x-2-y-x-2-xy-y-2-yy-sin-x-0-y-1-y-2-separation-of-variables-

Question Number 5097 by LMTV last updated on 12/Apr/16

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

{y y}^{'} + \sin x = 0

y^{'} = 1 - y^{2}

separation of variables

Answered by Yozzii last updated on 12/Apr/16

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

y^{^{'}} - \frac{x}{x^{2} - 1} y = \frac{x^{2}}{x^{2} - 1}

y^{^{'}} + P (x) y = Q (x)

I (x) = e^{- \int \frac{x}{x^{2} - 1} d x} = e^{- \frac{1}{2} \int \frac{2 x}{x^{2} - 1} d x} = e^{- 0.5 l n (x^{2} - 1)}

I (x) = \frac{1}{\sqrt{x^{2} - 1}}

U s i n g y I (x) = \int I (x) Q (x) d x

y I (x) = \int \frac{x^{2}}{{(x^{2} - 1)}^{3 / 2}} d x

x = c o s h u \Rightarrow d x = s i n h u d u

x^{2} - 1 = {c o s h}^{2} u - 1 = {s i n h}^{2} u .

y I (x) = \int \frac{{c o s h}^{2} u s i n h u}{{s i n h}^{3} u} d u

y I (x) = \int \frac{{c o s h}^{2} u}{{s i n h}^{2} u} d u = \int ({c o s e c h}^{2} u + 1) d u

D (c o t h u) = \frac{s i n h u s i n h u - c o s h u c o s h u}{{s i n h}^{2} u}

D (c o t h u) = \frac{- ({c o s h}^{2} u - {s i n h}^{2} u)}{{s i n h}^{2} u}

D (c o t h u) = - {c o s e c h}^{2} u

\Rightarrow c o t h u = - \int {c o s e c h}^{2} u d u

y I (x) = - c o t h u + u + C

u = {c o s h}^{- 1} x

\Rightarrow y = (C + {c o s h}^{- 1} x - c o t h ({c o s h}^{- 1} x)) \sqrt{x^{2} - 1}

y = (C + l n (x + \sqrt{x^{2} - 1}) - \frac{c o s h ({c o s h}^{- 1} x)}{s i n h ({c o s h}^{- 1} x)}) \sqrt{x^{2} - 1}

s i n h ({c o s h}^{- 1} x) = \frac{e^{l n (x + \sqrt{x^{2} - 1})} - e^{l n \frac{1}{x + \sqrt{x^{2} - 1}}}}{2}

s i n h ({c o s h}^{- 1} x) = \frac{{(x + \sqrt{x^{2} - 1})}^{2} - 1}{2 (x + \sqrt{x^{2} - 1})}

c o s h ({c o s h}^{- 1} x) = \frac{{(x + \sqrt{x^{2} - 1})}^{2} + 1}{2 (x + \sqrt{x^{2} - 1})}

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

c o t h ({c o s h}^{- 1} x) = \frac{x^{2} + 2 x \sqrt{x^{2} - 1} + x^{2} - 1 + 1}{x^{2} + 2 x \sqrt{x^{2} - 1} + x^{2} - 1 - 1}

c o t h ({c o s h}^{- 1} x) = \frac{x (x + \sqrt{x^{2} - 1})}{x^{2} + x \sqrt{x^{2} - 1} - 1}

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

= \frac{x}{x - x + \sqrt{x^{2} - 1}}

c o t h ({c o s h}^{- 1} x) = \frac{x}{\sqrt{x^{2} - 1}}

∴ y = (C + l n (x + \sqrt{x^{2} - 1})) - x

Commented by LMTV last updated on 12/Apr/16

sorry ToT change question \dots my miss \dots

Commented by Yozzii last updated on 12/Apr/16

A n s w e r b a s e d o n q u e s t i o n p o s t e d b e f o r e

c o r r e c t i o n : x^{2} y^{^{'}} = x^{2} + x y + y^{^{'}} .

Commented by Yozzii last updated on 12/Apr/16

P a s d e p r o b l \overset{`}{e m e} .

Answered by Yozzii last updated on 12/Apr/16

\frac{d y}{d x} = 1 - y^{2} \Rightarrow \int \frac{d y}{1 - y^{2}} = \int d x

\frac{1}{2} \int \frac{1}{1 - y} + \frac{1}{1 + y} d y = x + C

\frac{1}{2} (l n ∣ y + 1 ∣ - l n ∣ 1 - y ∣) = x + C

\frac{y + 1}{1 - y} = e^{2 x + 2 C} = {A e}^{2 x}

y + 1 = {A e}^{2 x} - {y A e}^{2 x}

y (1 + {A e}^{2 x}) = {A e}^{2 x} - 1

y = \frac{{A e}^{2 x} - 1}{{A e}^{2 x} + 1}

Answered by Yozzii last updated on 12/Apr/16

y^{'} = 1 + (\frac{y}{x}) + {(\frac{y}{x})}^{2} (x \neq 0)

L e t y = u x \Rightarrow y^{^{'}} = u + u^{^{'}} x {u = f (x)}

∴ u + u^{^{'}} x = 1 + u + u^{2}

u^{^{'}} x = 1 + u^{2}

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

{t a n}^{- 1} u = l n ∣ x ∣ + C

\Rightarrow u = t a n (C + l n ∣ x ∣)

S i n c e u = \frac{y}{x} \Rightarrow y = x t a n (C + l n ∣ x ∣)

w h e r e C i s c o n s t a n t .

Answered by Yozzii last updated on 12/Apr/16

{y y}^{^{'}} = - s i n x \Rightarrow \int y d y = - \int s i n x d x

0.5 y^{2} = - (- c o s x) + C

y^{2} = 2 c o s x + D

y = \pm \sqrt{D + 2 c o s x} .

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