# Thread: Use of Bernoulli's Equation

1. ## Use of Bernoulli's Equation

I have a problem I am working on:

xy - dy/dx = e^(-3x^2/2) * y^4

If I solve this using the Bernoulli equation then I have n = 4, 1-n = 3.

I make the substitution v = y^3 , y = v^1/3

I'm have a little problem with the next step of finding dx/dy. I'm pretty sure once I find this then I can substitute these values into the DE to get a linear DE that can then be solved.

2. First we write the DE as...

$\displaystyle y^{'} = xy - e^{-\frac{3}{2} x^{2}} y^{4} \rightarrow \frac{y^{'}}{y^{4}} = \frac{x}{y^{3}} - e^{-\frac{3}{2} x^{2}}$ (1)

... and then we set...

$\displaystyle \frac {1}{y^{4}} = z \rightarrow \frac{y^{'}}{y^{4}}= - \frac{z^{'}}{3}$ (2)

Now combining (1) and (2) we arrive to write...

$\displaystyle z^{'} = -3\cdot xz +3\cdot e^{-\frac{3}{2} x^{2}}$ (3)

... which is linear in $\displaystyle z$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$