McGill MATH 263: Ordinary Differential Equations Tutoring & Laplace Practice

Listed course

Course overview

McGill MATH 263, Ordinary Differential Equations for Engineers, covers first-order equations, higher-order linear equations, series solutions, Laplace transforms and linear systems. The original practice below develops two Laplace skills: carrying initial conditions through a calculation and recognizing a delayed input.

Open only to students in the Faculty of Engineering.

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Carry initial conditions through a Laplace transform
Check how a delayed input changes the transform

MATH 263 topics to review

First-order ordinary differential equationsSecond- and higher-order linear equationsSeries solutions at ordinary and regular singular pointsLaplace transformsLinear systems with a review of linear algebra

Common difficulties

Keeping the initial-value terms when transforming derivatives
Connecting partial fractions to functions of time
Shifting the whole input expression when a signal starts later

Worked example: keep the initial conditions in a Laplace solution

Let $t\ge0$ be dimensionless time and $y=y(t)$ an unknown real function. Primes denote derivatives with respect to $t$ . Solve the initial-value problem $\begin{gathered}y^{\prime\prime}+3y^{\prime}+2y=4,\\y(0)=1,\qquad y^{\prime}(0)=0.\end{gathered}$ The constant 4 is the forcing term. We will transform the equation, solve an algebraic equation and return to a function of time.

1. Define the transform and retain both initial terms

Write $Y(s)=\mathcal L\{y\}(s)$ , where $\mathcal L$ denotes the one-sided Laplace transform and $s>0$ is a real transform parameter. Its definition is $Y(s)=\int_0^\infty e^{-st}y(t)\,dt.$ Here $e$ is the base of the natural exponential; the integral combines values over nonnegative time with weight $e^{-st}$ . For this problem, the standard derivative identities give $\begin{aligned}\mathcal L\{y^{\prime}\}&=sY-y(0)=sY-1,\\\mathcal L\{y^{\prime\prime}\}&=s^2Y-sy(0)-y^{\prime}(0)\\&=s^2Y-s.\end{aligned}$ To keep the formulas shorter, we leave the argument $s$ out of $Y$ . The constant forcing has transform $\mathcal L\{4\}=4/s$ .

2. Collect the transformed equation

Transform each term and then collect the terms containing $Y$ : $\begin{gathered}(s^2Y-s)+3(sY-1)+2Y=\frac4s,\\(s^2+3s+2)Y=s+3+\frac4s.\end{gathered}$ The extra $s+3$ comes from the nonzero initial value. Since $s^2+3s+2=(s+1)(s+2)$ , $Y(s)=\frac{s^2+3s+4}{s(s+1)(s+2)}.$ Omitting the initial terms would solve a different problem.

3. Find the partial fractions

Let $A,B,C$ be real constants to determine. Write $Y(s)=\frac A s+\frac B{s+1}+\frac C{s+2}.$ Multiplying by the common denominator gives the polynomial identity $\begin{aligned}s^2+3s+4&=A(s+1)(s+2)\\&\quad+Bs(s+2)\\&\quad+Cs(s+1).\end{aligned}$ Substituting $s=0$ gives $4=2A$ , so $A=2$ . Substituting $s=-1$ gives $2=-B$ , so $B=-2$ . Substituting $s=-2$ gives $2=2C$ , so $C=1$ . These substitutions evaluate a polynomial identity, not the original transform integral outside its convergence range.

4. Return to time and interpret the result

The transform pairs $\mathcal L\{1\}=1/s$ and $\mathcal L\{e^{-at}\}=1/(s+a)$ , for a positive real constant $a$ , give $\begin{aligned}Y(s)&=\frac2s-\frac2{s+1}+\frac1{s+2},\\y(t)&=2-2e^{-t}+e^{-2t}.\end{aligned}$ By linearity, we can take the inverse Laplace transform of each term and add the results. The two exponential terms tend to zero, so $y(t)\to2$ as $t\to\infty$ . The constant function $y=2$ balances the forcing because $2(2)=4$ ; the exponential terms adjust the solution to its initial conditions.

5. Check the equation and the two conditions directly

Differentiate the result without using the transform: $\begin{aligned}y^{\prime}(t)&=2e^{-t}-2e^{-2t},\\y^{\prime\prime}(t)&=-2e^{-t}+4e^{-2t}.\end{aligned}$ Substitution into the original equation gives $\begin{gathered}y^{\prime\prime}+3y^{\prime}+2y=4\\+(-2+6-4)e^{-t}\\+(4-6+2)e^{-2t}\\=4.\end{gathered}$ Separately, $y(0)=2-2+1=1$ and $y^{\prime}(0)=2-2=0$ . Both conditions and the equation hold. The coefficients are continuous, and the coefficient of the highest derivative is nonzero. These conditions ensure that this initial-value problem has a unique solution.

Try it yourself

Keep the equation and $y(0)=1$ , but change the initial derivative to $y^{\prime}(0)=1$ . What changes in the transformed equation? Find the new solution and check both initial conditions.

