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$$y'=-\frac{x+y-2}{x-y+4}$$

I have used the substation:

$x=u-1$

$y=v+3$

And got $$\frac{dv}{du}=\frac{-1-\frac{v}{u}}{1-\frac{v}{u}}$$

$z=\frac{v}{u}$

And got:

$\frac{du}{u}\frac{1-z}{z^2-2z-1}$

$$u=e^{\frac{-z+2z+1}{4}+C}$$

What should I do next?

  • 1
    Assuming that all your work is correct, how about backsubbing $z$ in terms of $u$ and $v$ and from there you can get it into $x$ and $y$ – 2017-01-24
  • 0
    @imranfat So I will get $$x+1=e^{\frac{-\frac{y-3}{x+1}+2\frac{y-3}{x+1}+1}{4}+C}$$ – 2017-01-24
  • 0
    Essentially, yes, assuming that prior work is ok and that no further beautification of the right hand expression is needed. – 2017-01-24
  • 2
    But your prior work is wrong. You should get $$ \frac{z-1}{z^2 - 2z-1}\; dz = -\frac{du}{u}$$ and then $$\frac{1}{2} \ln(z^2-2z-1) = -\ln(u) + C$$. – 2017-01-24

2 Answers 2

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$$\frac { dv }{ du } =\frac { -1-\frac { v }{ u } }{ 1-\frac { v }{ u } } \\ z=\frac { v }{ u } \\ v=uz\\ z+{ u }z^{ \prime }=\frac { -1-z }{ 1-z } =\frac { 1+z }{ z-1 } \\ u{ z }^{ \prime }=\frac { 1+z }{ z-1 } -z=\frac { 1+z-{ z }^{ 2 }+z }{ z-1 } =\frac { -{ z }^{ 2 }+2z+1 }{ z-1 } =\frac { { z }^{ 2 }-2z-1 }{ 1-z } \\ \int { \frac { 1-z }{ { z }^{ 2 }-2z-1 } } dz=\int { \frac { du }{ u } } \\ \int { \frac { d\left( { z }^{ 2 }-2z-1 \right) }{ { z }^{ 2 }-2z-1 } } =-2\int { \frac { du }{ u } } \\ \ln { \left| { z }^{ 2 }-2z-1 \right| } =-2\ln { \left| u \right| } \\ { z }^{ 2 }-2z-1=\frac { C }{ { u }^{ 2 } } \\ \frac { { v }^{ 2 } }{ { u }^{ 2 } } -2\frac { v }{ u } -1=\frac { C }{ { u }^{ 2 } } \\ { v }^{ 2 }-2vu-{ u }^{ 2 }=C\\ \\ $$ then plug substition $x$ and $y$

  • 0
    Shouldnt it be $${ z }^{ 2 }-2z-1=\frac { C }{ { u }^{ 2 } }$$? – 2017-01-24
  • 1
    @gbox, it was a typo , you are absolutely right – 2017-01-24
  • 0
    Plugging in $x$ and $y$ and I am done? no need to write down $y$ explicitly – 2017-01-24
  • 1
    yes, you got it – 2017-01-24
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You could approach it this way:

$$ \frac{dy}{dx} = -\frac{x + y - 2}{x - y + 4} \\ \implies x\ dy - y\ dy + 4\ dy = -x\ dx - y \ dx - 2 \ dx \\ \implies (x \ dy + y \ dx) - x \ dx - 2 \ dx - y \ dy = 0 \\ \implies \int d(xy) - \int x \ dx - 2\int dx - \int y \ dy = 0 \\ \implies x^2 + y^2 - 2xy + 4x + c = 0 $$

which is a parabola.

  • 0
    Can you please elaborate on the first step? with technique have you used? – 2017-01-24
  • 0
    I've treated the differential as a fraction. In cases like this question, this treatment works. – 2017-01-24
  • 1
    Basically, this treatment works in solving first order homogeneous differential equations. – 2017-01-24