Second-Order Linear ODEs Practice Quiz with a Step-by-Step Interactive Lesson
Use the question set below to practice 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
Andra ordningens linjära ordinära differentialekvationer 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
Lös ODE:n \(\frac{d^2y}{dx^2} = 0\). Vad är den allmänna lösningen?
Correct answer: A. \(y = C_1 x + C_2\)
Explanation: Integrera en gång: \(y'=C_1\); integrera igen: \(y = C_1 x + C_2\).
Question 2Not answered
Lös ODE:n \(\dfrac{d^{2}y}{dx^{2}} + y = 0\). Vad är den allmänna lösningen?
Correct answer: A. \(y = C_{1}\cos x + C_{2}\sin x\)
Explanation: Karakteristisk ekvation: \(r^{2}+1=0 \Rightarrow r = \,\pm i\).
Alltså \(y = C_{1}\cos x + C_{2}\sin x\).
Question 3Not answered
Lös ODE:n \(\dfrac{d^{2}y}{dx^{2}} + \dfrac{dy}{dx} = 0\). Vad är den allmänna lösningen?