Bernoulli Substitution

If a first order equation isn’t separable or linear, then one of three types of substitutions might solve the equation.

The Essentials

The third kind of substitution is a Bernoulli substitution. A Bernoulli equation can fit into the form:

\[ \frac{dy}{dx} + p(x)y = f(x)y^n \]

If \( n = 0, 1 \) then the equation is linear and can be solved with an integrating factor, otherwise we use the Bernoulli substitution:

\[ v = y^{1-n} \]

We solve the substitution for y and differentiate to find \( \frac{dy}{dx} \). After we plug these in, we will have a first order linear differential equation.

Example

Solve the equation:

\[ x \frac{dy}{dx} = -4x^3 y^2 + y \]

First, we put the equation in standard form:

\[ \frac{dy}{dx} - \frac{1}{x} y = -4x^2 y^2 \]

Then, we determine our substitution:

\[ v = y^{1-n} \] \[ v = y^{-1} \] \[ y = v^{-1} \] \[ \frac{dy}{dx} = -v^{-2} \frac{dv}{dx} \]

Next, we perform the substitution and solve the resulting linear differential equation:

\[ -v^{-2} \frac{dv}{dx} - \frac{1}{x} \frac{1}{v} = -4x^2 v^{-2} \] \[ \frac{dv}{dx} + \frac{1}{x}v = 4x^2 \] \[ \mu = e^{\int x^{-1}dx} = x \] \[ \frac{d}{dx}(xv) = 4x^3 \] \[ xv = x^4 + C \] \[ v = \frac{x^4 + C}{x} \]

Finally, we undo the substitution and solve for y(x):

\[ \frac{1}{y} = \frac{x^4 + C}{x} \] \[ y(x) = \frac{x}{x^4 + C} \]

Practice

Evaluate these expressions:

\[ \frac{dy}{dx} + 6y = 6y^{2/3} \]

Solution:

\[ y(x) = (1 + Ce^{-2x})^3 \]