Solve-for-y-x-xy-y-2x-3-sin-2-y-x-

Question Number 67349 by Joel122 last updated on 26/Aug/19

Solve for y (x)

{x y}^{'} = y + 2 x^{3} \sin^{2} (\frac{y}{x})

Answered by mind is power last updated on 26/Aug/19

l e t u = \frac{y}{x}

\Rightarrow u^{^{'}} = \frac{y^{^{'}} x - y}{x^{2}} = \frac{y^{^{'}}}{x^{}} - \frac{y}{x^{2}}

{x y}^{,} = y + 2 x^{3} {s i n}^{2} (\frac{y}{x}) \Rightarrow \frac{y^{'}}{x^{}} - \frac{y}{x^{2}} = 2 {x s i n}^{2} (\frac{y}{x})

\Rightarrow u^{^{'}} = 2 {x s i n}^{2} (u)

\Rightarrow \frac{u^{^{'}}}{2 {s i n}^{2} (u)} = x

\Rightarrow \int \frac{d u}{2 {s i n}^{2} (u)} = \int x d x

\Rightarrow \int \frac{{s i n}^{2} (u) + {c o s}^{2} (u)}{2 {s i n}^{2} (u)} d u = \frac{x^{2}}{2} + c

\int \frac{1}{2} (1 + \cot^{2} (u)) d u = \frac{x^{2}}{2} + c

\Rightarrow - \frac{1}{2} c o t (u) = \frac{x^{2}}{2} + c

\Rightarrow t g (\frac{π}{2} - u) = - x^{2} + c

\Rightarrow u = \frac{π}{2} - \tan^{- 1} (- x^{2} + c)

\Rightarrow \frac{y}{x} = \frac{π}{2} - \tan^{- 1} (- x^{2} + c) \Rightarrow y = x (\frac{π}{2} - \tan^{- 1} (- x^{2} + c))

\Rightarrow \frac{y}{x}

Commented by Joel122 last updated on 26/Aug/19

t h a n k y o u v e r y m u c h

Commented by mind is power last updated on 26/Aug/19

y^{'} r e w e l c o m