First-Order ODEs Practice Quiz with a Step-by-Step Interactive Lesson
Use the question set below to practise 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 \(C\).
- Practise 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
First-Order 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.
Solve the ODE \(\frac{dy}{dx} = -5\). What is the general solution?
Correct answer: D. \(y = -5x + C\)
Explanation: Integrate: \(y = -5x + C\).
Solve the separable ODE \(\frac{dy}{dx} = x\,y\). What is the general solution?
Correct answer: B. \(y = Ce^{x^2/2}\)
Explanation: Separate: \(\frac{dy}{y} = x\,dx\). Integrate: \(\ln|y| = x^2/2 + C\) ⇒ \(y = Ce^{x^2/2}\).
Solve \(\frac{dy}{dx} = y\). What is the general solution?
Correct answer: D. \(y = Ce^x\)
Explanation: Separate: \(dy/y = dx\). Integrate: \(\ln|y| = x + C\) ⇒ \(y = Ce^x\).
Solve \(\frac{dy}{dx} = x\). What is the general solution?
Correct answer: B. \(y = x^2/2 + C\)
Explanation: Integrate: \(y = x^2/2 + C\).
Solve \(\frac{dy}{dx} = 4x\). What is the general solution?
Correct answer: B. \(y = 2x^2 + C\)
Explanation: Integrate: \(y = 2x^2 + C\).
Solve \(\frac{dy}{dx} = 6x^2\). What is the general solution?
Correct answer: B. \(y = 2x^3 + C\)
Explanation: Integrate: \(y = 2x^3 + C\).
Solve the linear ODE \(\frac{dy}{dx} + y = 0\). What is the general solution?
Correct answer: A. \(y = Ce^{-x}\)
Explanation: Rearrange: \(dy/y = -dx\). Integrate: \(\ln|y| = -x + C\) ⇒ \(y = Ce^{-x}\).
Solve \(\frac{dy}{dx} + 3y = 0\). What is the general solution?
Correct answer: A. \(y = Ce^{-3x}\)
Explanation: Rearrange: \(dy/y = -3\,dx\). Integrate: \(\ln|y| = -3x + C\) ⇒ \(y = Ce^{-3x}\).
Solve \(\frac{dy}{dx} = 2\) with initial condition \(y(0)=5\). What is the particular solution?
Correct answer: B. \(y = 2x + 5\)
Explanation: General: \(y=2x+C\). Use \(y(0)=5\) ⇒ \(C=5\). So \(y=2x+5\).
Solve \(\frac{dy}{dx} = x\) with \(y(1)=3\). What is \(y(x)\)?
Correct answer: A. \(y = x^2/2 + 2.5\)
Explanation: General: \(y=x^2/2+C\). Use \(y(1)=3\) ⇒ \(C=3-1/2=2.5\). So \(y=x^2/2+2.5\).
Result
Your score: 0 / 10
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