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Question Number 111105 by bobhans last updated on 02/Sep/20

y′′+2y′+y=e^(−2x) +2x+3

y+2y+y=e2x+2x+3

Commented by mohammad17 last updated on 02/Sep/20

r^2 +2r+1=0⇒(r+1)^2 =0⇒r_1 =r_2 =−1  y_h =c_1 e^(−x) +c_2 xe^(−x)     let:y_(p1) =ae^(−2x) ⇒y_(p1) ^′ =−2ae^(−2x) ⇒y_(p1) ^(′′) =4ae^(−2x)     (4ae^(−2x) −4ae^(−2x) +ae^(−2x) )=e^(−2x) ⇒a=1    ⇒y_(p1) =e^(−2x)     let:y_(p2) =(1/((1+D)^2 ))(2x+3)    y_(p2) ={1−2D+3D^2 +....}(2x−3)    y_(p2) =2x−3−4+0+...    y_(p2) =2x−7    y=y_h +y_(p1) +y_(p2) =c_1 e^(−x) +c_2 xe^(−x) +e^(−2x) +2x−7    by ⟨mohammad⟩“

r2+2r+1=0(r+1)2=0r1=r2=1yh=c1ex+c2xexlet:yp1=ae2xyp1=2ae2xyp1=4ae2x(4ae2x4ae2x+ae2x)=e2xa=1yp1=e2xlet:yp2=1(1+D)2(2x+3)yp2={12D+3D2+....}(2x3)yp2=2x34+0+...yp2=2x7y=yh+yp1+yp2=c1ex+c2xex+e2x+2x7bymohammad

Answered by mathmax by abdo last updated on 03/Sep/20

y^((2))  +2y^((1))  +y =e^(−2x)  +2x +3  h→r^2  +2r +1 =0 ⇒(r+1)^2  =0 ⇒r =−1 ⇒y_h =(ax+b)e^(−x)   =axe^(−x)  +be^(−x)  =au_1  +bu_2   W(u_1 ,u_2 ) = determinant (((xe^(−x)           e^(−x) )),(((1−x)e^(−x)     −e^(−x) )))=−xe^(−2x) +(x−1)e^(−2x)  =−e^(−2x)  ≠0  W_1 = determinant (((0                               e^(−x) )),((e^(−2x)  +2x+3      −e^(−x) )))=−e^(−x) (e^(−2x)  +2x+3)  W_2 = determinant (((xe^(−x)                 0)),(((1−x)e^(−x)      e^(−2x)  +2x+3)))=xe^(−x) (e^(−2x)  +2x+3)  V_1 =∫ (W_1 /W)dx =−∫   ((e^(−x) (e^(−2x)  +2x+3))/(−e^(−2x) )) dx=∫ e^x (e^(−2x)  +2x+3)dx  =∫ (e^(−x)  +2x e^x  +3e^x )dx =−e^(−x)  +3e^x  +2 ∫ xe^x  dx  =−e^(−x)  +3e^x  +2{ xe^x −e^x } =−e^(−x)  +e^x {1 +2x}  V_2 =∫ (W_2 /W)dx =∫  ((xe^(−x) (e^(−2x)  +2x+3))/(−e^(−2x) ))dx  =−∫  xe^x (e^(−2x)  +2x +3)dx =−∫  x( e^(−x)  +2xe^x  +3e^x )dx  =−∫ x e^(−x )  dx −∫  (2x^2  +3x) e^x  dx  =xe^(−x) −∫ e^(−x)   −{  (2x^2  +3x)e^x −∫  (4x+3)e^x }  =(x−1)e^(−x)  −(2x^2  +3x)e^x  +{  (4x+3)e^x −∫ 4e^x dx}  =(x−1)e^(−x)  −(2x^2  +3x)e^x +(4x+3)e^x −4e^x   =(x−1)e^(−x)  −(2x^2  −x +1)e^x  ⇒  y_p =u_1 v_1 +u_2 v_2 =xe^(−x) {−e^(−x)  +(2x+1)e^x }  +e^(−x) { (x−1)e^(−x) +(−2x^2  +x−1)e^x }  =−xe^(−2x) +2x^2  +x  +(x−1)e^(−2x)  −2x^2  +x−1  =−e^(−2x)   +2x−1  the general solution is  y =y_h  +y_p =axe^(−x)  +be^(−x)   −e^(−2x)  +2x−1

y(2)+2y(1)+y=e2x+2x+3hr2+2r+1=0(r+1)2=0r=1yh=(ax+b)ex=axex+bex=au1+bu2W(u1,u2)=|xexex(1x)exex|=xe2x+(x1)e2x=e2x0W1=|0exe2x+2x+3ex|=ex(e2x+2x+3)W2=|xex0(1x)exe2x+2x+3|=xex(e2x+2x+3)V1=W1Wdx=ex(e2x+2x+3)e2xdx=ex(e2x+2x+3)dx=(ex+2xex+3ex)dx=ex+3ex+2xexdx=ex+3ex+2{xexex}=ex+ex{1+2x}V2=W2Wdx=xex(e2x+2x+3)e2xdx=xex(e2x+2x+3)dx=x(ex+2xex+3ex)dx=xexdx(2x2+3x)exdx=xexex{(2x2+3x)ex(4x+3)ex}=(x1)ex(2x2+3x)ex+{(4x+3)ex4exdx}=(x1)ex(2x2+3x)ex+(4x+3)ex4ex=(x1)ex(2x2x+1)exyp=u1v1+u2v2=xex{ex+(2x+1)ex}+ex{(x1)ex+(2x2+x1)ex}=xe2x+2x2+x+(x1)e2x2x2+x1=e2x+2x1thegeneralsolutionisy=yh+yp=axex+bexe2x+2x1

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