Second-Order Linear ODEs Practice Quiz with a Step-by-Step Interactive Lesson
Use the question set below to practise second-order linear ordinary differential equations (second-order linear ODEs) with the most important skills for Differential Equations: writing the characteristic equation for constant-coefficient equations, classifying the roots (distinct real roots, repeated real root, complex conjugate roots), building the general solution using exponential solutions \(e^{rx}\) and (for complex roots) sine and cosine solutions, recognizing homogeneous vs nonhomogeneous linear ODEs, and using the Wronskian to check linear independence of solutions. If you want a refresher, click Start lesson to open a step-by-step guide with worked examples and quick checks.
Answer the question set and review your mistakes at the end.
How this second-order linear ODE practice works
1. Take the practice set: answer the second-order linear ODE questions below.
2. Open the lesson (optional): review the characteristic equation method, root cases, general solutions, Wronskians, and homogeneous vs nonhomogeneous forms with clear examples.
3. Retry: return to the question set and apply the solution templates immediately.
What you will learn in the second-order linear ODEs lesson
Standard form & characteristic equation
Recognize linear ODEs like \(y''+ay'+by=0\) (homogeneous) and \(y''+ay'+by=g(x)\) (nonhomogeneous)
Build the characteristic equation \(r^2+ar+b=0\) for constant coefficients
Connect solution templates to root types: real, repeated, or complex conjugate
Distinct real roots & repeated roots
If \(r_1≠ r_2\) are real: \(y=C_1 e^{r_1 x}+C_2 e^{r_2 x}\)
If the root is repeated \(r\): \(y=(C_1+C_2 x)e^{rx}\)
Solve common factoring cases like \(y''+10y'+21y=0\) and \(y''+6y'+8y=0\)
Complex roots & oscillations
If \(r=\alpha\pm i\beta\): \(y=e^{\alpha x}\bigl(C_1\cos(\beta x)+C_2\sin(\beta x)\bigr)\)
Pure oscillations when \(\alpha=0\): \(y=C_1\cos(\beta x)+C_2\sin(\beta x)\)
Connect \(\beta\) to frequency and solve equations like \(y''+16y=0\)
Wronskian & solution space
Compute the Wronskian \(W(y_1,y_2)=\begin{vmatrix}y_1&y_2\\y_1'&y_2'\end{vmatrix}\) to test linear independence
Know the dimension of the solution space for a homogeneous second-order linear ODE is \(2\)
Use given solutions (like \(e^{3x}\), \(e^x\)) to reconstruct the characteristic equation
Purpose: Master second-order linear ordinary differential equations with constant coefficients by using the characteristic equation method. You will learn to solve homogeneous ODEs of the form \(y''+ay'+by=0\) by converting them into the algebraic equation \(r^2+ar+b=0\), classify roots (two real roots, repeated root, complex conjugate roots), write the correct general solution, compute a Wronskian to confirm linear independence, and recognize when an equation is nonhomogeneous \(y''+ay'+by=g(x)\).
Success criteria
Recognize a second-order linear ODE and rewrite it in standard form \(y''+ay'+by=g(x)\).
Identify whether an equation is homogeneous (\(g(x)=0\)) or nonhomogeneous (\(g(x)≠ 0\)).
Write the characteristic equation \(r^2+ar+b=0\) for \(y''+ay'+by=0\).
Solve the characteristic equation and classify the roots as distinct real, repeated real, or complex conjugate.
Write the correct general solution for each root case.
Connect complex roots \(r=\alpha\pm i\beta\) to solutions \(e^{\alpha x}\cos(\beta x)\) and \(e^{\alpha x}\sin(\beta x)\).
Compute a Wronskian \(W(y_1,y_2)\) and interpret \(W≠ 0\) as linear independence.
Know the dimension of the solution space for a homogeneous second-order linear ODE is \(2\).
Key vocabulary
Second-order linear ODE: an equation involving \(y\), \(y'\), \(y''\) where \(y\) and its derivatives appear linearly, e.g. \(y''+ay'+by=g(x)\).
Characteristic equation: the polynomial equation \(r^2+ar+b=0\) associated with \(y''+ay'+by=0\).
Wronskian: \(W(y_1,y_2)=y_1y_2'-y_1'y_2\), used to test linear independence.
General solution: the family of all solutions, typically \(y=C_1y_1+C_2y_2\) for homogeneous second-order linear ODEs.
Quick pre-check
Pre-check 1: Which equation is nonhomogeneous?
