Difficulty: Introductory to Intermediate | Prerequisites: Calculus I and II (integration techniques, partial derivatives)
First-order ODEs are the foundation of the entire MAT 2388 course. Every technique here, from separation of variables to integrating factors, reappears in modified form when you move to higher-order equations, Laplace transforms, and systems. If you are behind, start here: the rest of the course builds directly on your ability to classify an equation and pick the right solution method.
First-order ODEs are equations involving y, y' and x. The course teaches you a decision tree: check whether the equation is separable, then whether it has homogeneous coefficients, then whether it is exact (or can be made exact with an integrating factor), and finally whether a Bernoulli substitution applies. Each method reduces the problem to integration, and the result is a general solution containing one arbitrary constant.
Differential equation
An equation involving an unknown function y, its derivatives, and functions of the independent variable. In simple terms, you are given a relationship between a function and its rate of change, and your job is to find the function itself.
Order
The highest-order derivative that appears in the equation. A first-order ODE contains y' but nothing higher.
Explicit solution
A solution written as y = f(x), with the dependent variable isolated on one side. Think of it as the "clean" form where y is fully solved for.
Implicit solution
A solution where y and x are mixed together in a single equation (e.g. x² + y² = C). You cannot easily isolate y, but the relationship still defines the solution.
General solution
The family of all solutions, written with an arbitrary constant C. Each value of C picks out one specific curve.
Separable equation
An ODE that can be rearranged so that all y-terms (with dy) sit on one side and all x-terms (with dx) sit on the other. Think of it as: can I split the equation into a y-only piece and an x-only piece?
Homogeneous coefficients
An equation in the form M(x,y)dx + N(x,y)dy = 0 where M and N have the same degree of homogeneity. In simple terms, when you replace every x with λx and every y with λy, the same power of λ factors out of both M and N.
Exact equation
An equation M dx + N dy = 0 where ∂M/∂y = ∂N/∂x. This means there exists some function u(x,y) whose total differential equals the left side, so the solution is just u(x,y) = C.
Integrating factor
A function μ that you multiply through the equation to make a non-exact equation become exact. Think of it as a correction multiplier that restores the symmetry the exactness test requires.
Bernoulli equation
A non-linear ODE of the form y' + p(x)y = g(x)yᵃ. The substitution u = y¹⁻ᵃ converts it into a linear equation, which you then solve with the earlier methods.
A separable equation is the simplest type of first-order ODE. The test: can you rearrange so that one side has only y (and dy) and the other has only x (and dx)?
Replace notation. Rewrite any y' as dy/dx. This is what lets you treat dy and dx as separate pieces you can move around.
Separate the variables. Algebra until every y-term sits with dy on one side and every x-term sits with dx on the other: g(y) dy = f(x) dx.
Integrate both sides. Each side produces its own constant of integration (c₁ and c₂). Combine them into a single constant c = c₂ – c₁ on one side.
Solve for y if possible. If you can isolate y, you have an explicit solution. If not, the implicit form is still a valid answer.
This is the first check in the classification flowchart. If the equation is separable, you never need the more advanced methods.
If the equation is not directly separable but can be written in the form M(x,y)dx + N(x,y)dy = 0, the next test is whether M and N have homogeneous coefficients of the same degree.
Substitute λ. Replace every x with λx and every y with λy in both M and N. Simplify until λ is factored out in front of each function.
Isolate the exponent. Compare the power of λ in front of M versus in front of N. Call them S and T.
Same degree test. If S = T, the equation has homogeneous coefficients and you can proceed.
Substitution. Set y = x·u(x) (or equivalently x = y·u(y)). Since y = xu, you also get dy = u dx + x du.
Substitute and separate. Plug in and simplify. The homogeneous property guarantees that the equation becomes separable in x and u. Solve by separation and then back-substitute u = y/x.
If S ≠ T, skip this method and move to the exactness test instead.
An equation M(x,y)dx + N(x,y)dy = 0 is exact when there exists a potential function u(x,y) such that ∂u/∂x = M and ∂u/∂y = N.
