# Tonnie's question at Yahoo! Answers regarding a Bernoulli equation

Staff member

#### MarkFL

Staff member
Hello Tonnie,

A first order ODE that can be written in the form:

$$\displaystyle \frac{dy}{dx}+P(x)y=Q(x)y^n$$

where $P(x)$ and $Q(x)$ are continuous on an interval $(a,b)$ and $n$ is a real number, is called a Bernoulli equation.

This equation was proposed for solution by James Bernoulli in 1695. It was solved by his brother John Bernoulli. James and John were two of eight mathematicians in the Bernoulli family. In 1696 Gottfried Leibniz showed that the Bernoulli equation can be reduced to a linear equation by making the substitution $v=y^{1-n}$.

We are given to solve:

(1) $$\displaystyle \frac{dy}{dx}+\frac{1}{x}y=y^2$$

Dividing through by $y^2$ (observing we are losing the trivial solution $y\equiv0$), we obtain:

(2) $$\displaystyle y^{-2}\frac{dy}{dx}+\frac{1}{x}y^{-1}=1$$

Using the substitution of Leibniz, i.e., $v=y^{-1}$, we find via the chain rule that:

$$\displaystyle \frac{dv}{dx}=-y^{-2}\frac{dy}{dx}$$

and (2) becomes:

(3) $$\displaystyle \frac{dv}{dx}-\frac{1}{x}v=-1$$

Now we have a linear equation in $v$. Computing the integrating factor, we find:

$$\displaystyle \mu(x)=e^{-\int\frac{dx}{x}}=\frac{1}{x}$$

Multiplying (3) by this integrating factor, we obtain:

$$\displaystyle \frac{1}{x}\frac{dv}{dx}-\frac{1}{x^2}v=-\frac{1}{x}$$

Rewriting the left hand side as the differentiation of a product, we have:

$$\displaystyle \frac{d}{dx}\left(\frac{v}{x} \right)=-\frac{1}{x}$$

Integrating with respect to $x$, there results:

$$\displaystyle \int\,d\left(\frac{v}{x} \right)=-\int\frac{1}{x}\,dx$$

$$\displaystyle \frac{v}{x}=-\ln|x|+C$$

$$\displaystyle v=x\left(C-\ln|x| \right)$$

Back-substituting for $v$, we have:

$$\displaystyle \frac{1}{y}=x\left(C-\ln|x| \right)$$

Hence:

$$\displaystyle y=\frac{1}{x\left(C-\ln|x| \right)}$$

To Tonnie and any other guests viewing this topic, I invite and encourage you to post other differential equations problems in our Differential Equations forum.

Best Regards,

Mark.