Answer to Question #131632 in Differential Equations for Neelam burman

Question

(D^2-2DD'+D^2)z=12xy

Answer 1

$$(D^2-2DD'+D'^2)z=12xy.$$

The auxiliary equation of the given equation is

$$m^2-2m+1=0$$

$$(m-1)^2=0$$

$$m_{1,2}=1$$

Then the complementary function of the given equation is

$$u=\varphi_1(y+1,x)+x\varphi_2(y+1,x)=\varphi_1(y,x)+x\varphi_2(y,x)$$

When $f(x, y)=V,$ where $V$ is function of $x$ and $y,$ then partial integral:

$$p=\dfrac{1}{F(D,D')}V$$

Then partial integral

$$p=\dfrac{1}{D^2-2DD'+D'^2 }(12xy)=$$

$$=\dfrac{1}{(D-D')^2 }(12xy)=\dfrac{12}{D^2 }(1-\dfrac{D'}{D})^{-2}(xy)=$$

$$=\dfrac{12}{D^2 }(1+\dfrac{2D'}{D}+\dfrac{3D'^2}{D^2}+...)(xy)=$$

$$=\dfrac{12}{D^2 }(xy+\dfrac{2}{D}(x)+0+...)=$$

$$=\dfrac{12}{D^2 }(xy)+\dfrac{24}{D^3 }(x)=12y(\dfrac{x^3}{6 })+24(\dfrac{x^4}{24 })=$$

$$=2x^3y+x^4$$

Then the general solution of the given equation is $z=u+p,$ such that

$$z=\varphi_1(y,x)+x\varphi_2(y,x)+x^4+2x^3y,$$

where $\varphi_1, \varphi_2$ are arbitrary functions.