Question

x^2-2xy+3y^2dx+y^2+6xy-x^2dy by using homogeneous differential equation​

Answer 1

Step-by-step explanation:

We have,

$( {x}^{2} - 2xy + 3 {y}^{2} )dx + ( {y}^{2} + 6xy - {x}^{2} )dy = 0$

$\implies \frac{dy}{dx} = \frac{ {x}^{2} - 2xy + 3{y}^{2} }{ {x}^{2} - 6xy - {y}^{2} }$

$Let \: \: y = vx \\ \implies \frac{dy}{dx} = v + x \frac{dv}{dx}$

$\implies \: v + x\frac{dv}{dx} = \frac{ {x}^{2} - 2v {x}^{2} + 3{v}^{2} {x}^{2} }{ {x}^{2} - 6vx- {x}^{2} {v}^{2} }$

$\implies \: v + x\frac{dv}{dx} = \frac{ {x}^{2} (1 - 2v + 3{v}^{2} ) }{ {x}^{2}(1 - 6v- {v}^{2} ) }$

$\implies \: x\frac{dv}{dx} = \frac{ (1 - 2v + 3{v}^{2} ) }{ (1 - 6v- {v}^{2} ) } - v$

$\implies \: x\frac{dv}{dx} = \frac{ (1 - 3v + 9{v}^{2} + {v}^{3} ) }{ (1 - 6v- {v}^{2} ) }$

$\implies \frac{1 - 6v - {v}^{2} }{ {v}^{3} + 9 {v}^{2} - 3v + 1} dv = \frac{dx}{x}$

Integrating both sides,

$\implies \int\frac{1 - 6v - {v}^{2} }{ {v}^{3} + 9 {v}^{2} - 3v + 1} dv = \int\frac{dx}{x}$

$\implies - \int\frac{ {v}^{2} + 6v - 1}{ {v}^{3} + 9 {v}^{2} - 3v + 1} dv = \int\frac{dx}{x}$

$Let \: \: {v}^{3} + 9 {v}^{2} - 3v + 1 = t \\ \implies(3 {v}^{2} + 18v - 3)dv = dt$

$\implies - \frac{1}{3} \int\frac{ 3{v}^{2} + 18v - 3}{ {v}^{3} + 9 {v}^{2} - 3v + 1} dv = \int\frac{dx}{x}$

$\implies - \frac{1}{3} \int\frac{ dt}{t} = \int\frac{dx}{x}$

$\implies - \frac{1}{3} ln(t) = ln(x) + c$

$\implies - \frac{1}{3} ln( {v}^{3} + 9 {v}^{2} - 3v + 1 ) = ln(x) + c$

$\implies - \frac{1}{3} ln( {( \frac{y}{x} )}^{3} + 9 {( \frac{y}{x}) }^{2} - 3( \frac{y}{x} ) + 1 ) = ln(x) + c$

$\implies - \frac{1}{3} ln( { y }^{3} + 9 x{y }^{2} - 3 {x}^{2} y + {x}^{3} ) + ln(x) = ln(x) + c$

$\implies - \frac{1}{3} ln( { y }^{3} + 9 x{y }^{2} - 3 {x}^{2} y + {x}^{3} ) = c$

$\implies ln( { y }^{3} + 9 x{y }^{2} - 3 {x}^{2} y + {x}^{3} ) = - 3 c$

$\implies { y }^{3} + 9 x{y }^{2} - 3 {x}^{2} y + {x}^{3} = {e}^ {- 3 c}$

$\implies { y }^{3} + 9 x{y }^{2} - 3 {x}^{2} y + {x}^{3} = k$