Organise into standard form. Write the equation as M dx + N dy = 0.
Compute partial derivatives. Find ∂M/∂y and ∂N/∂x.
Exactness test. If ∂M/∂y = ∂N/∂x, the equation is exact.
Set up u(x,y). Define a new function u(x,y) with u_x = M and u_y = N.
Integrate one partial. Integrate u_x = M with respect to x to get u(x,y) = ∫M dx + g(y), where g(y) is an unknown function of y alone (the "constant" of partial integration).
Determine g(y). Differentiate your result with respect to y and set it equal to N. Solve for g'(y), then integrate to find g(y).
Write the solution. The implicit solution is u(x,y) = C.
The order of integration does not matter: you could equally start from u_y = N and integrate with respect to y, then determine a function h(x).
When ∂M/∂y ≠ ∂N/∂x, the equation is not exact, but it can sometimes be made exact by multiplying through by an integrating factor μ.
Case 1: function of x only. Compute (M_y – N_x) / N. If this expression depends only on x, call it f(x). The integrating factor is μ = exp(∫f(x) dx). Multiply both M and N by μ, then proceed as an exact equation.
Case 2: function of y only. If Case 1 fails, compute (M_y – N_x) / M. If this depends only on y, call it g(y). The integrating factor is μ = exp(–∫g(y) dy). Multiply both M and N by μ, then proceed as exact.
Apply the factor. Replace M with μM and N with μN. The modified equation is now exact, so follow the exact ODE procedure from the previous section.
If neither case yields a function of a single variable, this approach does not apply directly and you may need to try a Bernoulli substitution or another technique.
A Bernoulli equation has the form y' + p(x)y = g(x)yᵃ, where a ≠ 0 and a ≠ 1 (otherwise it is already linear).
Identify the exponent a. Rearrange the equation until it matches y' + p(x)y = g(x)yᵃ.
Change of variable. Set u(x) = y¹⁻ᵃ. The derivative is u'(x) = (1 – a) y⁻ᵃ y'.
Isolate y'. From the u' definition: y' = yᵃ u' / (1 – a). Also isolate y' from the original Bernoulli equation.
Substitute. Set the two expressions for y' equal. The yᵃ terms cancel, leaving a linear ODE in u.
Solve the linear ODE. The resulting equation in u can be solved with separation of variables, an integrating factor, or whichever first-order method applies. Go back to the top of the decision flowchart if needed.
Back-substitute. Once you have u(x), recover y from y = u^(1/(1–a)).
When faced with a first-order ODE, work through these checks in order:
Separable? Can you write g(y) dy = f(x) dx? If yes, separate and integrate.
M dx + N dy = 0 form? If the equation fits this shape, continue to the next checks.
Homogeneous coefficients? Substitute λx and λy into M and N. If both scale with the same power of λ (S = T), set y = x·u and separate.
Exact? Compute ∂M/∂y and ∂N/∂x. If they are equal, find the potential function u(x,y).
Integrating factor? If not exact, check whether (M_y – N_x)/N is a function of x only (Case 1) or (M_y – N_x)/M is a function of y only (Case 2). Multiply by the appropriate μ and solve as exact.
Bernoulli? If the equation matches y' + p(x)y = g(x)yᵃ, use the substitution u = y¹⁻ᵃ to linearise, then restart at step 1 with the new equation.
Always start from the top. After a substitution (homogeneous or Bernoulli), the transformed equation re-enters the flowchart from step 1.
\text{Separable: } g(y)\,dy = f(x)\,dx \quad\Rightarrow\quad \int g(y)\,dy = \int f(x)\,dx + C\text{Homogeneous test: } M(\lambda x, \lambda y) = \lambda^S M(x,y), \quad N(\lambda x, \lambda y) = \lambda^T N(x,y), \quad S = T\text{Exactness test: } \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\text{Integrating factor (Case 1): } \mu = \exp\!\left(\int \frac{M_y - N_x}{N}\,dx\right)\text{Integrating factor (Case 2): } \mu = \exp\!\left(-\int \frac{M_y - N_x}{M}\,dy\right)\text{Bernoulli substitution: } u = y^{1-a}, \quad u' = (1-a)\,y^{-a}\,y'Students often forget to combine the two constants of integration into a single constant c = c₂ – c₁ after integrating a separable equation. Leaving two constants is not wrong, but it costs marks on exams expecting a clean general solution.
