# Bernoulli equation

• Apr 17th 2009, 01:00 PM
njr008
Bernoulli equation
I am having trouble expanding this Bernoulli equation correctly to solve it. Any help to get me this far would be wonderful.
x^2 dx/dy = 2y (e^1/x y^-1/2 -1)

For those who think it looks confusing what inside the brackets is supposed to read is e to the power of 1/x, y to the power of -1/2 is all one term and all of that has minus 1.
• Apr 17th 2009, 10:51 PM
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Quote:

Originally Posted by njr008
I am having trouble expanding this Bernoulli equation correctly to solve it. Any help to get me this far would be wonderful.
x^2 dx/dy = 2y (e^1/x y^-1/2 -1)

For those who think it looks confusing what inside the brackets is supposed to read is e to the power of 1/x, y to the power of -1/2 is all one term and all of that has minus 1.

So that it's easier to read, the DE is

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

I'll have a go of it now...

Edit: I'm going to assume that you've made a typo, because if it involves $\displaystyle \frac{dx}{dy}$ it is near impossible to get it to the required form

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

So I'll fix the typo so that it should say $\displaystyle \frac{dy}{dx}$ instead. Then it is easy to get to the required form

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

So, the original DE should be (I think...)

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

Some rearranging will give you

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

Here $\displaystyle n = \frac{1}{2}$ so $\displaystyle 1 - n = \frac{1}{2}$.

So make the substitution $\displaystyle v = y^{\frac{1}{2}} \implies y = v^2$

Now find $\displaystyle \frac{dy}{dx}$. Notice that $\displaystyle \frac{dy}{dx} = \frac{dy}{dv}\frac{dv}{dx}$.

So $\displaystyle \frac{dy}{dx} = 2v\frac{dv}{dx}$.

So, when substituting everything into the DE, it becomes

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

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

This is first order linear, so can be solved using an integrating factor.

$\displaystyle I = e^{\int{x^{-2}\,dx}} = e^{-\frac{1}{x}}$

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

$\displaystyle \frac{d}{dx}\left(e^{-\frac{1}{x}}v\right) = x^{-2}$

$\displaystyle e^{-\frac{1}{x}}v = \int{x^{-2}\,dx}$

$\displaystyle e^{-\frac{1}{x}}v = -\frac{1}{x} + C$

$\displaystyle v = -e^{\frac{1}{x}}\frac{1}{x} + Ce^{\frac{1}{x}}$

$\displaystyle y^{\frac{1}{2}} = -e^{\frac{1}{x}}\frac{1}{x} + Ce^{\frac{1}{x}}$

$\displaystyle y = \left(-e^{\frac{1}{x}}\frac{1}{x} + Ce^{\frac{1}{x}}\right)^2$.

Phew...
• Apr 17th 2009, 11:18 PM
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