solve the following initial values DEs 20y′′ + 4y′ +y = 0 ; y(0) = 3.2 and y′(0) = 0


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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
	$20y^(′′) +4y^′ +y =0 (20D^2 +4D+1)y=0 A.E. 20m^2 +4m+1=0 m= ((−4+^− (√(16−4.20.1)))/(2.20)) = ((−4+^− (√(16−80)))/(40))=((−4+^− (√(−8)))/(40)) = ((−4+^− 2(√2)i)/(40))= ((−2+^− (√2) i)/(20))= ((−1)/(10))+^− ((√2)/(20))i CF=e^(−(x/(10))) (C_1 cos ((√2)/(20)) x +C_2 sin ((√2)/(20))x) PI=0 y= CF+PI y=C_1 e^(−(x/(10))) cos ((√2)/(20))x +C_2 e^(−(x/(10))) sin((√2)/(20)) x Given : y(0)=3.2 and y^′ (0)=0 y(0)= e^(−0) {C_1 cos(0)+C_2 sin(0)} 3.2 = 1(C_1 .1+0) C_1 = 3.2 y^′ = C_1 {e^(−(x/(10))) .(−((√2)/(20)))sin((√2)/(20))x+cos((√2)/(20))x.(−(1/(10)))e^(−(x/(10))) }+ C_2 {e^(−(x/(10))) ((√2)/(20))cos((√2)/(20)) x+ sin((√2)/(20))x.(−(1/(10)))e^(−(x/(10))) } y^′ (0)=C_1 (0−(1/(10)))+C_2 (((√2)/(20))−0) 0 = −C_1 (1/(10))+C_2 ((√2)/(20)) ((3.2)/(10))= C_2 ((√2)/(20)) C_(2 ) ((√2)/2)= 3.2 C_2 = ((6.4)/(√2))=4.52 ∴ C_1 =3.2 and C_2 =4.52 y= e^((−x)/(10)) (3.2 cos((√2)/(20))x +4.52 sin ((√2)/(20))x) //. initial value around f(y)_(x→0) =3.2$
	${20 y}^{^{″}} + {4 y}^{^{'}} + y = 0$ $({20 D}^{2} + 4 D + 1) y = 0$ $A . E . {20 m}^{2} + 4 m + 1 = 0$ $m = \frac{- 4 \bar{+} \sqrt{16 - 4.20.1}}{2.20}$ $= \frac{- 4 \bar{+} \sqrt{16 - 80}}{40} = \frac{- 4 \bar{+} \sqrt{- 8}}{40}$ $= \frac{- 4 \bar{+} 2 \sqrt{2} i}{40} = \frac{- 2 \bar{+} \sqrt{2} i}{20} = \frac{- 1}{10} \bar{+} \frac{\sqrt{2}}{20} i$ $CF = e^{- \frac{x}{10}} (C_{1} \cos \frac{\sqrt{2}}{20} x + C_{2} \sin \frac{\sqrt{2}}{20} x)$ $PI = 0$ $y = CF + PI$ $y = C_{1} e^{- \frac{x}{10}} \cos \frac{\sqrt{2}}{20} x + C_{2} e^{- \frac{x}{10}} \sin \frac{\sqrt{2}}{20} x$ $Given : y (0) = 3.2 and y^{^{'}} (0) = 0$ $y (0) = e^{- 0} {C_{1} \cos (0) + C_{2} \sin (0)}$ $3.2 = 1 (C_{1} . 1 + 0)$ $C_{1} = 3.2$ $y^{^{'}} = C_{1} {e^{- \frac{x}{10}} . (- \frac{\sqrt{2}}{20}) \sin \frac{\sqrt{2}}{20} x + \cos \frac{\sqrt{2}}{20} x . (- \frac{1}{10}) e^{- \frac{x}{10}}} + C_{2} {e^{- \frac{x}{10}} \frac{\sqrt{2}}{20} \cos \frac{\sqrt{2}}{20} x + \sin \frac{\sqrt{2}}{20} x . (- \frac{1}{10}) e^{- \frac{x}{10}}}$ $y^{^{'}} (0) = C_{1} (0 - \frac{1}{10}) + C_{2} (\frac{\sqrt{2}}{20} - 0)$ $0 = - C_{1} \frac{1}{10} + C_{2} \frac{\sqrt{2}}{20}$ $\frac{3.2}{10} = C_{2} \frac{\sqrt{2}}{20}$ $C_{2} \frac{\sqrt{2}}{2} = 3.2$ $C_{2} = \frac{6.4}{\sqrt{2}} = 4.52$ $∴ C_{1} = 3.2 and C_{2} = 4.52$ $y = e^{\frac{- x}{10}} (3.2 \cos \frac{\sqrt{2}}{20} x + 4.52 \sin \frac{\sqrt{2}}{20} x) / / .$ $initial value around f {(y)}_{x \to 0} = 3.2$
	Commented by bemath last updated on 13/Jun/20

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