First-Order ODEs Practice Quiz with a Step-by-Step Interactive Lesson
Use the question set below to practice first-order ordinary differential equations with the most important ideas: interpreting \(\frac{dy}{dx}\) as a slope, identifying whether the equation is separable or linear, separating variables, building the integrating factor, applying initial conditions, and checking solutions by substitution and differentiation.
How this first-order ODE practice works
- 1. Take the practice set: answer the first-order ODE questions below.
- 2. Open the lesson (optional): review slope meaning, separable equations, integrating factors, initial-value problems, and common model examples.
- 3. Retry: return to the question set and apply the methods right away.
What you will learn in the first-order ODE lesson
Slope and basic setup
- Slope view: read \(\frac{dy}{dx}\) as the slope of the tangent.
- General form: write equations in terms of derivatives and isolate \(y'\) when needed.
- Practice: move terms carefully and reduce to the clearest solvable form.
Separable equations
- Recognize separable form: rewrite as \(y' = f(x)g(y)\) or \(\frac{dy}{g(y)}=f(x)\,dx\).
- Integrate: integrate both sides and add the constant \(C\).
- Practice examples: natural growth and decay models where separation is immediate.
Linear first-order ODEs
- Standard form: \(y' + P(x)y = Q(x)\).
- Integrating factor: \(\mu(x)=e^{\int P(x)\,dx}\).
- Workflow: multiply by \(\mu\), integrate, then solve for \(C\).
Initial-value problems (IVP)
- Apply initial data: use \(y(x_0)=y_0\) to pick the correct constant.
- Use conditions: convert context statements into valid \((x_0,y_0)\) information.
- Practice: check model solutions against the original equation and initial condition.
Practice set
EDO del primo ordine 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.
Risolvi l'EDO \(\frac{dy}{dx} = -5\). Qual è la soluzione generale?
Correct answer: D. \(y = -5x + C\)
Explanation: Integra: \(y = -5x + C\).
Risolvi l'EDO separabile \(\frac{dy}{dx} = x\,y\). Qual è la soluzione generale?
Correct answer: B. \(y = Ce^{x^2/2}\)
Explanation: Separa: \(\frac{dy}{y} = x\,dx\). Integra: \(\ln|y| = x^2/2 + C\) ⇒ \(y = Ce^{x^2/2}\).
Risolvi \(\frac{dy}{dx} = y\). Qual è la soluzione generale?
Correct answer: D. \(y = Ce^x\)
Explanation: Separa: \(dy/y = dx\). Integra: \(\ln|y| = x + C\) ⇒ \(y = Ce^x\).
Risolvi \(\frac{dy}{dx} = x\). Qual è la soluzione generale?
Correct answer: B. \(y = x^2/2 + C\)
Explanation: Integra: \(y = x^2/2 + C\).
Risolvi \(\frac{dy}{dx} = 4x\). Qual è la soluzione generale?
Correct answer: B. \(y = 2x^2 + C\)
Explanation: Integra: \(y = 2x^2 + C\).
Risolvi \(\frac{dy}{dx} = 6x^2\). Qual è la soluzione generale?
Correct answer: B. \(y = 2x^3 + C\)
Explanation: Integra: \(y = 2x^3 + C\).
Risolvi l'EDO lineare \(\frac{dy}{dx} + y = 0\). Qual è la soluzione generale?
Correct answer: A. \(y = Ce^{-x}\)
Explanation: Riscrivi: \(dy/y = -dx\). Integra: \(\ln|y| = -x + C\) ⇒ \(y = Ce^{-x}\).
Risolvi \(\frac{dy}{dx} + 3y = 0\). Qual è la soluzione generale?
Correct answer: A. \(y = Ce^{-3x}\)
Explanation: Riscrivi: \(dy/y = -3\,dx\). Integra: \(\ln|y| = -3x + C\) ⇒ \(y = Ce^{-3x}\).
Risolvi \(\frac{dy}{dx} = 2\) con condizione iniziale \(y(0)=5\). Qual è la soluzione particolare?
Correct answer: B. \(y = 2x + 5\)
Explanation: Generale: \(y=2x+C\). Usa \(y(0)=5\) ⇒ \(C=5\). Quindi \(y=2x+5\).
Risolvi \(\frac{dy}{dx} = x\) con \(y(1)=3\). Quanto vale \(y(x)\)?
Correct answer: A. \(y = x^2/2 + 2.5\)
Explanation: Generale: \(y=x^2/2+C\). Usa \(y(1)=3\) ⇒ \(C=3-1/2=2.5\). Quindi \(y=x^2/2+2.5\).
Result
Your score: 0 / 10
Review your result below.

