(s) = \frac{16}{s-1}. \]

Substituting \(y(0)=1\) and solving for \(F(s)\),

\[ F(s)(s+3) = \frac{16}{s-1}+1 = \frac{s+15}{s-1} \;\Longrightarrow\; F(s) = \frac{s+15}{(s+3)(s-1)}. \]

Using partial fractions, \(\dfrac{s+15}{(s+3)(s-1)} = \dfrac{A}{s+3} +\dfrac{B}{s-1}\), so \(s+15=A(s-1)+B(s+3)\). Setting \(s=1\): \(16=4B \Rightarrow B=4\). Setting \(s=-3\): \(12=-4A \Rightarrow A=-3\). So

\[ y = \mathcal{L}^{-1}\{F(s)\} = \mathcal{L}^{-1}\left\{\frac{-3}{s+3}\right\} + \mathcal{L}^{-1}\left\{\frac{4}{s-1}\right\} = \boxed{-3e^{-3t}+4e^{t}.} \]

Remark This method works for any linear differential equation with constant coefficients.

Example Solve the initial value problem \(y''+5y'+4y=0\), \(y(0)=1\), \(y'(0)=0\).

Taking the Laplace transform of both sides,

\[ \mathcal{L}\{y''\}+5\mathcal{L}\{y'\}+4\mathcal{L}\{y\} = \big(s^2F(s)-sy(0)-y'(0)\big) + 5\big(sF(s)-y(0)\big) + 4F(s) = 0. \]

Substituting \(y(0)=1\), \(y'(0)=0\),

\[ F(s)(s^2+5s+4) - s - 5 = 0 \;\Longrightarrow\; F(s) = \frac{s+5}{(s+4)(s+1)}. \]

Using partial fractions, \(\dfrac{s+5}{(s+4)(s+1)} = \dfrac{A}{s+4} +\dfrac{B}{s+1}\), so \(s+5=A(s+1)+B(s+4)\). Setting \(s=-4\): \(1=-3A \Rightarrow A=-\tfrac13\). Setting \(s=-1\): \(4=3B \Rightarrow B=\tfrac43\). So

\[ y = \mathcal{L}^{-1}\left\{\frac{-1/3}{s+4}\right\} + \mathcal{L}^{-1}\left\{\frac{4/3}{s+1}\right\} = \boxed{\frac43e^{-t}-\frac13e^{-4t}.} \]

Section 7.3

Translation Theorems

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The First Translation Theorem

Theorem (First Translation Theorem) If \(\mathcal{L}\{f(t)\}=F(s)\) exists for \(s>c\), and \(a\) is any real number, then \[ \mathcal{L}\{e^{at}f(t)\} = F(s-a), \qquad \text{for } s>a+c. \]

To see why, substitute directly into the definition:

\[ \mathcal{L}\{e^{at}f(t)\} = \int_0^\infty e^{-st}e^{at}f(t)\,dt = \int_0^\infty e^{-(s-a)t}f(t)\,dt = F(s-a). \]

Remark In words: multiplying \(f(t)\) by \(e^{at}\) shifts the transform horizontally by \(a\), replacing every \(s\) in \(F(s)\) with \(s-a\).

Example Since \(\mathcal{L}\{t^3\}=\dfrac{3!}{s^4}=\dfrac{6}{s^4}\), the First Translation Theorem gives \[ \boxed{\mathcal{L}\{e^{5t}t^3\} = \frac{6}{(s-5)^4}.} \]

Example Solve the initial value problem \(y'-y = 1+te^t\), \(y(0)=0\).

Taking the Laplace transform of both sides,

\[ \mathcal{L}\{y'\}-\mathcal{L}\{y\} = \big(sF(s)-y(0)\big)-F(s) = F(s)(s-1). \]

On the right, \(\mathcal{L}\{1\}=\dfrac1s\), and by the First Translation Theorem, \(\mathcal{L}\{te^t\} = \dfrac{1}{(s-1)^2}\) (since \(\mathcal{L}\{t\}=1/s^2\)). So

\[ F(s)(s-1) = \frac1s+\frac{1}{(s-1)^2} \;\Longrightarrow\; F(s) = \frac{1}{s(s-1)}+\frac{1}{(s-1)^3}. \]

Using partial fractions, \(\dfrac{1}{s(s-1)} = -\dfrac1s+\dfrac{1}{s-1}\), so

\[ F(s) = -\frac1s+\frac{1}{s-1}+\frac{1}{(s-1)^3}. \]

