solve-the-following-initial-values-DEs-20y-4y-y-0-y-0-3-2-and-y-0-0-

Question Number 98255 by john santu last updated on 12/Jun/20

solve the following initial values

DEs {20 y}^{″} + {4 y}^{'} + y = 0

; y (0) = 3.2 and y^{'} (0) = 0

Commented by bemath last updated on 13/Jun/20

auxilary equation euler - cauchy

{20 m}^{2} + 4 m + 1 = 0

m = - \frac{1}{10} \pm \frac{i}{5} ∴ y = e^{- \frac{x}{10}} [A \cos \frac{x}{5} + B \sin \frac{x}{5}]

\Rightarrow y (0) = 3.2 \Rightarrow A = 3.2

y^{'} (x) = e^{- \frac{x}{10}} (- \frac{3.2}{5} \sin \frac{x}{5} + \frac{B}{5} \sin \frac{x}{5})

y^{'} (0) = 0 \Rightarrow B = \frac{3.2}{2} = 1.6

hence y (x) = e^{- \frac{x}{10}} (3.2 \cos \frac{x}{5} + 1.6 \sin \frac{x}{5})

Answered by Mr.D.N. last updated on 12/Jun/20

{20 y}^{^{″}} + {4 y}^{^{'}} + y = 0

({20 D}^{2} + 4 D + 1) y = 0

A . E . {20 m}^{2} + 4 m + 1 = 0

m = \frac{- 4 \bar{+} \sqrt{16 - 4.20.1}}{2.20}

= \frac{- 4 \bar{+} \sqrt{16 - 80}}{40} = \frac{- 4 \bar{+} \sqrt{- 8}}{40}

= \frac{- 4 \bar{+} 2 \sqrt{2} i}{40} = \frac{- 2 \bar{+} \sqrt{2} i}{20} = \frac{- 1}{10} \bar{+} \frac{\sqrt{2}}{20} i

CF = e^{- \frac{x}{10}} (C_{1} \cos \frac{\sqrt{2}}{20} x + C_{2} \sin \frac{\sqrt{2}}{20} x)

PI = 0

y = CF + PI

y = C_{1} e^{- \frac{x}{10}} \cos \frac{\sqrt{2}}{20} x + C_{2} e^{- \frac{x}{10}} \sin \frac{\sqrt{2}}{20} x

Given : y (0) = 3.2 and y^{^{'}} (0) = 0

y (0) = e^{- 0} {C_{1} \cos (0) + C_{2} \sin (0)}

3.2 = 1 (C_{1} . 1 + 0)

C_{1} = 3.2

y^{^{'}} = C_{1} {e^{- \frac{x}{10}} . (- \frac{\sqrt{2}}{20}) \sin \frac{\sqrt{2}}{20} x + \cos \frac{\sqrt{2}}{20} x . (- \frac{1}{10}) e^{- \frac{x}{10}}} + C_{2} {e^{- \frac{x}{10}} \frac{\sqrt{2}}{20} \cos \frac{\sqrt{2}}{20} x + \sin \frac{\sqrt{2}}{20} x . (- \frac{1}{10}) e^{- \frac{x}{10}}}

y^{^{'}} (0) = C_{1} (0 - \frac{1}{10}) + C_{2} (\frac{\sqrt{2}}{20} - 0)

0 = - C_{1} \frac{1}{10} + C_{2} \frac{\sqrt{2}}{20}

\frac{3.2}{10} = C_{2} \frac{\sqrt{2}}{20}

C_{2} \frac{\sqrt{2}}{2} = 3.2

C_{2} = \frac{6.4}{\sqrt{2}} = 4.52

∴ C_{1} = 3.2 and C_{2} = 4.52

y = e^{\frac{- x}{10}} (3.2 \cos \frac{\sqrt{2}}{20} x + 4.52 \sin \frac{\sqrt{2}}{20} x) / / .

initial value around f {(y)}_{x \to 0} = 3.2

Commented by bemath last updated on 13/Jun/20

wrong in \sqrt{16 - 80} = \sqrt{- 8}

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