When testing for homogeneous coefficients, students sometimes substitute λ into only one of M and N, or forget to apply order of operations. Both functions must be tested, and the exponent on λ must be fully isolated.
A common error with exact equations is forgetting the unknown function g(y) (or h(x)) when integrating a partial derivative. Because you are integrating with respect to one variable, the "constant" is a function of the other variable, not a plain number.
With integrating factors, students mix up the two cases. Remember: divide by N first (Case 1, function of x), and only if that fails, divide by M (Case 2, function of y). Also note the negative sign in the Case 2 exponent.
⚠️ Classification is the most testable skill. Exam questions rarely tell you which method to use. You are expected to identify the type, then solve. Practise the flowchart until it is automatic.
⚠️ Exact equations appear on nearly every final. The partial derivative check (∂M/∂y vs ∂N/∂x) and the potential-function procedure are high-frequency targets.
⚠️ Integrating factors are a common "second step" question: the exam gives you a non-exact equation and expects you to find μ, then solve. Know both cases cold.
⚠️ Bernoulli equations test your ability to handle non-linear ODEs. The substitution u = y¹⁻ᵃ must be executed cleanly, including deriving u' and eliminating y entirely.
True or false: every separable equation is also exact. (True, because if g(y)dy – f(x)dx = 0 then M = –f(x), N = g(y), and M_y = N_x = 0.)
Fill in the blank: to test for homogeneous coefficients, replace x with ___ and y with ___ in both M and N. (λx, λy)
True or false: the integrating factor μ in Case 1 depends on y. (False, it depends only on x.)
Fill in the blank: for a Bernoulli equation y' + p(x)y = g(x)yᵃ, the substitution is u = ___. (y¹⁻ᵃ)
True or false: if ∂M/∂y ≠ ∂N/∂x, the equation has no solution. (False, it just means the equation is not exact; an integrating factor may still work.)
Q: Given the equation (2xy + 3) dx + (x² + 4y) dy = 0, determine whether it is exact and, if so, find the general solution.
A: M = 2xy + 3, N = x² + 4y. M_y = 2x, N_x = 2x. Since M_y = N_x, the equation is exact. Integrate u_x = M with respect to x: u = x²y + 3x + g(y). Differentiate with respect to y: u_y = x² + g'(y). Set equal to N: x² + g'(y) = x² + 4y, so g'(y) = 4y, giving g(y) = 2y². Solution: x²y + 3x + 2y² = C.
Q: Solve dy/dx = y/x (identify the method first).
A: This is separable. Rewrite as dy/y = dx/x. Integrate both sides: ln|y| = ln|x| + C. Exponentiate: y = Ax where A = ±eᶜ.
Q: Classify the equation y' + 2y = 5y³ and state the substitution you would use.
A: This is a Bernoulli equation with p(x) = 2, g(x) = 5, and a = 3. The substitution is u = y¹⁻³ = y⁻².
Q: If (M_y – N_x)/N = 3 (a constant), what is the integrating factor?
A: Since the expression is a function of x only (a constant counts), μ = exp(∫3 dx) = e³ˣ.
The classification skill you build here carries directly into higher-order ODEs (Doc 2): constant-coefficient and Euler-Cauchy equations reduce to characteristic equations, but the mindset of "identify the type, then apply the method" is the same. Integrating factors reappear in the variation of parameters method, and the concept of exactness connects to potential functions in multivariable calculus (gradient fields). Numerical methods (Doc 4) approximate the solutions you learn to find analytically here.
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first order ODE, separable differential equation, separation of variables, homogeneous coefficients, exact differential equation, potential function, integrating factor, Bernoulli equation, change of variable, implicit solution, explicit solution, general solution, particular solution, MAT 2388, ordinary differential equations, dy/dx, partial derivatives exactness test