Question #348972

(D^3+2D^2D'-DD'^2-2D'^3)z=(y+2)e^x

Expert's answer

"\\left(D^3+2D^2D^\\prime-DD^{\\prime2}-2D^{\\prime3}\\right)z=\\left(y+2\\right)e^x"

Find the complementary function "z=f_1\\left(y+m_1x\\right)+f_2\\left(y+m_2x\\right)+\\ldots+f_n\\left(y+m_nx\\right)"

"\\left(D^3+2D^2D^\\prime-DD^{\\prime2}-2D^{\\prime3}\\right)z=0" ,

The auxiliary equation is

"m^3+2m^2-m-2=0" ,

"m^2\\left(m+2\\right)-\\left(m+2\\right)=\\left(m^2-1\\right)\\left(m+2\\right)=\\left(m-1\\right)\\left(m+1\\right)\\left(m+2\\right)=0" ,

"m_1=-2,\\ m_2=-1,\\ m_3=1" ,

"C.I.=f_1\\left(y-2x\\right)+f_2\\left(y-x\\right)+f_3\\left(y+x\\right)" .

Then the particular integral (P.I.) is given by case "G\\left(x,y\\right)=e^{ax+by}x^ry^s" :

"P.I.=\\frac{1}{\\left(D^3+2D^2D^\\prime-DD^{\\prime2}-2D^{\\prime3}\\right)}\\left(y+2\\right)e^x="

"\\left(a=1,b=0\\right)=\\frac{e^x}{\\left(\\left(D+1\\right)^3+2\\left(D+1\\right)^2D^\\prime-\\left(D+1\\right)D^{\\prime2}-2D^{\\prime3}\\right)}\\left(y+2\\right)="

"=\\frac{e^x}{\\left({1+D}^3+3D^2+3D+2D^2D^\\prime+4DD^\\prime+2D^\\prime-DD^{\\prime2}-D^{\\prime2}-2D^{\\prime3}\\right)}\\left(y+2\\right)="

"{=e}^x\\left(1+D^3+3D^2+3D+2D^2D^\\prime+4DD^\\prime+2D^\\prime-DD^{\\prime2}-D^{\\prime2}-2D^{\\prime3}\\right)^{-1}\\left(y+2\\right)="

"{=e}^x\\left(1-D^3-3D^2-3D-2D^2D^\\prime-4DD^\\prime-2D^\\prime+DD^{\\prime2}+D^{\\prime2}+2D^{\\prime3}\\right)\\left(y+2\\right)="

"{=e}^x\\left(1-2D^\\prime\\right)\\left(y+2\\right)=e^x\\left(y+2-2\\right)=ye^x" .

Answer:

"z=f_1\\left(y-2x\\right)+f_2\\left(y-x\\right)+f_3\\left(y+x\\right)+ye^x" .

NB: "D=\\frac{\\partial}{\\partial x},\\ D^\\prime=\\frac{\\partial}{\\partial y},\\ \\frac{1}{1+x}=1-x+x^2-x^3+..."

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