The solution of the differential equation, [dy dx] = (x -- y)2, when y(1) = 1, is :

Question

The solution of the differential equation,

$${{dy} \over {dx}}$$ = (x -- y)2, when y(1) = 1, is :

Accepted Answer

x - y = t

$$ \Rightarrow $$ $${{dy} \over {dx}} = 1 - {{dt} \over {dx}}$$

$$ \Rightarrow $$  1 - $${{dt} \over {dx}}$$ = t2 $$ \Rightarrow $$  $$\int {{{dt} \over {1 - {t^2}}}} $$ = $$\int {1dx} $$

$$ \Rightarrow $$  $${1 \over 2}\ell n\left( {{{1 + t} \over {1 - t}}} \right) = x + \lambda $$

$$ \Rightarrow $$  $${1 \over 2}\ell n\left( {{{1 + x - y} \over {1 - x + y}}} \right) = x + \lambda $$ given y(1) = 1

$$ \Rightarrow $$  $${1 \over 2}\ell n(1) = 1 + \lambda  \Rightarrow \lambda  =  - 1$$

$$ \Rightarrow $$  $$\ell n\left( {{{1 + x - y} \over {1 - x + y}}} \right)$$ = 2(x - 1)

$$ \Rightarrow $$  $$ - \ell n\left( {{{1 - x + y} \over {1 + x - y}}} \right)$$ = 2(x - 1)

Solutions · 1