Question Number 100766 by john santu last updated on 28/Jun/20

Answered by Rio Michael last updated on 28/Jun/20

AE: m^2  +1 = 0 ⇒ m =±i  y_c  = (Acos x +B sin x)  let y_p  = λe^(2x)  ⇒ y′ = 2λe^(2x)  and y′′ = 4λe^(2x)   ⇒  4λe^(2x)  + λe^(2x )  = e^(2x) [      ⇒  5 λ = 1 ⇔ λ = (1/5)  ⇒ y_p  = (e^(2x) /5)  y_g  = y_c  + y_p  ⇒ y_g  = Acos x + Bsin x + (e^(2x) /5)

Answered by mathmax by abdo last updated on 28/Jun/20

y^(′′)  +y =e^(2x)      (he)→y^(′′)  +y =0 →r^2  +1=0⇒r =+^− i ⇒y_h  =acosx +bsinx  =au_1  +bu_2   W(u_1  ,u_2 ) = determinant (((cosx         sinx)),((−sinx       cosx)))=cos^2 x +sin^2 x =1  W_1 = determinant (((0                  sinx)),((e^(2x)                cosx)))=−e^(2x ) sinx  W_2 = determinant (((cosx       0)),((−sinx   e^(2x) )))=e^(2x)  cosx  v_1 =∫ (w_1 /w)dx =−∫  e^(2x)  cosx dx =−Re(∫ e^(2x +ix) dx) and  ∫ e^((2+i)x) ex =(1/(2+i)) e^((2+i)x)  =(e^(2x) /(2+i))(cosx +isinx)   =(1/5) e^(2x) (2−i)(cosx +isinx) =(e^(2x) /5){2cosx+2isinx −icosx+sinx} ⇒  v_1 =−(e^(2x) /5){2cosx+sinx}  v_2 =∫ (w_2 /w)dx =∫e^(2x)  cosx dx =(e^(2x) /5){2cosx +sinx} ⇒y_p =u_1 v_1  +u_2 v_2   =−(e^(2x) /5){2cos^2 x +cosx sinx} +(e^(2x) /5){ 2sinx cosx +sin^2 x}  =(e^(2x) /5){2sinx cosx+sin^2 x−2cos^2 x−cosx sinx}  =(e^(2x) /5){sinx cosx +1−3cos^2 x} =(e^(2x) /5){(1/2)sin(2x)+1−3×((1+cos(2x))/2)}  =(e^(2x) /5){(1/2)sin(2x)−(1/2)−(3/2)cos(2x)} =(e^(2x) /(10)){ sin(2x)−3cos(2x)−1}  y =y_h  +y_p =acosx +bsinx +(e^(2x) /(10)){ sin(2x)−3cos(2x)−1}

Answered by mathmax by abdo last updated on 28/Jun/20

Laplace method y^(′′)  +y =e^(2x)  ⇒L(y^(′′) )+L(y) =L(e^(2x) ) ⇒  x^2  L(y)−xy(0)−y^′ (0) +L(y) =L(e^(2x) ) ⇒(x^2  +1)L(y) =xy(0)+y^′ (0)+L(e^(2x) )  we have L(e^(2x) ) =∫_0 ^∞  e^(2t)  e^(−xt)  dt =∫_0 ^∞  e^((2−x)t)  dt =[(1/(2−x)) e^((2−x)t) ]_0 ^(+∞)   =−(1/(2−x)) =(1/(x−2)) ⇒(x^2  +1)L(y) =(1/(x−2)) +xy(o) +y^′ (0) ⇒  L(y) =(1/((x−2)(x^2  +1))) +(x/(x^2  +1))y(o) +((y^′ (0))/(x^2  +1)) ⇒  y(x) =L^(−1) ((1/((x−2)(x^2  +1))))+y(o)L^(−1) ((x/(x^2  +1)))+y^′ (o)L^(−1) ((1/(x^2 +1)))  we have L^(−1) ((x/(x^2  +1))) =cosx and L^(−1) ((1/(x^2  +1))) =sinx  f(x) =(1/((x−2)(x^2  +1))) =(1/((x−2)(x−i)(x+i))) =(a/(x−2)) +(b/(x−i)) +(c/(x+i))  L^(−1) (f(x)) =ae^(2x)  +b e^(ix)  +ce^(−ix)  →ae^(2x)  +αcosx +βsinx ⇒  a =(1/5) ⇒y(x) =(1/5)e^(2x)  +A cosx +B sinx

Answered by john santu last updated on 28/Jun/20