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Differential Equations

Question
CBSEENMA12033123

Show that the given differential equation is homogeneous and solve it:
(x-y) dy - (x+y) dx = 0



Solution
The given differential equation is
                       (x-y) dy - (x+y) dx = 0   or  (x-y) dy = (x+y) dy
or                    dy over dx space equals space fraction numerator straight x plus straight y over denominator straight x minus straight y end fraction                               ...(1)
It is a differential equation of the form dy over dx space equals space straight F left parenthesis straight x comma space straight y right parenthesis
Here,     straight F left parenthesis straight x comma space straight y right parenthesis space equals space fraction numerator straight x plus straight y over denominator straight x minus straight y end fraction
Replacing x by λx and y by λx comma space we space get comma
                                straight F left parenthesis λx comma space λy right parenthesis space equals space fraction numerator λx plus λy over denominator λx minus λy end fraction space equals space fraction numerator straight lambda left parenthesis straight x plus straight y right parenthesis over denominator straight lambda left parenthesis straight x minus straight y right parenthesis end fraction space equals space straight lambda degree space space left square bracket straight F left parenthesis straight x comma space straight y right parenthesis right square bracket

∴       F(x, y) is a homogeneous function of degree zero.
∴     given differential equation is a homogeneous differential equation.
Put y = v x so that dy over dx space equals space straight v plus straight x dv over dx
therefore space space space from space left parenthesis 1 right parenthesis comma space space space straight v plus straight x dv over dx space equals space fraction numerator straight x plus vx over denominator straight x minus vx end fraction space space or space space space straight x dv over dx equals fraction numerator 1 plus straight v over denominator 1 minus straight v end fraction minus straight v
therefore space space space straight x dv over dx space equals fraction numerator 1 plus straight v minus straight v plus straight v squared over denominator 1 minus straight v end fraction space space or space space space straight x dv over dx space equals space fraction numerator 1 plus straight v squared over denominator 1 minus straight v end fraction
Separating the variables,   fraction numerator 1 minus straight v over denominator 1 plus straight v squared end fraction dv space equals space 1 over straight x dx
Integrating ,  integral fraction numerator 1 minus straight v over denominator 1 plus straight v squared end fraction dv space equals space integral 1 over straight x dx
or   integral fraction numerator 1 over denominator 1 plus straight v squared end fraction dv space minus space 1 half integral fraction numerator 2 straight v over denominator 1 plus straight v squared end fraction dv space equals space integral 1 over straight x dx space space or space space tan to the power of negative 1 end exponent straight v minus 1 half log space left parenthesis 1 plus straight v squared right parenthesis space equals space log space open vertical bar straight x close vertical bar plus straight c
or  tan to the power of negative 1 end exponent straight y over straight x minus 1 half log space open parentheses 1 plus straight y squared over straight x squared close parentheses space equals space log space open vertical bar straight x close vertical bar plus straight c
or   tan to the power of negative 1 end exponent straight y over straight x minus 1 half log space open parentheses fraction numerator straight x squared plus straight y squared over denominator straight x squared end fraction close parentheses space equals space log space open vertical bar straight x close vertical bar space plus space straight c space is space the space required space solution. space

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