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Bernoulli Equation

Bernoulli ODE solver. Enter P(x), Q(x), n and an initial value to substitute and reduce it to a linear equation solved numerically, with steps—for ODE coursework.

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y′ + P(x)y = Q(x)yⁿ, z = y^(1−n)
coefficient of y
RHS coefficient
n ≠ 0,1
initial x
initial y(x₀)
target point

📖 Tutorial｜Bernoulli Equation

1. Definition
An equation of the form y′ + P(x)y = Q(x)yⁿ (n≠0,1) is a Bernoulli equation. Set z = y^(1−n) to linearize it.
2. Symbols
SymbolMeaning
npower (≠0,1)
znew variable y^(1−n)
1−nsubstitution exponent
3. How this tool works
  • Divide by yⁿ: y^(−n) y′ + P y^(1−n) = Q;
  • Set z = y^(1−n), so z′ = (1−n) y^(−n) y′;
  • The equation becomes z′ + (1−n)P z = (1−n)Q, linear;
  • Solve for z and substitute back y = z^(1/(1−n)).
4. Steps
  1. Enter P(x), Q(x) and power n;
  2. Enter initial value and target x₁;
  3. Click Calculate to see the substitution and numerical answer.
5. Example
Example: y′ = y², y(0)=1. n=2, z=y^(−1), z′=−1, z=−x+C; z(0)=1 gives C=1, y=1/(1−x). The exact solution has a pole at x=1, so pick a smaller x₁.
6. Pitfalls
n=0 or 1 already gives a linear equation; no substitution needed;
y=0 is a particular solution lost by substitution;
The back-substitution exponent is 1/(1−n).

❓ FAQ｜Bernoulli Equation

How does a Bernoulli equation relate to a linear one?
Setting z=y^(1−n) turns it into a first-order linear equation solved by the integrating factor.
Which values of n are allowed?
Any real n except 0 and 1; n=0 is linear, n=1 is separable.
Why is y=0 lost?
The substitution divides by y, so y=0 is not covered and must be checked separately.
Is the answer exact?
This tool uses Euler numerically; solve the linear equation and substitute back for an exact answer.
What if the numerical result diverges?
The solution has a pole there; use a smaller x₁.