Since \(\mathcal{L}^{-1}\{1/s^3\}=t^2/2\), the First Translation Theorem gives \(\mathcal{L}^{-1}\{1/(s-1)^3\} = \tfrac12t^2e^t\). Thus

\[ \boxed{y = -1+e^t+\frac12t^2e^t.} \]

The Unit Step Function and the Second Translation Theorem

Definition The unit step function is \[ \mathscr{U}(t-a) = \begin{cases} 0 & 0\le t\lt a \\ 1 & t\ge a\end{cases}. \]

Example Find \(\mathcal{L}\{f(t)\}\) where \[ f(t) = (2t-3)\mathscr{U}(t-1) = \begin{cases} 0 & 0\le t\le 1 \\ 2t-3 & t>1\end{cases}. \]

Since \(f(t)=0\) for \(t\lt 1\),

\[ \mathcal{L}\{f(t)\} = \int_1^\infty e^{-st}(2t-3)\,dt = 2\int_1^\infty te^{-st}\,dt - 3\int_1^\infty e^{-st}\,dt. \]

Evaluating each piece (using integration by parts on the first),

\[ \int_1^\infty e^{-st}\,dt = \frac{e^{-s}}{s}, \qquad \int_1^\infty te^{-st}\,dt = \frac{e^{-s}}{s}+\frac{e^{-s}}{s^2}. \]

\[ \mathcal{L}\{f(t)\} = 2\left(\frac{e^{-s}}{s}+\frac{e^{-s}}{s^2}\right) - 3\frac{e^{-s}}{s} = \boxed{\frac{2}{s^2}e^{-s} - \frac1s e^{-s}.} \]

More generally, for \(\mathcal{L}\{f(t-a)\mathscr{U}(t-a)\}\), split the integral at \(t=a\) (the piece below \(a\) vanishes) and substitute \(v=t-a\):

\[ \mathcal{L}\{f(t-a)\mathscr{U}(t-a)\} = \int_a^\infty e^{-st}f(t-a)\,dt = \int_0^\infty e^{-s(v+a)}f(v)\,dv = e^{-as}\int_0^\infty e^{-sv}f(v)\,dv = e^{-as}F(s). \]

Theorem (Second Translation Theorem) If \(\mathcal{L}\{f(t)\}=F(s)\) exists for \(s>c\ge0\), and \(a>0\), then \[ \mathcal{L}\{f(t-a)\mathscr{U}(t-a)\} = e^{-as}F(s), \qquad s>c. \]

Remark In words: delaying \(f(t)\) by \(a\) units and switching it on at \(t=a\) multiplies the transform by \(e^{-as}\).

Example Find \(\mathcal{L}\{f(t)\}\) where \[ f(t) = \begin{cases} 2 & 0\le t\lt 3 \\ -2 & t\ge3\end{cases} = 2 - 4\mathscr{U}(t-3). \]

By linearity and the Second Translation Theorem,

\[ \boxed{\mathcal{L}\{f(t)\} = \mathcal{L}\{2\} - 4\mathcal{L}\{\mathscr{U}(t-3)\} = \frac2s - \frac4s e^{-3s}.} \]

Example Solve the initial value problem \(y'+y=f(t)\), \(y(0)=0\), where \[ f(t) = \begin{cases} 0 & 0\le t\lt 1 \\ 5 & t\ge1\end{cases} = 5\mathscr{U}(t-1). \]

Taking the Laplace transform of both sides,

\[ \big(sF(s)-y(0)\big)+F(s) = F(s)(s+1) = 5\mathcal{L}\{\mathscr{U}(t-1)\} = \frac{5}{s}e^{-s}. \]

\[ F(s) = \frac{5e^{-s}}{s(s+1)} = 5e^{-s}\left(\frac1s-\frac{1}{s+1}\right). \]

Applying the Second Translation Theorem to each piece,

\[ \boxed{y = 5\mathscr{U}(t-1) - 5e^{-(t-1)}\mathscr{U}(t-1).} \]

Example Solve the initial value problem \(y''+4y = \sin(t)\mathscr{U}(t-2\pi)\), \(y(0)=1\), \(y'(0)=0\).

