3. Bernoulli's ODE

A first-order differential equation that can be transformed into a linear equation by a suitable change of variables.

$$\dfrac{dy}{dx} + f(x)y = g(x)y^n \quad \text{where} \quad n \neq 0, 1 \quad y(0) = 1$$

How to solve

Convert to linear ODE

$$\begin{equation} y^{-n}\dfrac{dy}{dx} + f(x)y^{1-n} = g(x) \end{equation}$$

we need the $\; y^{1-n} \;$ term to be linear so set

$$z = y^{1-n}, \quad y = z^{\textstyle \frac{1}{1 - n}}, \quad \dfrac{dz}{dy} = (1 - n)y^{-n}$$

we will also need

$$\dfrac{dz}{dx} = \dfrac{d(y^{1 - n})}{dx} = (1 - n)y^{-n}\dfrac{dy}{dx}$$

note the expression $\; y^{-n}\dfrac{dy}{dx} \;$ in the above equation is the same as the first term in $(1)$

lets solve for this term

$$y^{-n}\dfrac{dy}{dx} = \dfrac{1}{1 - n}\dfrac{dz}{dx}$$

now lets get rid of all the $y$ terms in $(1)$ by subbing in $\; \dfrac{1}{1 - n}\dfrac{dz}{dx} \;$ and $\; z = y^{1 - n} \;$

this gives us

$$\dfrac{1}{1 - n}\dfrac{dz}{dx} + f(x)z = g(x)$$

$$\boxed{\dfrac{dz}{dx} + (1 - n)f(x)z = (1 - n)g(x)}$$

this is in the same form as a first order ODE.

Note: it can be useful to memorise this equation

Example

Solve

$$\dfrac{dy}{dx} + y = xy^3 \quad \text{where} \quad y(0) = 1$$

Answer

$n = 3, \quad z = y^{-2}, \quad y = z^{- \textstyle \frac{1}{2}}, \quad f(x) = 1, \quad g(x) = x$

Sub into Bernoulli's ODE

$$\dfrac{dz}{dx} -2z = -2x$$

Solve for integrating factor (IF)

$$\mu(x) = e^{\textstyle \int{-2 \, \; dx}} = e^{-2x}$$

multiply ODE by $\mu(x)$

$$e^{-2x}\dfrac{dz}{dx} -2ze^{-2x} = xe^{-2x}$$

integrate both sides

$$z = x + \frac{1}{2} + Ce^{2x}$$

so

$$y = (x + \frac{1}{2} + Ce^{2x})^{- \textstyle \frac{1}{2}}$$

since $\; y(0) = 1$, $\; C = \dfrac{1}{2}$

therefore

$$y = (x + \frac{1}{2} + \frac{1}{2}e^{2x})^{- \textstyle \frac{1}{2}}$$