Get the answer to "y(x2+xy−2y2)dx+x(3y2−xy−x2)dy=0; when x=1,y=1"

Name: y(x2+xy−2y2)dx+x(3y2−xy−x2)dy=0; when x=1,y=1
Uploaded: 2022-02-23T17:22:57+00:00
Description: y(x2+xy−2y2)dx+x(3y2−xy−x2)dy=0; when x=1,y=1

Helen Lewis

Answered question

2021-12-31

y (x^{2} + x y - 2 y^{2}) d x + x (3 y^{2} - x y - x^{2}) d y = 0; w h e n x = 1, y = 1

Answer & Explanation

Ronnie Schechter

Beginner2022-01-01Added 27 answers

\Rightarrow x (3 y^{2} - x y - x^{2}) d y = - y (x^{2} + x y - 2 y^{2}) d x

\Rightarrow \frac{d y}{d x} = \frac{- y (x^{2} + x y - 2 y^{2})}{x (3 y^{2} - x y - x^{2})}

= \frac{- x^{2} y - x y + 2 y^{3}}{3 x y^{2} - x^{2} y - x^{3}}

= \frac{\frac{- x^{2} y - x y + 2 y^{3}}{x^{3}}}{\frac{3 x y^{2} - x^{3} y - x^{2}}{x^{3}}}

\Rightarrow \frac{d y}{d x} = \frac{(- \frac{y}{x} - \frac{y^{2}}{x^{2}} + 2 \frac{y^{3}}{x^{3}})}{(3 \frac{y^{2}}{x^{2}} - \frac{y}{x} - 1)}

Which is in the form:

\frac{d y}{d x} = F (\frac{y}{x}) = F (\frac{y}{x})

i.e. the above differential eq is homogeneous equation

\Rightarrow y = v x

\Rightarrow \frac{d y}{d x} = v + \frac{d v}{d x}

\Rightarrow v + \frac{d v}{d x} = \frac{(2 v^{3} - v^{2} - v)}{3 v^{2} - v - 1}

\Rightarrow \frac{d v}{d x} = \frac{2 v^{3} - v^{2} - v}{3 v^{2} - v - 1} - v

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censoratojk

Beginner2022-01-02Added 46 answers

Given differential equation:
$y (x^{2} + x y - 2 y^{2}) d x + x (3 y^{2} - x y - x^{2}) d y = 0$
$\Rightarrow \frac{d y}{d x} = - \frac{y (x^{2} + x y - 2 y^{2})}{x (3 y^{2} - x y - x^{2})}$
$= - (\frac{\frac{y}{x} + {(\frac{y}{x})}^{2} - 2 {(\frac{y}{x})}^{3}}{3 {(\frac{y}{x})}^{2} - \frac{y}{x} - 1})$ (i)
Put $\frac{y}{x} = v$
$\Rightarrow y = v x$
$\Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$
The equation (i) becomes
$v + x \frac{d v}{d x} = - \frac{v + v^{2} - 2 v^{3}}{3 v^{2} - v - 1}$
$\Rightarrow x \frac{d v}{d x} = - (\frac{v + v^{2} - 2 v^{3}}{3 v^{2} - v - 1}) + v)$
$= - (\frac{v + v^{2} - 2 v^{3} + 3 v^{3} - v^{2} - v}{3 v^{2} - v - 1})$
$x \frac{d v}{d x} = - (\frac{v^{3}}{3 v^{2} - v - 1})$
$\Rightarrow \frac{3 v^{2} - v - 1}{v^{3}} d v = \frac{- d x}{x}$
$\Rightarrow (\frac{3}{v} - \frac{1}{v^{2}} - \frac{1}{v^{3}}) d v = - \frac{d x}{x}$
on integration, we get

Vasquez

Expert2022-01-09Added 669 answers

Please help me with solve the following differential equations properly. It's Advanced Math. Thank you

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