Answer to Question #348972 in Differential Equations for Beauti

Question

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

Answer 1

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

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