First-Order ODEs Practice Quiz with a Step-by-Step Interactive Lesson
Use the quiz at the top of the page 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 quiz: answer the first-order ODE questions at the top of the page.
2. Open the lesson (optional): review slope meaning, separable equations, integrating factors, initial-value problems, and common model examples.
3. Retry: return to the quiz 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.
Back to the quiz
When you are ready, return to the quiz at the top of the page and keep practicing first-order ODEs.
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First-Order ODEs Lesson
Step-by-step guide
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First-Order ODEs Lesson
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Lesson Overview
Lesson overview
Purpose: Solve first-order ODEs by identifying separable and linear forms, integrating correctly, using initial conditions, and checking by substitution/differentiation.
Success criteria
Identify the equation type (separable, linear, or initial-value problem).
Apply separation or the integrating factor workflow correctly.
Use initial conditions to find \(C\).
Check your solution by differentiation and substitution.
Choose the right method quickly on quiz questions.
Identify the equation type (separable, linear, or initial-value problem).
Apply separation or the integrating factor workflow correctly.
Use initial conditions to find \(C\).
Check your solution by differentiation and substitution.
Choose the right method quickly on quiz questions.
Key vocabulary
General solution: the family \(y(x)=\text{solution}(C)\).
Particular solution: the specific solution after using an initial condition.
Separable: equations that can be written as \(\frac{dy}{dx}=f(x)g(y)\).
First-order linear: \(y'+P(x)y=Q(x)\).
Quick pre-check
Pre-check 1: If \(y'=\dfrac{1}{x}\), is the equation separable?
A separable equation can be rewritten as \(y' = f(x)g(y)\).
Initial value problems: turning a family into one curve
Learning goal: Use an initial condition to find the constant \(C\) and write the particular solution.
Key idea
A general solution contains an arbitrary constant \(C\). An initial condition like \(y(x_0)=y_0\) chooses the unique member of the family (when it exists).
Worked example
Example: Solve \(\displaystyle \frac{dy}{dx}=2\) with \(y(0)=5\).
Integrate: \[
\ty = 2x + C.
\]
Use the initial condition \(y(0)=5\):
\[
\t5 = 2(0) + C \Rightarrow C=5.
\]
So the particular solution is
\[
\ty = 2x + 5.
\]
Try it
Try it 1: Solve the ODE \(\displaystyle \frac{dy}{dx}=5x^4\). What is the general solution?
Hint: \(\int 5x^4dx = x^5 + C\).
Try it 2: Solve \(\displaystyle \frac{dy}{dx}=-3\). What is the general solution?
Hint: If the derivative is constant \(-3\), the solution is a line with slope \(-3\).
Summary
General solution \(\Rightarrow\) contains \(C\). Initial condition \(\Rightarrow\) find \(C\).
Always check by differentiating after you apply the initial condition.
Applications & Big Picture
Why first-order ODEs matter (and a final check)
Learning goal: Connect first-order ODE techniques to modeling and build a reliable solve + check routine.
Where first-order ODEs show up
Growth/decay: \(y'+ky\) models populations and radioactive decay.
Motion with constant acceleration: velocity and position come from integrating derivatives.
Electrical circuits: first-order linear ODEs in simple RC circuits (current/voltage relationships).
Cooling: Newton’s law of cooling uses \(T' = -k(T-T_{\text{env}})\).
Worked example
Example: Show that \(y=\sin x + C\) satisfies \(\displaystyle \frac{dy}{dx}=\cos x\).
Differentiate:
\[
\t\frac{dy}{dx}=\frac{d}{dx}(\sin x + C)=\cos x.
\]
This matches the right-hand side exactly, so \(y=\sin x + C\) is a solution.
Try it
Try it 1: What is the general solution of \(\displaystyle \frac{dy}{dx}=\cos x\)?
Hint: \(\int \cos x\,dx=\sin x + C\).
Try it 2: Which function satisfies \(\displaystyle \frac{dy}{dx}=\cos x\)?
Hint: Differentiate each option: \((\sin x)'=\cos x\).
Final recap
General vs particular: solve first, then use the initial condition to find \(C\).
Separable ODEs: rearrange into \(g(y)\,dy=f(x)\,dx\), integrate, and solve for \(y\) if possible.
Linear ODEs: \(y'+P(x)y=Q(x)\), integrating factor \(\mu=e^{\int P\,dx}\), then \((\mu y)'=\mu Q\).
Always check: differentiate your solution and substitute back into the original ODE.
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 ODE type (direct integration, separable, or linear integrating factor).
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