# Bernoulli's equation

• Jan 9th 2008, 04:37 AM
Bernoulli's equation
solve the equation:
$\displaystyle (3\sin{y} - 5x)dx +2x^2\cot{y}dy=0$

this is a tough problem for me that i cant start...
i can't find any separable nor any linearity in this solution

ill try to solve
solution:
$\displaystyle (3\sin{y} - 5x)dx +2x^2\frac{\cos{y}}{\sin{y}}dy=0$

substitute let w = sin(y)
dw = cos(y)dy so substitute

$\displaystyle (3w - 5x)dx + 2x^2\frac{dw}{w} = 0$

$\displaystyle \frac{3w - 5x}{2x^2}dx = \frac{dw}{w}$

$\displaystyle 2x^2dw - 5xwdx = -3w^2dx$

$\displaystyle \frac{dw}{dx} - \frac{5w}{2x} = \frac{-3w^2}{2x^2}$
....
now im stuck
• Jan 9th 2008, 05:30 AM
TwistedOne151
Further transform
Let us define another change of variable for our dependent variable: choose $\displaystyle u=\frac{5}{3w}$.
Then $\displaystyle w=\frac{5}{3u}$, $\displaystyle dw=-\frac{5}{3u^2}du$, and thus our equation becomes
$\displaystyle -\frac{5}{3u^2}\frac{du}{dx} = \frac{25}{6ux} - \frac{25}{6u^2x^2}$

$\displaystyle \frac{du}{dx} = -\frac{5}{2x}u + \frac{5}{2x^2}$

$\displaystyle \frac{du}{dx} + \frac{5}{2x}u = \frac{5}{2x^2}$
This is a nonhomogenous first order linear differential equation. We can see by substitution that $\displaystyle u=\frac{5}{3x}$ is a solution. Thus, we find and add to that the general solution for the corresponding homogenous equation
$\displaystyle \frac{du}{dx} + \frac{5}{2x}u = 0$, which is separable:
$\displaystyle \frac{du}{u} = \frac{-5}{2x}$
$\displaystyle \int\frac{du}{u} = \int\frac{-5}{2x}$
$\displaystyle \ln{u} = \frac{-5}{2}\ln{x}+c$
$\displaystyle u = \frac{C}{x^{\frac{5}{2}}}$
Thus the solution to $\displaystyle \frac{du}{dx} + \frac{5}{2x}u = \frac{5}{2x^2}$ is $\displaystyle u=\frac{5}{3x}+\frac{C}{x^{\frac{5}{2}}}$, and from there you can convert back into w, and then into $\displaystyle y=\arcsin{w}$.

--Kevin C.