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Question Number 111105 by bobhans last updated on 02/Sep/20
y″+2y′+y=e−2x+2x+3
Commented by mohammad17 last updated on 02/Sep/20
r2+2r+1=0⇒(r+1)2=0⇒r1=r2=−1yh=c1e−x+c2xe−xlet:yp1=ae−2x⇒yp1′=−2ae−2x⇒yp1″=4ae−2x(4ae−2x−4ae−2x+ae−2x)=e−2x⇒a=1⇒yp1=e−2xlet:yp2=1(1+D)2(2x+3)yp2={1−2D+3D2+....}(2x−3)yp2=2x−3−4+0+...yp2=2x−7y=yh+yp1+yp2=c1e−x+c2xe−x+e−2x+2x−7by⟨mohammad⟩‘‘
Answered by mathmax by abdo last updated on 03/Sep/20
y(2)+2y(1)+y=e−2x+2x+3h→r2+2r+1=0⇒(r+1)2=0⇒r=−1⇒yh=(ax+b)e−x=axe−x+be−x=au1+bu2W(u1,u2)=|xe−xe−x(1−x)e−x−e−x|=−xe−2x+(x−1)e−2x=−e−2x≠0W1=|0e−xe−2x+2x+3−e−x|=−e−x(e−2x+2x+3)W2=|xe−x0(1−x)e−xe−2x+2x+3|=xe−x(e−2x+2x+3)V1=∫W1Wdx=−∫e−x(e−2x+2x+3)−e−2xdx=∫ex(e−2x+2x+3)dx=∫(e−x+2xex+3ex)dx=−e−x+3ex+2∫xexdx=−e−x+3ex+2{xex−ex}=−e−x+ex{1+2x}V2=∫W2Wdx=∫xe−x(e−2x+2x+3)−e−2xdx=−∫xex(e−2x+2x+3)dx=−∫x(e−x+2xex+3ex)dx=−∫xe−xdx−∫(2x2+3x)exdx=xe−x−∫e−x−{(2x2+3x)ex−∫(4x+3)ex}=(x−1)e−x−(2x2+3x)ex+{(4x+3)ex−∫4exdx}=(x−1)e−x−(2x2+3x)ex+(4x+3)ex−4ex=(x−1)e−x−(2x2−x+1)ex⇒yp=u1v1+u2v2=xe−x{−e−x+(2x+1)ex}+e−x{(x−1)e−x+(−2x2+x−1)ex}=−xe−2x+2x2+x+(x−1)e−2x−2x2+x−1=−e−2x+2x−1thegeneralsolutionisy=yh+yp=axe−x+be−x−e−2x+2x−1
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