Question

Is the following equation Exact? (3xy-y^{2})dx+x(x-y)dy=0 ​

Answer 1

Step-by-step explanation:

We have,

$\left( 3xy - {y}^{2} \right)dx + x \left(x - y \right)dy = 0$

$\implies x \left(x - y \right)dy = - \left( 3xy - {y}^{2} \right)dx$

$\implies x \left(x - y \right)dy = \left( {y}^{2} - 3xy\right)dx$

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

$\bf{Put \: \: \: y = vx}$

$\bf{ \mapsto\: \: \: \dfrac{dy}{dx} = v + x \dfrac{dv}{dx}}$

So,

$\implies v + x\dfrac{dv}{dx} = \dfrac{ {v}^{2} {x}^{2} - 3v {x}^{2} }{ {x}^{2} - v {x}^{2} }$

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

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

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

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

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

$\implies \dfrac{ 1 - v }{ v(v - 2) } \: dv = 2 \dfrac{dx}{x}$

Integrating both sides,

$\displaystyle \implies \int \dfrac{ 1 - v }{ v(v - 2) } \: dv = 2 \int \dfrac{dx}{x}$

$\displaystyle \implies - \int \dfrac{ v - 1 }{ v(v - 2) } \: dv = 2 \int \dfrac{dx}{x}$

$\displaystyle \implies - \dfrac{1}{2} \int \dfrac{ 2v -2 }{ v(v - 2) } \: dv = 2 \int \dfrac{dx}{x}$

$\displaystyle \implies - \dfrac{1}{2} \int \dfrac{ v + v -2 }{ v(v - 2) } \: dv = 2 \int \dfrac{dx}{x}$

$\displaystyle \implies - \dfrac{1}{2} \int \dfrac{ dv }{ v - 2 } - \dfrac{1}{2} \int \dfrac{ dv }{ v }= 2 \int \dfrac{dx}{x}$

$\displaystyle \implies - \dfrac{1}{2} \ln | v - 2 | - \dfrac{1}{2} \ln | v | = 2 \ln |x| + \ln |c|$

$\displaystyle \implies - \dfrac{1}{2} \ln | v( v - 2) | = 2 \ln |x| + \ln |c|$

$\displaystyle \implies \ln | v( v - 2) | = 4 \ln |x| + 2\ln |c|$

$\displaystyle \implies \ln \left| \dfrac{y}{x} \left( \dfrac{y}{x} - 2 \right) \right| = 4 \ln |x| + \ln |c^{2} |$

$\displaystyle \implies \ln \left| \dfrac{y}{x} \left( \dfrac{y - 2x}{x} \right) \right| = 4 \ln |x| + \ln |c^{2} |$

$\displaystyle \implies \ln \left| \dfrac{y(y - 2x)}{{x}^{2} } \right| = 4 \ln |x| + \ln |c^{2} |$

$\displaystyle \implies \ln \left| y(y - 2x) \right| - 2 \ln |x| = 4 \ln |x| + \ln |c^{2} |$

$\displaystyle \implies \ln \left| y(y - 2x) \right| = 6 \ln |x| + \ln |c^{2} |$

Put c² = C,

$\displaystyle \implies \ln \left| y(y - 2x) \right| = 6 \ln |x| + \ln | </p><p>C|$

$\displaystyle \implies \ln \left| y(y - 2x) \right| = \ln |C {x}^{6} |$

$\displaystyle \implies y(y - 2x) = C {x}^{6}$