Can't Make Sense of Bernoulli Equation

[snowJT]

it...just...does...not...make...the...slightest...sense...to...me...

Here it goes...

$$y' + \frac{y}{x} = 3x^2y^2$$

This is a Bernoulli equation with $$P = \frac{1}{2}$$, $$Q = 3x^2$$, and $$n = 2$$. We first divide through by $$y^2$$, obtaining...

$$\frac{1}{y^2} \frac{dy}{dx} + \frac{y^-^1}{x} = 3x^2$$

We substitute $$z = y^-^1$$ and

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

Substituting... $$-\frac{dz}{dx} + \frac{z}{x} = 3x^2 \Longrightarrow \frac{dz}{x} - \frac{z}{x} = -3x^2$$

I'll stop there... two things I don't get... how does $$P = \frac{1}{2}$$ and how does that substitution work...

If you can help, I would really appreciate it.

 

[HallsofIvy]

Yes, for that sentence to make any sense, you would have to know what a "Bernoulli equation" was, wouldn't you? Not remembering exactly and not having a differential equations textbook immediately handy, I "googled" on "Bernoulli differential equation" and get (from Wikipedia)
"an equation of the form y'+ P(x)y= Q(x)yⁿ for n not equal to 1"

There's obviously an error- either in your source or in your copying. That is a Bernoulli equation with P(x)= 1/x, not 1/2.

Dividing a Bernoulli equation by yⁿ and then making the substitution z= y^1-n always reduces the equation to a linear equation.

As for "how the substitution works" I don't know what more to say. It is shown pretty clearly there. What are you not understanding? (Other than that "P(x)= 1/2" should be P(x)= 1/x".)

Do you see how they go from z= y^-1 to dz/dx= -y^-2dy/dx? That's straight forward "chain rule". Then, of course, y^-2dy/dx in the differential equation becomes -dz/dx. y^-1/x is simply z/x. The differential equation y^-2dy/dx+ y^-1/x= 3x² becomes -dz/dx+ (1/x)z= 3x².

 

[tacman]

Thank you.The Bernoulli equation can seem confusing at first, but with some understanding and practice, it can become easier to comprehend. Let's break down the steps and try to make sense of it.

First, let's look at the equation given: y' + \frac{y}{x} = 3x^2y^2. This is in the form of a Bernoulli equation, which has the general form of y' + P(x)y = Q(x)y^n. In this case, P(x) = \frac{1}{x} and Q(x) = 3x^2.

Now, to solve a Bernoulli equation, we use a substitution to transform it into a linear equation. In this case, we choose z = y^-1, which means that y = z^-1. This substitution allows us to rewrite the equation as \frac{dz}{dx} = -y^-2\frac{dy}{dx} \Longrightarrow \frac{dy}{dx} = -y^2\frac{dz}{dx}. This is known as the Bernoulli substitution.

Now, substituting y = z^-1 into our original equation, we get -\frac{dz}{dx} + \frac{z}{x} = 3x^2. This is now a linear equation, which we can solve using standard techniques.

As for why P = \frac{1}{2}, this is because we are dividing the original equation by y^2, which is equivalent to multiplying by y^-2. In other words, we are dividing the coefficient of y' by the power of y in the original equation, which gives us P = \frac{1}{2}.

I hope this explanation helps to make sense of the Bernoulli equation for you. It may take some practice and understanding of substitution and linear equations, but with time, it will become clearer. Keep working at it and don't hesitate to ask for help if you need it. Good luck!