If-d-2-y-dx-2-4y-0-y-0-1-and-y-pi-6-3-2-1-2-How-to-find-y-x-

Question Number 15179 by Joel577 last updated on 08/Jun/17

If \frac{d^{2} y}{{d x}^{2}} + 4 y = 0

y (0) = 1 and y (\frac{π}{6}) = \frac{\sqrt{3}}{2} + \frac{1}{2}

How to find y (x) ?

Answered by sma3l2996 last updated on 08/Jun/17

y^{″} (x) + 4 y = 0

r^{2} + 4 = 0 \Leftrightarrow r_{1} = 2 i, r = - 2 i

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

y (0) = A + B = 1 \Leftrightarrow B = 1 - A

y (\frac{π}{6}) = {A e}^{\frac{π}{3} i} + (1 - A) e^{- \frac{π}{3} i} = A (e^{\frac{π}{3} i} - e^{- \frac{π}{3} i}) + \frac{1}{2} - i \frac{\sqrt{3}}{2}

= 2 i A s i n (\frac{π}{3}) + \frac{1}{2} - i \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} + \frac{1}{2}

2 i A \frac{\sqrt{3}}{2} + \frac{1}{2} - i \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} - \frac{1}{2} = 0

2 i A = i + 1 \Leftrightarrow A = \frac{1}{2} - i \frac{1}{2} = \frac{\sqrt{2}}{2} e^{- i \frac{π}{4}}

B = \frac{1}{2} + i \frac{1}{2} = \frac{\sqrt{2}}{2} e^{i \frac{π}{4}}

y (x) = \frac{\sqrt{2}}{2} (e^{i (2 x - \frac{π}{4})} + e^{- i (2 x - \frac{π}{4})}) = \frac{\sqrt{2}}{2} \times 2 c o s (2 x - \frac{π}{4})

y (x) = \sqrt{2} c o s (2 x - \frac{π}{4})

Commented by Joel577 last updated on 08/Jun/17

t h a n k y o u v e r y m u c h

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