Hint: A nonhomogeneous equation has a nonzero right-hand side \(g(x)≠ 0\).
Pre-check 2: What is the dimension of the solution space for a homogeneous second-order linear ODE?
Hint: A second-order linear homogeneous ODE has two linearly independent solutions \(y_1,y_2\), so the general solution is \(C_1y_1+C_2y_2\).
Characteristic Equation
Standard form and the characteristic equation method
Learning goal: Convert \(y''+ay'+by=0\) into an algebra problem \(r^2+ar+b=0\), then use the roots to build the general solution.
Key idea
For a homogeneous second-order linear ODE with constant coefficients, \[ y''+ay'+by=0, \] we try a solution of the form \(y=e^{rx}\). Then \(y'=re^{rx}\) and \(y''=r^2e^{rx}\). Substitute: \[ r^2e^{rx}+ar e^{rx}+b e^{rx}=0 \quad\Rightarrow\quad (r^2+ar+b)e^{rx}=0. \] Because \(e^{rx}≠ 0\), we get the characteristic equation \[ r^2+ar+b=0. \] Solve this quadratic. The type of roots determines the correct general solution.
Root cases (memorize these templates)
Two distinct real roots \(r_1≠ r_2\): \(y=C_1e^{r_1x}+C_2e^{r_2x}\).
Write the characteristic equation: \[ r^2+10r+21=0. \] Factor: \[ (r+3)(r+7)=0 \quad\Rightarrow\quad r_1=-3,\; r_2=-7. \] Two distinct real roots, so \[ y=C_1e^{-3x}+C_2e^{-7x}. \]
Try it
Try it 1: What is the characteristic equation for \(y''+8y'+16y=0\)?
Hint: Replace \(y\to 1\), \(y'\to r\), \(y''\to r^2\) to get \(r^2+ar+b=0\).
Try it 2: If the characteristic equation has a repeated root \(r=3\), what is the general solution?
Hint: A repeated root needs the extra factor \(x\) for the second independent solution.
Summary
For \(y''+ay'+by=0\), solve \(r^2+ar+b=0\).
Use the root templates to write the general solution quickly and correctly.
Distinct Real Roots
Two distinct real roots: factoring and fast solutions
Learning goal: Solve constant-coefficient ODEs when the characteristic equation factors into \((r-r_1)(r-r_2)=0\).
Key idea
If the characteristic equation has two different real roots \(r_1≠ r_2\), then \(e^{r_1x}\) and \(e^{r_2x}\) are linearly independent solutions, and the general solution is \[ y=C_1e^{r_1x}+C_2e^{r_2x}. \] Many quiz problems are designed so the quadratic factors nicely.
Try it 1: Solve \(\displaystyle y''-6y'+8y=0\). What is the general solution?
Hint: The characteristic equation is \(r^2-6r+8=0=(r-2)(r-4)\).
Try it 2: If the solutions are \(y=e^{3x}\) and \(y=e^{x}\), what is the characteristic equation?
Hint: \(e^{rx}\) corresponds to a root \(r\). So the roots are \(3\) and \(1\).
Summary
Distinct real roots \(\Rightarrow\) sum of exponentials \(C_1e^{r_1x}+C_2e^{r_2x}\).
If you are given exponential solutions, read off the roots and build the characteristic polynomial.
Repeated Roots
Repeated roots: why the \(x e^{rx}\) term appears
Learning goal: Recognize repeated-root quadratics and write the correct general solution without missing the \(x\) factor.
Key idea
If the characteristic equation has a repeated root \(r\), you only get one exponential solution \(e^{rx}\) at first. To get a second linearly independent solution, you multiply by \(x\): \[ y_1=e^{rx},\qquad y_2=xe^{rx}. \] So the general solution is \[ y=(C_1+C_2x)e^{rx}. \]
Do not forget the \(x\) term - it makes the second solution independent.
Complex Roots
Complex roots: sinusoidal solutions and oscillation frequency
Learning goal: Convert complex roots into real-valued solutions using sines and cosines, and connect \(\beta\) to oscillations.
Key idea
If the characteristic equation has complex conjugate roots \[ r=\alpha\pm i\beta, \] then the real-valued general solution is \[ y=e^{\alpha x}\bigl(C_1\cos(\beta x)+C_2\sin(\beta x)\bigr). \] The parameter \(\beta\) is the angular frequency of oscillation. If \(\alpha=0\), the motion is a pure oscillation with constant amplitude. If \(\alpha<0\), the oscillations decay (damping). If \(\alpha>0\), the oscillations grow.
