Question

solve (3xy-2aysquare) dx+(xsquare -2axy) dy=0​

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

$\rm \: (3xy - 2a {y}^{2} )dx + ( {x}^{2} - 2axy)dy = 0$

$\rm : \implies \: (3xy - 2a {y}^{2} )dx = - ( {x}^{2} - 2axy)dy$

$\rm : \implies \: \rm \: (3xy - 2a {y}^{2} )dx = ( 2axy - {x}^{2} )dy$

$\rm : \implies \: \dfrac{dy}{dx} \: = \: \dfrac{3xy - 2a {y}^{2}}{2axy - {x}^{2}}$

Since, the degree of each term is same, therefore its a homogeneous differential equation.

To solve such equation,

$\rm : \implies \: Put \: y \: = \: vx$

So, equation can be rewritten as

$\rm : \implies \: \dfrac{d}{dx}(vx) = \dfrac{3x(vx) - 2a {v}^{2} {x}^{2} }{2ax(vx) - {x}^{2} }$

$\rm : \implies \: v \: + \: x\dfrac{dv}{dx} = \dfrac{3v - 2a {v}^{2} }{2av - 1}$

$\rm : \implies \: \: \: x\dfrac{dv}{dx} = \dfrac{3v - 2a {v}^{2} }{2av - 1} - v$

$\rm : \implies \: \: x\dfrac{dv}{dx} = \dfrac{3v - 2a {v}^{2} - 2a {v}^{2} + v }{2av - 1}$

$\rm : \implies \: \: x\dfrac{dv}{dx} = \dfrac{4v - 4a {v}^{2} }{2av - 1}$

$\rm : \implies \: \: x\dfrac{dv}{dx} = - \dfrac{4(a {v}^{2} - v) }{2av - 1}$

$\rm : \implies \: \dfrac{2av - 1}{a {v}^{2} - v} dv \: = \: - 4 \: \dfrac{dx}{x}$

Now, integrating both sides, we get

$\rm : \implies \: \int \: \dfrac{2av - 1}{a {v}^{2} - v} dv \: = \: - 4 \: \int\dfrac{dx}{x}$

$\rm : \implies \: log(a {v}^{2} - v) = - 4 log(x) + log(c)$

$\begin{gathered} \because\:{\underline{\boxed{\bf{\blue{{\tt \: \int \:\dfrac{1}{x} dx\: = \: log(x) + c }}}}}} \\ \end{gathered}$

$\begin{gathered} \because\:{\underline{\boxed{\bf{\blue{{\tt \: \int \:\dfrac{f'(x)}{f(x)} dx\: = \: log(f(x)) + c }}}}}} \\ \end{gathered}$

$\rm : \implies \: log(a {v}^{2} - v) = log(x) ^{ - 4} + log(c)$

$\rm : \implies \: log(a {v}^{2} - v) = log( {x}^{ - 4} c)$

$\rm : \implies \: {av}^{2} - v \: = \: c {x}^{ - 4}$

On substituting the value of v, we get

$\rm : \implies \: a\dfrac{ {y}^{2} }{ {x}^{2} } - \dfrac{y}{x} = c {x}^{ - 4}$

$\rm : \implies \: \dfrac{a {y}^{2} - xy }{ {x}^{2} } = \dfrac{c}{ {x}^{4} }$

$\rm : \implies \: {ax}^{2} {y}^{2} - {yx}^{3} = c$