Question Number 103769 by bemath last updated on 17/Jul/20

2y′′−y′+y = cos 3x

Answered by mathmax by abdo last updated on 17/Jul/20

let solve by laplace transform  e ⇒2L(y^(′′) )−L(y^′ )+L(y) =L(cos(3x)) ⇒  2{x^2 L(y)−xy(o)−y^′ (o)}−(xL(y)−y(o)) +L(y) =L(cos(3x)) ⇒  (2x^2 −x+1)L(y)−(2x−1)y(o)−2y^′ (o) =L(cos(3x)) ⇒  (2x^2 −x+1)L(y) =(2x−1)y(0)+2y^′ (o)+L(cos(3x))  L(cos(3x))=∫_0 ^∞  cos(3t)e^(−xt) dt =Re(∫_0 ^∞ e^((−x+3i)t) dt) and  ∫_0 ^∞  e^((−x+3i)t)  dt =[(1/(−x+3i))e^((−x+3i)t) ]_0 ^∞  =−(1/(−x+3i)) =(1/(x−3i)) =((x+3i)/(x^2  +9)) ⇒  L(cos(3x)) =(x/(x^2  +9))  e ⇒L(y) =((2x−1)/((2x^2 −x+1)))y(o)+(2/(2x^2 −x+1))y^′ (o) +(x/((x^2 +9)(2x^2 −x+1))) ⇒  y(x) =y(o)L^(−1) (((2x−1)/(2x^2 −x+1)))+2y^′ (0)L^(−1) ((1/(2x^2 −x+1)))+L^(−1) ((x/((x^2  +9)(2x^2 −x+1))))  f(x) =((2x−1)/(2x^2 −x+1))  Δ =1−8 =−7 ⇒x_1 =((1+i(√7))/4)  and x_2 =((1−i(√7))/2)  f(x) =((2x−1)/(2(x−x_1 )(x−x_2 ))) =(a/(x−x_1 )) +(b/(x−x_2 ))  eazy to find a and b ⇒  L^(−1) (f) =a e^(x_1 x)  +b e^(x_2 x)  →e^(x/2) { a^′  cos(((√7)/2)x) +b^′ sin(((√7)/2)x)}  L^(−1) ((1/(2x^2 −x+1))) =e^(x/2) (α cos(((√7)/2)x) +βsin(((√7)/2)x)}  g(x) =(x/((x^2  +9)(2x^2 −x+1))) =(a/(x−3i)) +(b/(x+3i)) +(c/(x−x_1 )) +(d/(x−x_2 )) ⇒  L^(−1) (g) =a e^(3ix)  +b e^(−3ix)  + e^(x/2) { α cos(((√7)/2)x) +βsis(((√7)/2)x)}  =a^′  cos(3x) +b^′  sin(3x) +e^(x/2) {α cos(((√7)/2)x) +β sin(((√7)/2)x)}...