Difficulty: Intermediate | Prerequisites: Standard form for linear ODEs, integrating factors, basic integration techniques.
Tags: Bernoulli equation, separable ODE, substitution v = y^(1-n), implicit solution, interval of validity, initial value problem, IVP, MATH 441, ordinary differential equations
Bernoulli equations look nonlinear because of a y^n term, but a substitution (v = y^(1-n)) converts them into linear ODEs you already know how to solve. Separable equations split cleanly into a function of y on one side and a function of x on the other, letting you integrate both sides directly. Both types appear heavily on MATH 441 exams.
Bernoulli equation
An ODE of the form y' + p(t)y = q(t) y^n, where n ≠ 0 and n ≠ 1. The y^n term is what makes it nonlinear. In simple terms, it is a linear ODE with a power of y bolted onto the right-hand side.
Bernoulli substitution
The change of variable v = y^(1-n), which transforms the Bernoulli equation into a first-order linear ODE in v. Think of it as a trick that absorbs the nonlinearity into a new variable.
Separable equation
An ODE that can be written as f(y) dy = g(x) dx, so that each side depends on only one variable. In simple terms, you can physically move all the y's to one side and all the x's to the other, then integrate.
Implicit solution
A solution expressed as a relationship F(x, y) = C rather than y = f(x). Many separable equations produce implicit solutions because the integral on the y-side cannot be inverted algebraically.
Interval of validity
The range of x (or t) values on which the solution is defined and the original ODE's assumptions hold. You find it by checking where denominators are nonzero and where the implicit curve passes through the initial point.
Write the ODE in standard Bernoulli form: y' + p(t)y = q(t) y^n.
Identify n. If n = 0 or n = 1, the equation is already linear; no substitution needed.
Example: t²y' + 2ty - y³ = 0 becomes y' + (2/t)y = y³/t², so p(t) = 2/t, q(t) = 1/t², and n = 3.
Define v = y^(1-n). For n = 3, this gives v = y^(-2) = 1/y².
Differentiate: v' = (1-n) y^(-n) y'. For n = 3, v' = -2 y^(-3) y'.
Substitute into the original ODE. Multiply through by (1-n) y^(-n) to get a linear ODE in v.
For the example: v' - (4/t) v = -2/t².
Use the integrating factor method on the linear ODE in v.
Integrating factor: μ = e^(∫ -4/t dt) = e^(-4 ln t) = 1/t⁴.
Multiply through, integrate, and solve for v.
Result: (1/t⁴) v = 2/(5t⁵) + c, so v = 2/(5t) + ct⁴.
Substitute back using v = 1/y²:
1/y² = 2/(5t) + ct⁴
y² = 1 / (2/(5t) + ct⁴)
The ODE y' = f(x, y) is separable if f(x, y) can be factored as h(x) · k(y), or equivalently if you can rearrange to get all y-terms (including dy) on one side and all x-terms (including dx) on the other.
Example: y' = (1 + 3x²) / (3y² - 6y). Rewrite as (3y² - 6y) dy = (1 + 3x²) dx.
Integrate both sides independently:
∫ (3y² - 6y) dy = y³ - 3y²
∫ (1 + 3x²) dx = x + x³
Combine with constant: y³ - 3y² = x + x³ + c.
Plug in the initial values to find c.
y(0) = 1: 1 - 3 = 0 + 0 + c, so c = -2.
Implicit solution: y³ - 3y² = x + x³ - 2.
The original ODE has y' = (1 + 3x²) / (3y² - 6y). The denominator 3y² - 6y = 3y(y - 2) must not be zero, so y ≠ 0 and y ≠ 2.
Substitute y = 0 into the implicit solution: 0 = x + x³ - 2, i.e. x + x³ - 2 = 0.
Substitute y = 2: 8 - 12 = x + x³ - 2, i.e. x + x³ + 2 = 0.
The interval of validity is the largest interval containing x = 0 that avoids all real roots of both cubic equations.
On an exam, stating this condition clearly (without necessarily computing the roots numerically) is typically sufficient.
