(d^2 y/dx^2 ) + 2 (dy/dx) +2y = cos 4x? by UC method


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Question Number 103468 by bemath last updated on 15/Jul/20

$\frac{d^{2} y}{{d x}^{2}} + 2 \frac{d y}{d x} + 2 y = \cos 4 x ?$ $b y U C m e t h o d$
	Answered by bramlex last updated on 15/Jul/20
	$Homogenous solution ρ^2 +2ρ+2 = 0 (ρ+1)^2 +1 = 0 ρ = −1± i y_h = e^(−x) (C_1 cos x+C_2 sin x) particular solution y_p = Acos 4x+Bsin 4x y′_p =−4Asin 4x+4Bcos 4x y′′_p = −16Acos 4x−16Bsin 4x comparing coefficient −16Acos 4x−16Bsin 4x−8Asin 4x +8Bcos 4x+2Acos 4x+2Bsin 4x = cos 4x (−14A+8B)cos 4x +(−14B−8A)sin 4x = cos 4x ⇒ { ((−14B−8A=0 ⇒A=−(7/4))),((−14A+8B=1 ⇒B=1−14((7/4)))) :} B = 1−((49)/2)= −((47)/2) general solution y=e^(−x) (C_1 cos x+C_2 sin x)−(7/4)cos 4x−((47)/2) sin 4x .(⊕)$
	$Homogenous solution$ $ρ^{2} + 2 ρ + 2 = 0$ ${(ρ + 1)}^{2} + 1 = 0$ $ρ = - 1 \pm i$ $y_{h} = e^{- x} (C_{1} \cos x + C_{2} \sin x)$ $particular solution$ $y_{p} = Acos 4 x + Bsin 4 x$ $y_{p}^{'} = - 4 Asin 4 x + 4 Bcos 4 x$ $y_{p}^{″} = - 16 Acos 4 x - 16 Bsin 4 x$ $comparing coefficient$ $- 16 Acos 4 x - 16 Bsin 4 x - 8 Asin$ $4 x + 8 Bcos 4 x + 2 Acos 4 x + 2 Bsin$ $4 x = \cos 4 x$ $(- 14 A + 8 B) \cos 4 x$ $+ (- 14 B - 8 A) \sin 4 x = \cos 4 x$ $\Rightarrow {\begin{cases} - 14 B - 8 A = 0 \Rightarrow A = - \frac{7}{4} \\ - 14 A + 8 B = 1 \Rightarrow B = 1 - 14 (\frac{7}{4}) \end{cases}$ $B = 1 - \frac{49}{2} = - \frac{47}{2}$ $general solution$ $y = e^{- x} (C_{1} \cos x + C_{2} \sin x) - \frac{7}{4} \cos$ $4 x - \frac{47}{2} \sin 4 x . (\oplus)$
	Answered by mathmax by abdo last updated on 16/Jul/20
	$let use laplace transform y^(′′) +2y^′ +2y =cos(4x) ⇒L(y^(′′) )+2L(y^′ )+2L(y) =L(cos(4x)) ⇒ x^2 L(y)−xy(o)−y^′ (0)+2(xL(y)−y(o))+2L(y) =L(cos(4x)) ⇒ (x^2 +2x+2)L(y)−(x+2)y(o)−y^′ (0) =L(cos(4x)) L(cos(4x)) =∫_0 ^∞ cos(4t)e^(−xt) dt =Re(∫_0 ^∞ e^((−x+4i)t) dt) ∫_0 ^∞ e^((−x+4i)t) dt =[(1/(−x+4i))e^((−x+4i)t) ]_0 ^∞ =((−1)/(−x+4i)) =(1/(x−4i)) =((x+4i)/(x^2 +16)) ⇒ L(cos(4x)) =(x/(x^2 +16)) e ⇒(x^2 +2x+2)L(y) =(x+2)y(o)+y^′ (0)+ ⇒L(y) =((x+2)/(x^2 +2x+2))y(o)+((y^′ (0))/(x^2 +x+2)) +(x/((x^2 +16)(x^2 +2x+2))) ⇒ y(x) =y(o)L^(−1) (((x+2)/(x^2 +2x+2))) +y^′ (0)L^(−1) ((1/(x^2 +x+2))) +L^(−1) ((x/((x^2 +16)(x^2 +2x+2)))) let decompose f(x) =((x+2)/(x^2 +2x+2)) Δ^′ =−1 ⇒x_1 =−1+i and x_2 =−1−i ⇒f(x) =((x+2)/((x−x_1 )(x−x_2 ))) =(1/(2i))(x+2){(1/(x−x_1 ))−(1/(x−x_2 ))} =(1/(2i)){((x+2)/(x−x_1 ))−((x+2)/(x−x_2 ))} =(1/(2i)){((x−x_1 +2+x_1 )/(x−x_1 ))−((x−x_2 +2+x_2 )/(x−x_2 ))} =(1/(2i)){((2+x_1 )/(x−x_1 ))−((2+x_2 )/(x−x_2 ))} ⇒L^(−1) (f) =((2+x_1 )/(2i))e^(x_1 x) −((2+x_2 )/(2i))e^(x_2 x) L^(−1) ((1/(x^2 +x+2))) =L^(−1) ((1/(2i))((1/(x−x_1 ))−(1/(x−x_2 )))) =(1/(2i)) e^(x_1 x) −(1/(2i))e^(x_2 x) =(1/(2i)){ e^(−x) e^(ix) −e^(−x) e^(−ix) } =e^(−x) ×((e^(ix) −e^(−ix) )/(2i)) =e^(−x) sinx....be continued...$
	$let use laplace transform$ $y^{^{″}} + {2 y}^{^{'}} + 2 y = \cos (4 x) \Rightarrow L (y^{^{″}}) + 2 L (y^{^{'}}) + 2 L (y) = L (\cos (4 x)) \Rightarrow$ $x^{2} L (y) - xy (o) - y^{^{'}} (0) + 2 (xL (y) - y (o)) + 2 L (y) = L (\cos (4 x)) \Rightarrow$ $(x^{2} + 2 x + 2) L (y) - (x + 2) y (o) - y^{^{'}} (0) = L (\cos (4 x))$ $L (\cos (4 x)) = \int_{0}^{\infty} \cos (4 t) e^{- xt} dt = Re (\int_{0}^{\infty} e^{(- x + 4 i) t} dt)$ $\int_{0}^{\infty} e^{(- x + 4 i) t} dt = {[\frac{1}{- x + 4 i} e^{(- x + 4 i) t}]}_{0}^{\infty} = \frac{- 1}{- x + 4 i} = \frac{1}{x - 4 i} = \frac{x + 4 i}{x^{2} + 16} \Rightarrow$ $L (\cos (4 x)) = \frac{x}{x^{2} + 16}$ $e \Rightarrow (x^{2} + 2 x + 2) L (y) = (x + 2) y (o) + y^{^{'}} (0) +$ $\Rightarrow L (y) = \frac{x + 2}{x^{2} + 2 x + 2} y (o) + \frac{y^{^{'}} (0)}{x^{2} + x + 2} + \frac{x}{(x^{2} + 16) (x^{2} + 2 x + 2)} \Rightarrow$ $y (x) = y (o) L^{- 1} (\frac{x + 2}{x^{2} + 2 x + 2}) + y^{^{'}} (0) L^{- 1} (\frac{1}{x^{2} + x + 2}) + L^{- 1} (\frac{x}{(x^{2} + 16) (x^{2} + 2 x + 2)})$ $let decompose f (x) = \frac{x + 2}{x^{2} + 2 x + 2}$ $Δ^{^{'}} = - 1 \Rightarrow x_{1} = - 1 + i and x_{2} = - 1 - i \Rightarrow f (x) = \frac{x + 2}{(x - x_{1}) (x - x_{2})}$ $= \frac{1}{2 i} (x + 2) {\frac{1}{x - x_{1}} - \frac{1}{x - x_{2}}} = \frac{1}{2 i} {\frac{x + 2}{x - x_{1}} - \frac{x + 2}{x - x_{2}}}$ $= \frac{1}{2 i} {\frac{x - x_{1} + 2 + x_{1}}{x - x_{1}} - \frac{x - x_{2} + 2 + x_{2}}{x - x_{2}}}$ $= \frac{1}{2 i} {\frac{2 + x_{1}}{x - x_{1}} - \frac{2 + x_{2}}{x - x_{2}}} \Rightarrow L^{- 1} (f) = \frac{2 + x_{1}}{2 i} e^{x_{1} x} - \frac{2 + x_{2}}{2 i} e^{x_{2} x}$ $L^{- 1} (\frac{1}{x^{2} + x + 2}) = L^{- 1} (\frac{1}{2 i} (\frac{1}{x - x_{1}} - \frac{1}{x - x_{2}}))$ $= \frac{1}{2 i} e^{x_{1} x} - \frac{1}{2 i} e^{x_{2} x} = \frac{1}{2 i} {e^{- x} e^{ix} - e^{- x} e^{- ix}} = e^{- x} \times \frac{e^{ix} - e^{- ix}}{2 i}$ $= e^{- x} sinx . . . . be continued . . .$
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