Since \(\sin(t)\) is periodic with period \(2\pi\), \(\sin(t-2\pi)=\sin(t)\), so we can write the forcing term as \(\sin(t-2\pi)\mathscr{U}(t-2\pi)\) — exactly the form needed for the Second Translation Theorem, with \(f(t)=\sin(t)\), \(F(s) = \dfrac{1}{s^2+1}\). Taking the Laplace transform of both sides,

\[ \big(s^2F(s)-s\cdot1-0\big)+4F(s) = (s^2+4)F(s)-s = e^{-2\pi s}\cdot\frac{1}{s^2+1}. \]

\[ F(s) = \frac{s}{s^2+4} + \frac{e^{-2\pi s}}{(s^2+1)(s^2+4)}. \]

Using partial fractions, \(\dfrac{1}{(s^2+1)(s^2+4)} = \dfrac{1/3}{s^2+1} - \dfrac{1/3}{s^2+4}\), so

\[ F(s) = \frac{s}{s^2+4} + \frac{e^{-2\pi s}}{3}\cdot\frac{1}{s^2+1} - \frac{e^{-2\pi s}}{3}\cdot\frac{1}{s^2+4}. \]

Taking the inverse transform of each piece (applying the Second Translation Theorem to the last two, and using \(\sin(t-2\pi)=\sin(t)\), \(\sin(2(t-2\pi))=\sin(2t)\)),

\[ \boxed{y = \cos(2t) + \frac13\sin(t)\mathscr{U}(t-2\pi) - \frac16\sin(2t)\mathscr{U}(t-2\pi).} \]

Section 7.4

Derivatives of Transforms

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Differentiating a Laplace Transform

Suppose \(F(s) = \mathcal{L}\{f(t)\}\). Since \(F(s)\) is defined by an integral in \(t\) with parameter \(s\), we can differentiate under the integral sign with respect to \(s\):

\[ \frac{d}{ds}F(s) = \frac{d}{ds}\int_0^\infty e^{-st}f(t)\,dt = \int_0^\infty \frac{\partial}{\partial s}\big(e^{-st}\big)f(t)\,dt = \int_0^\infty -te^{-st}f(t)\,dt = -\mathcal{L}\{tf(t)\}. \]

So \(\mathcal{L}\{tf(t)\} = -F'(s)\). Differentiating a second time,

\[ \frac{d^2}{ds^2}F(s) = \int_0^\infty \frac{\partial^2}{\partial s^2} \big(e^{-st}\big)f(t)\,dt = \int_0^\infty t^2e^{-st}f(t)\,dt = \mathcal{L}\{t^2f(t)\}. \]

Continuing this pattern gives the general result.

Theorem Suppose \(f(t)\) is piecewise continuous on \([0,\infty)\) and \(\mathcal{L}\{f(t)\}=F(s)\). Then for any integer \(n\ge1\), \[ \mathcal{L}\{t^nf(t)\} = (-1)^n\frac{d^n}{ds^n}F(s). \]

Example Find \(\mathcal{L}\{t\sin(4t)\}\).

We know \(\mathcal{L}\{\sin(4t)\} = F(s) = \dfrac{1}{s^2+16}\). By the theorem with \(n=1\),

\[ \mathcal{L}\{t\sin(4t)\} = -F'(s) = -\frac{d}{ds}\left(\frac{1}{s^2+16}\right) = -\left(\frac{-2s}{(s^2+16)^2}\right). \]

Thus

\[ \boxed{\mathcal{L}\{t\sin(4t)\} = \frac{2s}{(s^2+16)^2}.} \]

Example Find \(\mathcal{L}\{t^2\cos(2t)\}\).

We know \(\mathcal{L}\{\cos(2t)\} = F(s) = \dfrac{s}{s^2+4}\). By the theorem with \(n=2\), we need \(F''(s)\). First,

\[ F'(s) = \frac{(s^2+4)-s(2s)}{(s^2+4)^2} = \frac{4-s^2}{(s^2+4)^2}. \]

Differentiating again (quotient rule, with \(u=4-s^2\), \(v=(s^2+4)^{-2}\), so \(u'=-2s\), \(v'=-4s(s^2+4)^{-3}\)),

\[ F''(s) = -2s(s^2+4)^{-2} + (4-s^2)\big({-4s(s^2+4)^{-3}}\big) = \frac{-2s(s^2+4) - 4s(4-s^2)}{(s^2+4)^3} = \frac{2s^3-24s}{(s^2+4)^3}. \]

So \(F''(s) = \dfrac{2s(s^2-12)}{(s^2+4)^3}\), and since \((-1)^2=1\),

\[ \boxed{\mathcal{L}\{t^2\cos(2t)\} = \frac{2s(s^2-12)}{(s^2+4)^3}.} \]