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

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

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 .(⊕)

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...