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Question Number 16514 by Sai dadon. last updated on 23/Jun/17

H e l p p l z

G i v e n 2 d^{2} x / {d t}^{2} + 2 d x / d t + x = k s i n t

w h e r e k i s c o n s t a n t s h o w t h a t i f x = n

a n d d x / d t = 0 w h e n t = 0 t h e n

x = e^{- 1 / 2 t} {(n + 2 / 5 k) c o s (t / 2) +) + (n + 4 / 5 k) s i n (t / 2)}

- 1 / 5 k s i n t - 2 / 5 k c o s t .

Commented by Sai dadon. last updated on 23/Jun/17

I n e e d h e l p g u y s .

Answered by ajfour last updated on 23/Jun/17

c h a r a c t e r i s t i c e q u a t i o n i s :

2 λ^{2} + 2 λ + 1 = 0

λ = \frac{- 2 \pm \sqrt{4 - 8}}{4}

\Rightarrow λ_{1} = - \frac{1}{2} + \frac{i}{2}, λ_{2} = - \frac{1}{2} - \frac{i}{2}

x_{h} = e^{- t / 2} [A \cos (t / 2) + B \sin (t / 2)]

l e t x_{p} = K \cos t + M \sin t

x_{p}^{^{'}} = - K \sin t + M \cos t

x_{p}^{^{″}} = - K \cos t - M \sin t = - x_{p}

s u b s t i t u t i n g x = x_{p} i n

2 x^{^{″}} + 2 x^{'} + x = k \sin t

- 2 x_{p} + 2 x_{p}^{^{'}} + x_{p} = k \sin t

2 x_{p}^{^{'}} - x_{p} = k \sin t

2 (- K \sin t + M \cos t)

- M \sin t - K \cos t = k \sin t

\Rightarrow 2 K + M = - k a n d

2 M - K = 0 o r K = 2 M

s o, 5 M = - k

M = - \frac{k}{5} a n d K = - \frac{2}{5}

x_{p} = - \frac{2 k}{5} \cos t - \frac{k}{5} \sin t

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

x = e^{- t / 2} [A \cos (t / 2) + B \sin (t / 2)]

- \frac{2 k}{5} \cos t - \frac{k}{5} \sin t \dots (i)

w i t h x = n f o r t = 0 i n (i) g i v e s

n = A - \frac{2 k}{5} \Rightarrow A = n + \frac{2 k}{5}

\frac{d x}{d t} = - \frac{1}{2} e^{- t / 2} [A \cos (t / 2) + B \sin (t / 2)]

+ e^{- t / 2} [- \frac{A}{2} \sin (t / 2) + \frac{B}{2} \cos (t / 2)]

+ \frac{2 k}{5} \sin t - \frac{k}{5} \cos t

a t t = 0 w e h a v e \frac{d x}{d t} = 0, s o

0 = - \frac{A}{2} + \frac{B}{2} - \frac{k}{5}

B = \frac{2 k}{5} + A = \frac{2 k}{5} + n + \frac{2 k}{5}

B = n + \frac{4 k}{5}, w e t h e n h a v e

x = e^{- t / 2} {(n + \frac{2 k}{5}) \cos \frac{t}{2} + (n + \frac{4 k}{5}) \sin \frac{t}{2}}

- \frac{k}{5} \sin t - \frac{2 k}{5} \cos t .

Commented by Sai dadon. last updated on 23/Jun/17

T h a n k y o u a j .

G o d b l e s s y o u .