Question

solve using bernaulli equation 3dy/dx+xy=xy^-2​

Answer 1

Step-by-step explanation:

We have

$3 \frac{dy}{dx} + xy = xy^{ - 2}$

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

$\frac{1}{y} = v \\ \implies - \frac{1}{ {y}^{2} } \frac{dy}{dx} = \frac{dv}{dx}$

$\implies - 3 \frac{dv}{dx} + xv = x$

$\implies \frac{dv}{dx} - \frac{x}{3}.v = \frac{x}{3}$

$\implies \: v. {e}^{ - \frac{ {x}^{2} }{6} } = - \int \frac{x}{3} . {e}^{ - \frac{ {x}^{2} }{6} } dx$

$\implies \: v {e}^{ - \frac{ {x}^{2} }{6} } = {e}^{ - \frac{ {x}^{2} }{6} } + c$

$\implies \: v = 1 + c {e}^{ \frac{ {x}^{2} }{6} }$

$\implies \frac{1}{y} = 1 + c {e}^{ - \frac{ {x}^{2} }{6} }$