Bernoulli form:
y' + p(t) y = q(t) y^n
Bernoulli substitution:
v = y^(1-n), v' = (1-n) y^(-n) y'
Resulting linear ODE (general):
v' + (1-n) p(t) v = (1-n) q(t)
Separable equation pattern:
f(y) dy = g(x) dx → ∫ f(y) dy = ∫ g(x) dx + c
Bernoulli equations appear in population dynamics models where growth rate depends on a power of the population size (for instance, logistic-type models with harvesting). Separable equations are everywhere in physics and engineering: Newton's law of cooling, radioactive decay, and simple circuit models all reduce to separable form. If you have ever seen "separate and integrate" in a physics derivation, that is this technique.
Students frequently forget the factor of (1-n) when deriving the linear ODE in v. For n = 3, the chain rule gives v' = -2 y^(-3) y', and that -2 must propagate through the entire equation.
A common error with separable equations is dividing both sides by an expression involving y without noting where that expression is zero. Any y-value that makes the denominator zero is excluded from the solution, and those excluded values determine the interval of validity.
Students sometimes assume the implicit solution y³ - 3y² = x + x³ - 2 must be solved explicitly for y. It does not. Implicit solutions are perfectly valid; just confirm the initial condition satisfies the relation.
When n is negative in a Bernoulli equation, the substitution still works. The formula v = y^(1-n) applies regardless of the sign of n.
⚠️ Bernoulli problems will give you the ODE in a non-standard arrangement. Your first job is always to get it into y' + p(t)y = q(t) y^n form.
⚠️ For separable equations with an IVP, the exam will usually ask for the interval of validity. Do not skip this step.
⚠️ Watch the algebra when substituting back from v to y. Errors in taking reciprocals or square roots are the most common way to lose marks.
⚠️ If the problem says "solve" without specifying explicit or implicit, an implicit solution is acceptable.
Fill in the blank: In a Bernoulli equation y' + p(t)y = q(t) y^n, the substitution is v = ________.
True or false: A separable equation always produces an explicit solution y = f(x).
True or false: For the Bernoulli equation with n = 3, the substitution gives v = y^(-2).
Fill in the blank: The interval of validity for a separable IVP is found by checking where the ________ of the original ODE is nonzero.
True or false: If n = 1 in y' + p(t)y = q(t) y^n, you should use the Bernoulli substitution.
Answers: 1. y^(1-n). 2. False (often implicit). 3. True. 4. Denominator. 5. False (n = 1 is already linear).
Q: Write the Bernoulli equation t²y' + 2ty - y³ = 0 (t > 0) in standard form and identify n.
A: Divide by t²: y' + (2/t)y = y³/t². Here n = 3.
Q: What substitution converts this into a linear ODE, and what is the resulting equation?
A: v = y^(1-3) = y^(-2). After substitution, v' - (4/t)v = -2/t².
Q: For the separable IVP y' = (1 + 3x²)/(3y² - 6y), y(0) = 1, find the implicit solution.
A: Separate and integrate: y³ - 3y² = x + x³ + c. Apply y(0) = 1: c = -2. Solution: y³ - 3y² = x + x³ - 2.
Q: What values of y must be excluded from the solution, and why?
A: y = 0 and y = 2, because 3y² - 6y = 3y(y - 2) = 0 at those values, making the denominator of y' zero.
Q: How do you determine the interval of validity for the implicit solution?
A: Substitute y = 0 and y = 2 into y³ - 3y² = x + x³ - 2 to find the corresponding x-values. The valid interval is the largest interval around x = 0 that avoids those x-values.
Bernoulli equations connect directly to the integrating factor method, since the substitution reduces them to a linear ODE solved by that technique. Separable equations are the simplest class of nonlinear ODEs and reappear in the mixing/tank problems covered in the applied ODE notes. The concept of implicit solutions comes back in exact equations (covered in the next set of notes), where solutions are almost always left in implicit form.
Bernoulli equation, Bernoulli substitution, v = y^(1-n), separable ODE, separation of variables, implicit solution, interval of validity, integrating factor, first-order nonlinear ODE, direct integration, IVP, MATH 441 exam 1