FRBook a Lesson
Concordia University · ENGR 213

Concordia ENGR 213: ordinary differential equations tutoring

One-to-one online tutoring for Concordia ENGR 213. Get help choosing an ODE method, finding missing solution terms and applying initial conditions. Try the original examples and guided practice below.

Book a LessonText usEmail us
Online 60 min: $55 CAD · PricingMidterm & final exam helpTry a free practice questionFirst-order linear equations: integrating-factor guide
Listed course

Course overview

Concordia ENGR 213 introduces the theory and application of ordinary differential equations in engineering. The course moves from first-order models to higher-order linear equations, series solutions and linear systems. Start with a repeated-root or system example below, or use the guided questions to practise choosing a particular solution when the right-hand side is nonzero.

ENGR 213 topics to review

First-order equations and initial-value modelsHigher-order linear equations and forcing methodsComplex numbers and power-series solutionsLinear systems, eigenvalues and eigenvectors

Common difficulties

  • Keeping two independent solution terms when a characteristic root repeats
  • Distinguishing the initial value from the initial derivative
  • Keeping the state-variable order consistent in a linear system

Choose the differential-equation step to practise

Use the equation you are studying to choose a starting point. Follow your section’s topics and permitted methods.

Preparing for this course

Review the product rule, exponential derivatives, basic integration and multiplication of a matrix by a column vector. In these exercises, all variables are dimensionless. First solve the short system diagnostic; the worked example then explains why a repeated root needs a second solution term.

Registration requirements

Concordia lists MATH 204 as a course to complete beforehand or take concurrently, and MATH 205 as a course to complete beforehand. Confirm your own eligibility and any equivalent preparation with Concordia.

Check the official course catalogue

Check a second-order equation written as a system

Let tt be real and y=y(t)y=y(t) be a real, twice-differentiable function. Primes denote derivatives with respect to tt; all variables are dimensionless. Consider y′′+3y′+2y=0,y(0)=2,y′(0)=−1.\begin{gathered}y^{\prime\prime}+3y^{\prime}+2y=0,\\ y(0)=2,\quad y^{\prime}(0)=-1.\end{gathered} Define the state variables z1=yz_1=y and z2=y′z_2=y^{\prime}, and the column vector z=(z1,z2)T\mathbf z=(z_1,z_2)^{\mathsf T}, where T\mathsf T means transpose. A student writes z′=(01−3−2)z,z(0)=(22).\begin{gathered}\mathbf z^{\prime}=\begin{pmatrix}0&1\\-3&-2\end{pmatrix}\mathbf z,\\ \mathbf z(0)=\begin{pmatrix}2\\2\end{pmatrix}.\end{gathered} Correct the matrix and initial vector. Then find z1′(0)z_1^{\prime}(0) and z2′(0)z_2^{\prime}(0) without solving the equation.

Show a hint

Differentiate each definition first: what are z1′z_1^{\prime} and z2′z_2^{\prime}? Isolate y′′y^{\prime\prime} in the original equation. Each matrix row must give the derivative of one state variable, in the order (z1,z2)(z_1,z_2).

Show the reasoning

  1. Write the derivatives of the two state variables

    By definition, z1′=y′=z2z_1^{\prime}=y^{\prime}=z_2. The original equation gives z2′=y′′=−2y−3y′=−2z1−3z2.\begin{aligned}z_2^{\prime}&=y^{\prime\prime}\\&=-2y-3y^{\prime}\\&=-2z_1-3z_2.\end{aligned}

    Therefore the correct matrix is A=(01−2−3),z′=Az.\begin{gathered}A=\begin{pmatrix}0&1\\-2&-3\end{pmatrix},\\\mathbf z^{\prime}=A\mathbf z.\end{gathered} Here AA denotes the constant coefficient matrix. Its second row puts the coefficient of yy first and the coefficient of y′y^{\prime} second. The student reversed them.

  2. Transfer each initial condition

    The first component is z1(0)=y(0)=2z_1(0)=y(0)=2. The second is z2(0)=y′(0)=−1z_2(0)=y^{\prime}(0)=-1, so z(0)=(2−1).\mathbf z(0)=\begin{pmatrix}2\\-1\end{pmatrix}. Repeating the value 2 in both positions would impose a different initial derivative.

  3. Check the initial rates in both forms

    Matrix multiplication gives z′(0)=Az(0)=(−1−4+3)=(−1−1).\begin{aligned}\mathbf z^{\prime}(0)&=A\mathbf z(0)\\ &=\begin{pmatrix}-1\\-4+3\end{pmatrix}\\ &=\begin{pmatrix}-1\\-1\end{pmatrix}.\end{aligned} Thus z1′(0)=−1z_1^{\prime}(0)=-1 and z2′(0)=−1z_2^{\prime}(0)=-1.

    The original differential equation also gives y′′(0)=−3(−1)−2(2)=−1.\begin{aligned}y^{\prime\prime}(0)&=-3(-1)-2(2)\\&=-1.\end{aligned} The two forms agree. The vector z(0)\mathbf z(0) records two initial values; z′(0)\mathbf z^{\prime}(0) records their rates of change.

