splve-the-d-e-xy-x-2-y-2-y-with-x-gt-0-

Question Number 29442 by prof Abdo imad last updated on 08/Feb/18

s p l v e t h e d . e {x y}^{^{'}} = \sqrt{x^{2} + y^{2}} + y w i t h x > 0

Answered by mrW2 last updated on 09/Feb/18

{x y}^{^{'}} = \sqrt{x^{2} + y^{2}} + y

y^{^{'}} = \sqrt{1 + {(\frac{y}{x})}^{2}} + (\frac{y}{x})

t = \frac{y}{x}

y = t x

y^{'} = t + x \frac{d t}{d x}

\Rightarrow t + x \frac{d t}{d x} = \sqrt{1 + t^{2}} + t

\Rightarrow x \frac{d t}{d x} = \sqrt{1 + t^{2}}

\Rightarrow \int \frac{d t}{\sqrt{1 + t^{2}}} = \int \frac{d x}{x}

\Rightarrow \ln (t + \sqrt{1 + t^{2}}) = \ln x + C

\Rightarrow \ln \frac{t + \sqrt{1 + t^{2}}}{x} = C

\Rightarrow \frac{t + \sqrt{1 + t^{2}}}{x} = c

\Rightarrow \frac{y}{x} + \sqrt{1 + {(\frac{y}{x})}^{2}} = c x

\Rightarrow y + \sqrt{x^{2} + y^{2}} = {c x}^{2}

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