Differential Equations

Solve the following initial value problem: - Wired Faculty

Question

Solve the following initial value problem:
$dy over dx space equals space fraction numerator straight y left parenthesis straight x plus 2 straight y right parenthesis over denominator straight x left parenthesis 2 straight x plus straight y right parenthesis end fraction. space space space straight y left parenthesis 1 right parenthesis space equals space 2$

Solution

The given differential equation is

dy over dx space equals space fraction numerator straight y left parenthesis straight x plus 2 straight y right parenthesis over denominator straight x left parenthesis 2 straight x plus straight y right parenthesis end fraction

Put y = vx so that

dy over dx space equals space straight v plus straight x dv over dx

therefore space space space space space space space straight v plus straight x dv over dx space equals space fraction numerator vx left parenthesis straight x plus 2 vx right parenthesis over denominator straight x left parenthesis 2 straight x plus vx right parenthesis end fraction

therefore space space space space straight v plus straight x dv over dx space equals space fraction numerator straight v left parenthesis 1 plus 2 straight v right parenthesis over denominator 2 plus straight v end fraction therefore space space space space space straight x dv over dx space equals space fraction numerator straight v left parenthesis 1 plus 2 straight v right parenthesis over denominator 2 plus straight v end fraction minus straight v space equals space space fraction numerator straight v plus 2 straight v squared minus 2 straight v minus straight v squared over denominator 2 plus straight v end fraction therefore space space space space space straight x dv over dx space equals space fraction numerator straight v squared minus straight v over denominator straight v plus 2 end fraction

Separating the variables and integrating, we get,

integral fraction numerator straight v plus 2 over denominator straight v squared minus straight v end fraction dv space equals space integral 1 over straight x dx

therefore space space space integral open square brackets fraction numerator straight v plus 2 over denominator straight v left parenthesis straight v minus 1 right parenthesis end fraction close square brackets dv space equals space integral 1 over straight x dx therefore space space space space integral open square brackets fraction numerator 0 plus 2 over denominator straight v left parenthesis 0 minus 1 right parenthesis end fraction plus fraction numerator 1 plus 2 over denominator left parenthesis 1 right parenthesis thin space left parenthesis straight v minus 1 right parenthesis end fraction close square brackets dv space equals space integral 1 over straight x dx therefore space space space minus integral 2 over straight v dv plus integral fraction numerator 3 over denominator straight v minus 1 end fraction dv space equals space integral 1 over straight x dx therefore space space space minus 2 space logv plus space 3 space log space left parenthesis straight v minus 1 right parenthesis space equals space log space straight x space plus straight c therefore space space space minus 2 log space straight y over straight x plus 3 space log space open parentheses straight y over straight x minus 1 close parentheses space equals space logx plus space straight c

Now,

straight y left parenthesis 1 right parenthesis space equals space 2 space space space space space space space space rightwards double arrow space space space space straight y space equals space 2 space space space space space space when space space space space straight x space equals space 1

therefore space space space space space space minus 2 space log space 2 over 1 plus 3 space log space open parentheses 2 over 1 minus 1 close parentheses space equals space log space 1 space plus space straight c comma space space space space space space therefore space space space minus 2 space log space 2 space plus space 3 space log space 1 space equals space log space 1 space plus straight c therefore space space space space minus 2 space log space 2 space plus space 0 space equals space 0 plus space straight c space space space space space space space space space space space rightwards double arrow space space space space straight c space equals space minus 2 space log space 2 therefore space space space solution space is space space minus 2 space log space straight y over straight x plus 3 space log space open parentheses fraction numerator straight y minus straight x over denominator straight x end fraction close parentheses space equals space log space straight x space plus space 2 space log space 2