Question

solve 2y'-(y/x)=y²/x² using Bernoulli equation​

Answer 1

Step-by-step explanation:

We have,

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

$\implies \frac{2}{ {y}^{2} } \frac{dy}{dx} - \frac{1}{yx} = \frac{ 1 }{ {x}^{2} }$

$Let \: - \frac{1}{y} = t \\ \implies \frac{1}{ {y}^{2} } \frac{dy}{dx} = \frac{dt}{dx}$

$\implies 2 \frac{dt}{dx} + \frac{t}{x} = \frac{ 1 }{ {x}^{2} }$

$\implies \frac{dt}{dx} + \frac{t}{2x} = \frac{ 1 }{ {2x}^{2} }$

$l.F. = {e}^{ \int \frac{dx}{2x} } = {e}^{\frac{1}{2} ln(x) } = \sqrt{x}$

Now,

$\sqrt{x} .t = \int \frac{ \sqrt{x} }{2 {x}^{2} } dx$

$\implies\sqrt{x} .t = \frac{1}{2} \int \frac{ 1}{ {x}^{ \frac{3}{2} } } dx$

$\implies\sqrt{x} .t = \frac{1}{2} (\frac{ {x}^{ - \frac{3}{2} + 1 } }{ - \frac{3}{2} + 1} )$

$\implies2\sqrt{x} .t = (\frac{ {x}^{ - \frac{1}{2} } }{ - \frac{1}{2} } ) + c$

$\implies2\sqrt{x} .t = - \frac{2}{ \sqrt{x} } + c$

$\implies2\sqrt{x}. \frac{1}{y} = - \frac{2}{ \sqrt{x} } + c$

$\implies\frac{1}{y} = - \frac{1}{ x} + \frac{c}{2 \sqrt{x} }$