Find-the-solution-of-the-d-e-sinhx-dy-dx-2-2-dy-dx-sinhx-0-which-satisfies-y-0-at-x-0-

Question Number 2052 by Yozzi last updated on 01/Nov/15

F i n d t h e s o l u t i o n o f t h e d . e

(s i n h x) {(\frac{d y}{d x})}^{2} + 2 \frac{d y}{d x} - s i n h x = 0

w h i c h s a t i s f i e s y = 0 a t x = 0 .

Commented by prakash jain last updated on 01/Nov/15

\int \cosh x \cdot \coth x d x

= \coth x \cdot \sinh x + \int {csch}^{2} x \cdot \sinh x d x

= \coth x \cdot \sinh x + \int csch x d x

= \cosh x + \ln \tanh \frac{x}{2} + C

Commented by Rasheed Soomro last updated on 04/Nov/15

F i n d t h e s o l u t i o n o f t h e d . e

(s i n h x) {(\frac{d y}{d x})}^{2} + 2 \frac{d y}{d x} - s i n h x = 0

w h i c h s a t i s f i e s y = 0 a t x = 0 .

- - - - - - - - - - + + + + - - - - - - -

L e t \frac{d y}{d x} = u

(s i n h x) u^{2} + 2 u - s i n h x = 0

u = \frac{- 2 \pm \sqrt{{(2)}^{2} - 4 (s i n h x) (- s i n h x)}}{2 s i n h x}

= \frac{- 2 \pm 2 \sqrt{1 + {s i n h}^{2} x}}{2 s i n h x}

= \frac{- 1 \pm c o s h x}{s i n h x}

u = \frac{c o s h x - 1}{s i n h x} ∣ u = - \frac{1 + c o s h x}{s i n h x}

u = \frac{c o s h x}{s i n h x} - \frac{1}{s i n h x} ∣ u = - \frac{1 + c o s h x}{s i n h x}

\frac{d y}{d x} = c o t h x - c s c h x ∣ \frac{d y}{d x} = - \frac{1}{s i n h x} - \frac{c o s h x}{s i n h x}

y = \int (c o t h x - c s c h x) d x ∣ y = - \int (c s c h x + c o t h x) d x

y = \int (c o t h x) d x - \int (c s c h x) d x ∣ y = - \int (c s c h x) d x - \int (c o t h x) d x

\underline{= l n ∣ s i n h x ∣ - l n ∣ t a n h \frac{x}{2} ∣ + C} ∣ = - l n ∣ s i n h x ∣ - l n ∣ t a n h \frac{x}{2} ∣ + C

F u r t h e r e v a l u a t i o n

S u g g e s t e d b y M r . Y o z z i :

y = l n \frac{s i n h x}{t a n h \frac{x}{2}} + C ∣ y = - l n ∣ (s i n h x) (t a n h \frac{x}{2}) ∣ + C

y = l n {2 s i n h \frac{x}{2} c o s h \frac{x}{2} \times \frac{c o s h \frac{x}{2}}{s i n h \frac{x}{2}}} + C ∣ y = - l n {2 s i n h \frac{x}{2} c o s h \frac{x}{2} \times \frac{s i n h \frac{x}{2}}{c o s h \frac{x}{2}}} + C

= l n {2 {c o s h}^{2} \frac{x}{2}} + C ∣ y = - l n {2 {s i n h}^{2} \frac{x}{2}} + C [C i s n o t d e f i n e d]

S e e c o m m e n t (02 - 11 - 2015) b y Y o z z i f o r f u r t h e r p r o c e s s .

Commented by Yozzi last updated on 01/Nov/15

{T h e r e}^{'} s a n e r r o r i n y o u r e v a l u a t i o n M r . S o o m r o .

\sqrt{1 + {s i n h}^{2} x} = \pm c o s h x \neq {c o s h}^{2} x .

I a p p r e c i a t e t h e i n t e g r a l

\int c o s h x c o t h x d x t h a t a r o s e h o w e v e r .

E r r o r s c o u l d l e a d o n e t o l e a r n i n g n e w

t h i n g s .

Commented by Rasheed Soomro last updated on 03/Nov/15

TH a n k S f o r m e n t i o n i n g m y c a l c u l a t i o n m i s t a k e .

I a m g o i n g t o c o r r e c t a c c o r d i n g t o y o u r s u g g e s t i o n .

T h e i n t e g r a l \int c o s h x c o t h x d x w a s d e r i v e d m i s t a k e n l y

(T h e m i s t a k e m e n t i o n e d b y y o u) . I t h i n k t h a t i t w a s

n o t c o r r e c t .

Commented by Yozzi last updated on 02/Nov/15

I f y o u g o f u r t h e r o n s i m p l i f y i n g

t h e f i r s t a n s w e r y o u g e t

y = l n ∣ \frac{s i n h x}{t a n h \frac{x}{2}} ∣ + C

y = l n ∣ 2 c o s h \frac{x}{2} s i n h \frac{x}{2} \times \frac{c o s h (x / 2)}{s i n h (x / 2)} ∣ + C

y = l n ∣ 2 {c o s h}^{2} \frac{x}{2} ∣ + C

y = 0 a t x = 0

∴ C = - l n 2

∴ y = l n ∣ 2 {c o s h}^{2} 0.5 x ∣ - l n 2

y = l n ∣ \frac{2}{2} {c o s h}^{2} 0.5 x ∣

y = l n ∣ {c o s h}^{2} 0.5 x ∣

y = 2 l n ∣ c o s h 0.5 x ∣

c o s h x > 0 \forall x \in R

∴ y = 2 l n c o s h 0.5 x .

T h e s e c o n d s o l u t i o n t u r n s o u t n o t

t o b e a s o l u t i o n .

C h e c k i n g y = 2 l n c o s h 0.5 x .

y^{^{'}} = 2 \times \frac{0.5 s i n h 0.5 x}{c o s h 0.5 x} = t a n h 0.5 x

Missing \left or extra \right

{s i n h x t a n h}^{2} 0.5 x + 2 t a n h 0.5 x - s i n h x

= s i n h x (- {s e c h}^{2} 0.5 x) + 2 t a n h 0.5 x

= \frac{2 s i n h 0.5 x c o s h 0.5 x}{- {c o s h}^{2} 0.5 x} + 2 t a n h 0.5 x

= - 2 t a n h 0.5 x + 2 t a n h 0.5 x

= 0 a s r e q u i r e d .

T h e o t h e r s o l u t i o n i s i n v a l i d h o w e v e r .

s i n c e l n 0 i s u n d e f i n e d .

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