Question Number 102527 by bemath last updated on 09/Jul/20

2y′′−y′+y = cos 3x

Answered by Ar Brandon last updated on 09/Jul/20

Let y=acos3x+bsin3x ⇒y′=−3asin3x+3bcos3x , y′′=−9acos3x−9bsin3x ⇒−2×9(acos3x+bsin3x)−3(bcos3x−asin3x)+acos3x+bsin3x ⇒(−18a−3b+a)cos3x+(−18b+3a+b)sin3x=cos3x ⇒ { ((−17a−3b=1)),((3a−17b=0)) :} ⇒−17(((17b)/3))−3b=1⇒b=−(3/(298)) , a=((−17)/(298)) Particular integral is; y_(PI) =((−1)/(298))(17cos3x+3sin3x) auxillary equation; 2m^2 −m+1=0 ⇒ m=((1±(√(−7)))/4) ⇒m=(1/4)±((7i)/4) ⇒Complementary function is; y_(CF) =e^(x/4) (c∙cos((7x)/4)+d∙sin((7x)/4)) ∴ y=e^(x/4) (c∙cos((7x)/4)+d∙sin((7x)/4))−(1/(298))(17cos3x+3sin3x) Well this is what I think. I haven′t yet mastered D.E so I may have erred at some point. I′ll like to know you think.

Commented bybemath last updated on 10/Jul/20

cooll 😎😎😎

Commented byAr Brandon last updated on 10/Jul/20

Alright 😆

Answered by mathmax by abdo last updated on 10/Jul/20

(he) →2y^(′′) −y^′ +y =0 →2r^2 −r +1 =0 Δ =1−8=−7 ⇒r_1 =((1+i(√7))/2) and r_2 =((1−i(√7))/2) ⇒y_h =a e^(r_1 x) +b e^(r_2 x) =e^(x/2) { α cos(((√7)/2)x) +βsin(((√7)/2)x)} =α u_1 +β u_2 W(u_1 ,u_2 ) = determinant (((e^(x/2) cos(((√7)/2)x) e^(x/2) sin(((√7)/2)x))),(((1/2)e^(x/2) cos(((√7)/2)x)−((√7)/2)e^(x/2) sin(((√7)/2)x) (1/2)e^(x/2) sin(((√7)/2)x)+((√7)/2) e^(x/2) cos(((√7)/2)x)))) =(1/2)e^x cos(((√7)/2)x)sin(((√7)/2)x) +((√7)/2) e^x cos^2 (((√7)/2)x)−(1/2) e^x cos(((√7)/2)x)sin(((√7)/2)x) +((√7)/2) e^x sin^2 (((√7)/2)x) =((√7)/2) e^x W_1 = determinant (((o e^(x/2) sin(((√7)/2)x))),((cos(3x) (1/2)e^(x/2) sin(((√7)/2)x)+((√7)/2)e^(x/2) cos(((√7)/2)x)))) =−e^(x/2) cos(3x)sin(((√7)/2)x) W_2 = determinant (((e^(x/2) cos(((√7)/2)x) 0)),((.... cos(3x)))) =e^(x/(2 )) cos(3x)cos(((√7)/2)x) v_1 =∫(w_1 /w)dx =−∫ ((e^(x/2) cos(3x)sin(((√7)/2)x))/(((√7)/2) e^x )) =−(2/(√7)) ∫ e^(−(x/(2 ))) cos(3x)sin(((√7)/2)x)dx v_2 =∫ (w_2 /w) dx =∫ ((e^(x/(2 )) cos(3x)cos(((√7)/2)x))/(((√7)/2)e^x )) dx =(2/(√7)) ∫ e^(−(x/(2 ))) cos(3x)cos(((√7)/2)x)dx v_1 and v_2 are eazy to slve ⇒y_p =u_1 v_(1 ) +u_2 v_2 the general solution is y =y_h +y_p

Answered by mathmax by abdo last updated on 10/Jul/20

let use laplace transform e ⇒2L(y^(′′) )−L(y^′ ) +L(y) =L(cos(3x)) ⇒ 2{x^2 L(y)−x y(0)−y^′ (0))−{xL(y)−y(o)}+L(y) =L(cos(3x)) ⇒ (2x^2 −x+1)L(y) −2xy(o)−2y^′ (o)+y(o) =L(cos(3x)) ⇒ (2x^2 −x+1)L(y) =2xy(0)+2y^′ (o)−y(o) +L(cos(3x)) L(cos(3x)) =∫_0 ^∞ cos(3t) e^(−xt) dt =Re(∫_0 ^∞ e^((−x+3i)t) dt) we have ∫_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 ⇒(2x^2 −x+1)L(y) =(x/(x^2 +9)) +2y(o)x +2y^′ (0)−y(0) ⇒ L(y) =(x/((x^2 +9)(2x^2 −x+1))) +2y(o)(x/(2x^2 −x+1)) +((2y^′ (o)−y(o))/(2x^2 −x+1)) ⇒ y(x) =L^(−1) ((x/((x^2 +9)(2x^2 −x+1))))+2y(o)L^(−1) ((x/(2x^2 −x+1)))+(2y^′ (0)−y(o))L^(−1) ((1/(2x^2 −x+1))) ....be continued...