First-Order ODEs Practice Questions with Answers and a Guided 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.
Answer the question set and review your mistakes at the end.
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.
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 \(y' = f(x)g(y)\).
Pre-check 2: In \(y'+3y=0\), what is \(P(x)\)?
Compare to \(y'+P(x)y=Q(x)\) with \(Q(x)=0\).
ODE Basics
Reading and checking first-order ODEs
Learning goal: Interpret derivative notation, integrate a simple derivative, and verify a proposed solution.
Key idea
A first-order ODE contains a first derivative such as \(y'\) or \(\frac{dy}{dx}\). It describes how the unknown function changes rather than giving the function directly.
How to check a solution
Differentiate the proposed function and substitute both \(y\) and \(y'\) into the original equation. The two sides must agree on the stated interval.
Integrate both sides: \[y=\int 2x\,dx=x^2+C.\] Checking gives \(y'=2x\), so the family is correct.
Try it
Try it 1: Which function is the general solution of \(y'=2y\)?
Hint: Differentiate \(Ce^{2x}\) and compare the result with \(2y\).
Try it 2: Solve \(\displaystyle \frac{dy}{dx}=x^2\) with \(y(1)=3\).
Hint: Integrate to get \(y=x^3/3+C\), then use \(y(1)=3\).
Summary
A first-order ODE involves the first derivative of an unknown function.
Direct integration reverses differentiation.
Always check a solution in the original equation.
Separable Equations
Separate the variables, then integrate
Learning goal: Recognize \(y'=f(x)g(y)\), move the \(y\)-terms and \(x\)-terms to opposite sides, and integrate.
Key idea
For \(\frac{dy}{dx}=f(x)g(y)\), rewrite the equation as \(\frac{1}{g(y)}\,dy=f(x)\,dx\) when valid, then integrate both sides and solve for \(y\) if possible.
Separate and integrate: \[\frac{1}{y}\,dy=x\,dx,\qquad \ln|y|=\frac{x^2}{2}+C.\] Therefore \(y=Ce^{x^2/2}\), including the zero solution through \(C=0\).
Try it
Try it 1: Solve \(\displaystyle \frac{dy}{dx}=\frac{2}{x}\).
Hint: \(\int \frac{2}{x}\,dx=2\ln|x|+C\).
Try it 2: If \(\frac{dy}{dx}=2xy\), \(y(0)=4\), what is the constant in \(y=Ce^{x^2}\)?
Hint: Substitute \(x=0\) and \(y=4\) into \(y=Ce^{x^2}\).
Summary
Recognize a product of an \(x\)-function and a \(y\)-function.
Separate before integrating.
Use initial data after finding the general solution, and check for solutions lost during division.
Exponential Models
The model \(y'=ky\)
Learning goal: Recognize and solve exponential growth and decay equations.
Key idea
When a quantity changes at a rate proportional to its current value, \(y'=ky\), its solution is \(y=Ce^{kx}\). Positive \(k\) gives growth and negative \(k\) gives decay.
Worked example
Example: Solve \(y'=-2y\) with \(y(0)=10\).
The general form is \(y=Ce^{-2x}\). Since \(y(0)=C=10\), the particular solution is \(y=10e^{-2x}\).
Try it
Try it 1: Which equation represents exponential decay at rate \(3\)?
Hint: Decay means the proportionality constant is negative.
Try it 2: If \(y'=2y\) and \(y(0)=5\), what is \(C\) in \(y=Ce^{2x}\)?
Hint: At \(x=0\), \(e^{2x}=1\), so \(y(0)=C\).
Summary
\(y'=ky\) has the solution \(y=Ce^{kx}\).
The sign of \(k\) determines growth or decay.
An initial value determines \(C\).
Linear ODEs
Use an integrating factor
Learning goal: Put a linear equation in standard form and apply the integrating-factor method.
Key idea
For \(y'+P(x)y=Q(x)\), choose \(\mu(x)=e^{\int P(x)\,dx}\). Multiplication by \(\mu\) turns the left side into \((\mu y)'\), which can be integrated directly.
Worked example
Example: Solve \(y'+y=e^x\).
Here \(\mu=e^x\). Then \((e^xy)'=e^{2x}\), so \(e^xy=\frac12e^{2x}+C\) and \(y=\frac12e^x+Ce^{-x}\).
Try it
Try it 1: What is an integrating factor for \(y'+y=x\)?
Hint: \(P(x)=1\), so \(\mu=e^{\int 1\,dx}\).
Try it 2: After multiplying \(y'+2y=x\) by \(e^{2x}\), which equation do you integrate?
Hint: The left side becomes \((e^{2x}y)'\); expand that product derivative to identify the matching form.
Summary
First write \(y'+P(x)y=Q(x)\).
Use \(\mu=e^{\int P(x)\,dx}\).
Integrate \((\mu y)'=\mu Q\), solve for \(y\), and check.
Method Check
Separable, linear, or direct integration?
Learning goal: Classify a first-order ODE before doing algebra.
Key idea
A quick structure check usually reveals the shortest valid method:
Direct integration: \(y'=f(x)\).
Separable: \(y'=f(x)g(y)\).
Linear: \(y'+P(x)y=Q(x)\).
Some equations fit more than one category. Choose the form that makes the solution clearest, and verify the result afterward.
Worked example
Example: Classify \(y'+3y=x\).
It is linear with \(P(x)=3\) and \(Q(x)=x\). It is not immediately separable because the right side cannot be written as a product \(f(x)g(y)\).
Try it
Try it 1: Which equation is not separable as written?
Hint: Look for a product of a function of \(x\) and a function of \(y\).
Try it 2: What is the general solution of \(y'=1\)?
Hint: Integrating \(1\) with respect to \(x\) gives \(x+C\).
Summary
Inspect the equation before choosing a method.
Direct, separable, and linear forms cover many first-order problems.
Classification is useful only if the final solution also passes substitution.
Initial Value Problems
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).
