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Question Number 76842 by jagoll last updated on 31/Dec/19

w h a t i s s o l u t i o n y^{^{″}} + y = 0 .

Commented by mathmax by abdo last updated on 31/Dec/19

t h e c a r a c e r i s t c e q u a t i o n i s r^{2} + 1 = 0 \Rightarrow r = i o r r = - i \Rightarrow

y (x) = α e^{i x} + β e^{- i x} \Rightarrow y (x) = a c o s x + b s i n x

Answered by mr W last updated on 31/Dec/19

l e t y^{'} = \frac{d y}{d x} = u

y^{″} = \frac{d (y^{'})}{d x} = \frac{d u}{d x} = \frac{d u}{d y} \times \frac{d y}{d x} = \frac{d u}{d y} \times u

\Rightarrow \frac{d u}{d y} \times u + y = 0

\Rightarrow u d u = - y d y

\Rightarrow \int u d u = - \int y d y

\Rightarrow \frac{u^{2}}{2} = - \frac{y^{2}}{2} + \frac{c_{1}^{2}}{2}

\Rightarrow u = \pm \sqrt{c_{1}^{2} - y^{2}}

\Rightarrow \frac{d y}{d x} = \pm \sqrt{c_{1}^{2} - y^{2}}

\Rightarrow \frac{d y}{\sqrt{c_{1}^{2} - y^{2}}} = \pm d x

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

\Rightarrow \sin^{- 1} \frac{y}{c_{1}} = \pm x + c_{2}

\Rightarrow y = c_{1} \sin (\pm x + c_{2})

\Rightarrow y = c_{1} \sin (x + c_{2})

Commented by jagoll last updated on 31/Dec/19

t h a n k s y o u s i r

Commented by jagoll last updated on 31/Dec/19

s i r w h y n o t c_{1} a s c o n s t a n f i r s t i n t e g r a t e ?

b u t \frac{1}{2} c_{1}^{2}

Commented by mr W last updated on 31/Dec/19

i t m a k e s n o r e a l d i f f e r e n c e, b u t

l o o k s b e t t e r i n l a t e r o p e r a t i o n s .

Commented by jagoll last updated on 31/Dec/19

o o o k s i r t h a n k s y o u

Answered by Rio Michael last updated on 18/Jan/20

i c o u l d a l s o s o l v e t h i s b y

\frac{d^{2} y}{{d x}^{2}} + y = 0

t h e a u x i l i a r y e q u a t i o n i s

r^{2} + 1 = 0 \Rightarrow r = i o r r = - i

o r r = 0 + i a n d r = 0 - i

t h e g e n e r a l s o l u t i o n

i s o f t h e f i r m y = e^{p x} (A c o s q x + B s i n q x)

w h e r e p = 0 a n d q = 1

y = A c o s x + B s i n x