Worked example
Example: Solve \(\displaystyle y''+16y=0\).
Characteristic equation: \[ r^2+16=0 \Rightarrow r^2=-16 \Rightarrow r=\pm 4i. \] Here \(\alpha=0\), \(\beta=4\), so \[ y=C_1\cos(4x)+C_2\sin(4x). \]
Try it
Try it 1: Which ODE yields oscillations with frequency \(2\) (i.e., \(\beta=2\))?
Hint: Pure oscillations come from \(y''+\beta^2 y=0\), so \(\beta^2=4\Rightarrow \beta=2\).
Try it 2: What is the nature of the roots for \(y''+2y'+10y=0\)?
Hint: For \(r^2+2r+10=0\), the discriminant is \(2^2-4\cdot 1\cdot 10=-36<0\).
For \(y''+\beta^2y=0\), the oscillation frequency parameter is \(\beta\).
Related Patterns
Quick connections: first-order linear ODEs and easy second-order reductions
Learning goal: Recognize easy cases like \(y''+y'=0\) or \(y''-4y'=0\) and solve them efficiently using the characteristic equation.
Key idea
Some second-order linear ODEs are especially fast because the characteristic equation factors with \(r=0\) as a root. Examples include \(y''+y'=0\) and \(y''-4y'=0\). Treat them like any other constant-coefficient ODE: \[ y''+ay'+by=0\quad\Rightarrow\quad r^2+ar+b=0. \]
Try it 2: Which function solves the first-order ODE \(y'+3y=0\)?
Hint: \(y'+3y=0\Rightarrow y=Ce^{-3x}\).
Summary
Even "simple-looking" second-order linear ODEs are solved by the same characteristic equation method.
First-order linear homogeneous \(y'+ky=0\Rightarrow y=Ce^{-kx}\).
Wronskian
Wronskian: checking linear independence of solutions
Learning goal: Compute the Wronskian quickly and use it to confirm that two solutions form a fundamental set.
Key idea
For two differentiable functions \(y_1(x)\) and \(y_2(x)\), the Wronskian is \[ W(y_1,y_2)(x)= \begin{vmatrix} y_1 & y_2\\ y_1' & y_2' \end{vmatrix} = y_1y_2'-y_1'y_2. \] If \(W(y_1,y_2)(x_0)≠ 0\) at some point \(x_0\), then \(y_1\) and \(y_2\) are linearly independent (and can be used to build the general solution \(y=C_1y_1+C_2y_2\)).
Worked example
Example: Compute the Wronskian of \(y_1=e^{x}\) and \(y_2=e^{2x}\) at \(x=0\).
First compute derivatives: \(y_1'=e^x\), \(y_2'=2e^{2x}\). Then \[ W=y_1y_2'-y_1'y_2 = e^x(2e^{2x})-(e^x)(e^{2x})= (2-1)e^{3x}=e^{3x}. \] At \(x=0\), \(W(0)=e^{0}=1\).
Try it
Try it 1: What is the Wronskian of \(y_1=e^x\) and \(y_2=e^{2x}\) at \(x=0\)?
Hint: \(W=e^x(2e^{2x})-(e^x)(e^{2x})=e^{3x}\), so \(W(0)=1\).
Try it 2: If the general solution is \(y=C_1e^{x}+C_2e^{-x}\), what is a possible ODE?
Hint: Roots \(r=1\) and \(r=-1\) give \((r-1)(r+1)=r^2-1=0\Rightarrow y''-y=0\).
Summary
\(W(y_1,y_2)(x)=y_1y_2'-y_1'y_2\).
If \(W(x_0)≠ 0\), the functions are linearly independent and can form the general solution.
Big Picture & Practice
Why second-order linear ODEs matter (and final practice set)
Learning goal: Connect the solution templates to real applications (oscillations, damping) and finish with final checks that match common quiz patterns.
Wronskian: \(W≠ 0\) confirms linear independence of two solutions.
Next step: Close this lesson and try your quiz again. If you miss a question, reopen the book and review the page that matches the root case (distinct real, repeated, or complex) you need.
Practice set
Second-Order Linear ODEs practice questions with instant score
Answer all 10 questions below, then get your final score and a mistake review at the end so you know exactly what to improve.
0/10answered
Question 1Not answered
Solve the ODE \(\frac{d^2y}{dx^2} = 0\). What is the general solution?
