Question Number 114225 by mohammad17 last updated on 17/Sep/20

y^(′′) −3y^′ +2y=xsin(x)

Answered by mathmax by abdo last updated on 18/Sep/20

h→r^2 −3r+2=0→Δ =9−8=1 ⇒r_1 =((3+1)/2)=2 and r_2 =((3−1)/2)=1 ⇒  y_h =ae^(2x)  +be^x  =au_1  +bu_2   W(u_1 ,u_2 ) = determinant (((e^(2x)          e^x )),((2e^(2x)        e^x )))=e^(3x) −2e^(3x)  =−e^(3x)  ≠0  W_1 = determinant (((0          e^x )),((xsinx   e^x )))=−xe^x  sinx  W_2 = determinant (((e^(2x)           0)),((2e^(2x)      xsinx)))=xe^(2x)  sinx  V_1 =∫ (W_1 /W)dx =∫  ((−xe^x  sinx)/(−e^(3x) )) dx =∫ x e^(−2x)  sinx dx  =Im(∫  x e^(−2x+ix) dx) but ∫ x e^((−2+i)x)  dx =(x/(−2+i)) e^((−2+i)x)   −∫ (1/(−2+i)) e^((−2+i)x)  dx =−x((2+i)/5) e^((−2+i)x)  +(1/(2−i))×(1/(−2+i)) e^((−2+i)x)   =−(x/5)(2+i)e^((−2+i)x)  −(((2+i)^2  )/((5)^2 )) e^((−2+i)x)   ={−(x/5)(2+i)−((3+4i)/(25))}e^(−2x) (cosx +isinx)  =((−5x(2+i)−3−4i)/(25)) e^(−2x) (cosx +isinx)  =((−10x−5ix−3−4i)/(25)) e^(−2x) (cosx +isinx)=....  V_2 =∫ (W_2 /W)dx =∫  ((xe^(2x)  sinx)/(−e^(3x) )) dx =−∫ x e^(−x)  sinx dx  =−Im(∫ x e^(−x+ix) dx) but ∫ x e^((−1+i)x) dx  =(x/(−1+i)) e^((−1+i)x)  +∫ (1/(−1+i)) e^((−1+i)x)  dx  =e^(−x) (cosx +isinx){((−x)/(1−i))−(1/(1−i))}   =e^(−x) (cosx +sinx){((−x(1+i))/2)−((1+i)/2)} =....  ⇒y_p =u_1 v_1  +u_2 v_2 =e^(2x)  ∫ xe^(−2x)  sinx dx −e^x  ∫ x e^(−x)  sinx dx  the general solution is y =y_h  +y_p