Show the explained solution

Now $\mathcal L\{y^{\prime\prime}\}=s^2Y-s-1$ . Collecting terms gives $\begin{gathered}(s^2+3s+2)Y\\=s+4+\frac4s,\\Y(s)=\frac{(s+2)^2}{s(s+1)(s+2)}\\=\frac{s+2}{s(s+1)}\\=\frac2s-\frac1{s+1}.\end{gathered}$ Thus $y(t)=2-e^{-t}$ . It satisfies $y(0)=1$ and $y^{\prime}(0)=1$ . Its derivatives are $y^{\prime}=e^{-t}$ and $y^{\prime\prime}=-e^{-t}$ , so $y^{\prime\prime}+3y^{\prime}+2y=4$ . The limiting value is still 2, but the initial slope and the decaying terms have changed.

If the initial terms were difficult, write the derivative identities before substituting any numbers. If the inversion was difficult, recombine your partial fractions and compare their numerator with the transformed equation. Then use the delay diagnostic to test a different Laplace skill.

Preparing for this course

Review exponential derivatives, integration by parts and partial fractions. The worked example explains the Laplace notation before using it. Then try the delay check below. The related calculus and linear-algebra pages are optional refreshers; McGill’s registration requirements appear separately.

Registration requirements

McGill lists MATH 262 as a corequisite of MATH 263. The MATH 263 entry does not separately name a prerequisite. The course is restricted to Engineering students and excludes students taking or having taken MATH 315 or MATH 325. Confirm your registration eligibility with McGill.

Check the official course catalogue

Check a delayed exponential input

Let $t\ge0$ be dimensionless time, $s>0$ a real transform parameter, and $g(t)$ an input. Write $G(s)=\mathcal L\{g\}(s)$ for its one-sided Laplace transform. The unit step $u(t-2)$ is 0 before $t=2$ and 1 from $t=2$ onward. A student claims that $G(s)=\frac{e^{-2s}}{s+1}$ corresponds to $g(t)=u(t-2)e^{-t}$ . Correct the input, then find its value at $t=3$ . Explain what must change inside the exponential when the start is delayed.

Show a hint

The factor $e^{-2s}$ delays the original input by 2. At $t=3$ , how much time has passed since the new start? To verify a proposed input directly, begin the transform integral at 2 and substitute $r=t-2$ , where $r$ is elapsed time after the start.

Show the reasoning

Restart the exponential at the new origin
The unshifted input $e^{-t}$ has transform $1/(s+1)$ . A delay of 2 gives $g(t)=u(t-2)e^{-(t-2)}.$ Equivalently, $g(t)=0$ for $0\le t<2$ , and $g(t)=e^{-(t-2)}$ for $t\ge2$ . The argument $t-2$ measures time since activation. With the stated step convention, the input starts at $g(2)=1$ . Its value at the single switching point does not affect the transform.
Verify the delay using the defining integral
The input is zero before 2, so its transform is $\begin{gathered}\mathcal L\{g\}(s)\\=\int_2^\infty e^{-st}e^{-(t-2)}\,dt.\end{gathered}$ Set $r=t-2$ . Then $t=r+2$ , $dt=dr$ , and the lower limit becomes 0: $\begin{gathered}\mathcal L\{g\}(s)\\=e^{-2s}\int_0^\infty e^{-(s+1)r}\,dr\\=\frac{e^{-2s}}{s+1}.\end{gathered}$ The integral converges for the stated $s>0$ .
Compare values one time unit after activation
At $t=3$ , the corrected input is $g(3)=e^{-(3-2)}=e^{-1}$ . The student’s expression gives $e^{-3}$ . In fact, $u(t-2)e^{-t}=e^{-2}g(t)$ , so it also reduces the amplitude by $e^{-2}$ . A pure delay preserves the original starting height and shifts the elapsed-time argument.

How tutoring can help

Start with a MATH 263 problem you’re stuck on or a basic concept you’d like to understand better. Your course outline, if you have it, can help us choose examples. A tutor can help you decide whether Laplace transforms fit the problem, check each term containing an initial value and work through a partial-fraction decomposition. For a delayed input, a tutor can compare its formula before and after it starts, then help you check the transform and the resulting solution separately.

Exam preparation for MATH 263

Practise choosing a method from the equation and the given conditions. For Laplace transforms, keep the initial-value terms, find the inverse Laplace transform of each partial-fraction term, and check the result in the original equation. These exercises cover selected skills; follow your section’s current instructions for the topics and methods in each assessment.

We work through permitted practice and comparable problems, without completing graded assignments or exams on your behalf.

Useful links for preparation or further study—not transfer equivalencies or a required registration sequence.

Independent practice

Free practice and study guides for MATH 263

For first-order linear equations, this lesson explains how to put the equation into the form with coefficient 1 on the first derivative, use an integrating factor and check an initial value. The Laplace example on this page uses a different method.

18 min

Integrating Factors for First-Order Linear ODEs

See why an integrating factor works, normalize a linear ODE, apply an initial condition and keep the correct interval. One worked example, a graph and three explained questions.

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MATH 263 · official course reference