Tonnie's question at Yahoo Answers regarding a Bernoulli equation

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SUMMARY

The discussion focuses on solving the Bernoulli differential equation given by y' + y/x = y^2. This first-order ordinary differential equation (ODE) can be transformed into a linear equation using the substitution v = y^{-1}. The solution is derived through the integration of the transformed equation, resulting in y = 1/(x(C - ln|x|)), where C is a constant. The historical context includes contributions from James and John Bernoulli, as well as Gottfried Leibniz, who provided the substitution method.

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MarkFL
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Here is the question:

Determine the solution to the following differential equation.?

Determine the solution to the following differential equation.

y' + y/x = y^2

Here is a link to the question:

Determine the solution to the following differential equation.? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello Tonnie,

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

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

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

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

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

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

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

and (2) becomes:

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

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

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

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

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

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

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

Integrating with respect to $x$, there results:

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

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

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

Back-substituting for $v$, we have:

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

Hence:

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

To Tonnie and any other guests viewing this topic, I invite and encourage you to post other differential equations problems in our http://www.mathhelpboards.com/f17/ forum.

Best Regards,

Mark.
 

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