Worked example: a repeated root and two initial conditions

Let t∈Rt\in\mathbb R be a dimensionless independent variable and y=y(t)y=y(t) a real, twice-differentiable function. Primes mean derivatives with respect to tt, and ee is the base of the natural exponential. Solve y′′+4y′+4y=0,y(0)=1,y′(0)=0.\begin{gathered}y^{\prime\prime}+4y^{\prime}+4y=0,\\ y(0)=1,\quad y^{\prime}(0)=0.\end{gathered}

1. Recognize the repeated root

This equation is second-order: its highest derivative is the second derivative. It is linear with constant coefficients because its left-hand side adds constant multiples of the unknown function and its derivatives. It is homogeneous because, with those terms on the left, the right-hand side is zero. Testing y=erty=e^{rt}, where rr is a real constant, gives (r2+4r+4)ert=0(r^2+4r+4)e^{rt}=0. The exponential is nonzero, so r2+4r+4=(r+2)2=0.r^2+4r+4=(r+2)^2=0. The characteristic root r=−2r=-2 occurs twice. This test finds e−2te^{-2t}; it does not yet give two independent solution terms.

2. Find the second independent solution term

Write y=e−2tu(t)y=e^{-2t}u(t), with a new unknown function uu. This substitution is reversible for every real tt. The product rule gives y′=e−2t(u′−2u),y′′=e−2tu′′−4e−2tu′+4e−2tu.\begin{aligned}y^{\prime}&=e^{-2t}(u^{\prime}-2u),\\y^{\prime\prime}&=e^{-2t}u^{\prime\prime}\\&\quad-4e^{-2t}u^{\prime}\\&\quad+4e^{-2t}u.\end{aligned} Substitution in the equation cancels the uu and u′u^{\prime} terms, leaving e−2tu′′=0e^{-2t}u^{\prime\prime}=0. Therefore u′′=0u^{\prime\prime}=0, so u=a+btu=a+bt for arbitrary real constants a,ba,b. Hence y=(a+bt)e−2t.y=(a+bt)e^{-2t}. This explains the extra factor tt instead of simply memorizing it.

3. Use both initial conditions

At zero, y(0)=a=1y(0)=a=1. Differentiating the family gives y′=(b−2a−2bt)e−2t.y^{\prime}=(b-2a-2bt)e^{-2t}. Thus y′(0)=b−2a=0y^{\prime}(0)=b-2a=0, so b=2b=2. The solution is y(t)=(1+2t)e−2t,t∈R.\begin{gathered}y(t)=(1+2t)e^{-2t},\\t\in\mathbb R.\end{gathered}

4. Check the equation and the initial values

The derivatives are y′=−4te−2ty^{\prime}=-4te^{-2t} and y′′=(8t−4)e−2ty^{\prime\prime}=(8t-4)e^{-2t}. In y′′+4y′+4yy^{\prime\prime}+4y^{\prime}+4y, factor out e−2te^{-2t}. The coefficient of tt is 8−16+8=08-16+8=0, and the constant term is −4+4=0-4+4=0, so the equation holds. Also y(0)=1y(0)=1 and y′(0)=0y^{\prime}(0)=0. The formula and the equation are defined for all real tt; there is no denominator that becomes zero, and the coefficient of the highest derivative stays nonzero, so neither restricts the interval.

Try it yourself

Keep the same equation, but replace the conditions with y(0)=0y(0)=0 and y′(0)=1y^{\prime}(0)=1. Find a,ba,b in y=(a+bt)e−2ty=(a+bt)e^{-2t}, then verify both conditions.

Show the explained solution

Now a=0a=0, and b−2a=1b-2a=1 gives b=1b=1. Hence y=te−2ty=te^{-2t} for all real tt. Its derivative is y′=(1−2t)e−2ty^{\prime}=(1-2t)e^{-2t}, so y(0)=0y(0)=0 and y′(0)=1y^{\prime}(0)=1. The second derivative is y′′=(4t−4)e−2ty^{\prime\prime}=(4t-4)e^{-2t}; the equation checks because (4t−4)+4(1−2t)+4t=0(4t-4)+4(1-2t)+4t=0.

Before using a formula on another problem, identify its order, linearity and coefficients. Repeated-root reasoning applies here because the equation is homogeneous with constant coefficients. Other equation types require their own method.

Complex numbers: powers and all roots

Choose the correct argument, calculate powers, and find every distinct root. Start with a quadrant check if your signs do not agree.

How tutoring can help

Start with an ENGR 213 problem you’re stuck on, or revisit the basic concepts with your tutor before choosing a method. Your course outline, if you have it, can help us choose examples. Your tutor can check the equation type, a missing homogeneous or particular term, and how you apply the initial conditions. For a forced constant-coefficient equation, compare your trial with the characteristic roots before expanding derivatives. For a system, check the state-variable order and each matrix row. Then try a changed equation and explain the choice yourself.

Exam preparation for ENGR 213

Practise identifying the equation type before choosing a method. Then check the solution in the original equation and apply every stated condition. Use the topics and permitted methods specified for your own assessment; the examples here cover only two selected skills.

Use your section’s outline to choose the chapters to review. 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.

MATH 204 tutoringVectors and MatricesMATH 205 tutoringDifferential and Integral Calculus II
Independent practice

Free practice and study guides for ENGR 213

For a first-order linear equation, this lesson shows how to put the equation into the form with coefficient 1 on the first derivative, use an integrating factor and find the interval where the solution is valid.

Need help with ENGR 213?

What happens after I request tutoring?

Send your topic and preferred times. We’ll confirm tutoring coverage and the schedule by text or email before payment.

Request ENGR 213 tutoring

Nablio is an independent service and is not affiliated with this institution.

ENGR 213 · official course referenceENGR 213 · official course outline