Question

(D² + 3DD '+ 2D'²) z = x+y

Answer 1

SOLUTION

Let us use some known facts. Let the given differential equation be

$$F(D,D')=f(x,y)$$

Factorize $F(D,D')$ into linear factors. Then use the following results:

 Corresponding to each non-repeated factor $\left(bD-aD'-c\right)$ , the part of C.F. is taken as

$$\exp\left(\frac{cx}{b}\right)\cdot\varphi(by+ax),\quad if\quad b\neq0$$

In our case,

$$\left(D^2+3DD'+2D'^2\right)z=x+y\longrightarrow z(x,y)=C.F.+P.I.$$

0 STEP: We factor the expression

$$\left(D^2+3DD'+2D'^2\right)=\left(D^2+DD'\right)+\left(2DD'+2D'^2\right)=$$

$$=D\left(D+D'\right)+2D'\left(D+D'\right)=\left(D+D'\right)\left(D+2D'\right)$$

Conclusion,

$$\boxed{\left(D^2+3DD'+2D'^2\right)z=\left(D+D'\right)\left(D+2D'\right)z}$$

1 STEP: Let find C.F.

$$\left\{\begin{array}{c} \left(D+D'\right)z\\ \left(bD-aD'-c\right)z \end{array}\right.\rightarrow \left\{\begin{array}{c} a=-1\\ b=1\\ c=0 \end{array}\right.\rightarrow$$

$$(C.F.)_1=\exp\left(\frac{0\cdot x}{1}\right)\cdot\varphi(1\cdot y+(-1)x)$$

$$\boxed{(C.F.)_1=\varphi_1(y-x),\quad where\,\,\,\varphi_1\,\,\,is \,\,\,arbitrary\,\,\,function}$$

$$\left\{\begin{array}{c} \left(D+2D'\right)z\\ \left(bD-aD'-c\right)z \end{array}\right.\rightarrow \left\{\begin{array}{c} a=-2\\ b=1\\ c=0 \end{array}\right.\rightarrow$$

$$(C.F.)_2=\exp\left(\frac{0\cdot x}{1}\right)\cdot\varphi(1\cdot y+(-2)x)$$

$$\boxed{(C.F.)_2=\varphi_2(y-2x),\quad where\,\,\,\varphi_2\,\,\,is \,\,\,arbitrary\,\,\,function}$$

Then,

$$C.F.=(C.F.)_1+(C.F.)_2=\varphi_1(y-x)+\varphi_2(y-2x)$$

2 STEP: Let find P.I.

$$P.I.=\frac{1}{(D+D')(D+2D')}(x+y)=\frac{1}{D\left(1+\frac{D'}{D}\right)D\left(1+\frac{2D'}{D}\right)}(x+y)=$$

$$=\frac{1}{D^2}\left(1+\frac{D'}{D}\right)^{-1}\left(1+\frac{2D'}{D}\right)^{-1}(x+y)=$$

$$=\frac{1}{D^2}\left[1-\frac{D'}{D}+\ldots\right]\left[1-\frac{2D'}{D}+\ldots\right](x+y)=$$

$$=\left[\frac{1}{1+x}=1-x+x^2-x^3+\ldots,|x|<1\right]=$$

$$=\frac{1}{D^2}\left[1-\frac{D'}{D}-\frac{2D'}{D}+\frac{2D'^2}{D^2}+\ldots\right](x+y)=$$

$$=\frac{1}{D^2}\left[1-\frac{3D'}{D}+\frac{2D'^2}{D^2}+\ldots\right](x+y)=$$

$$=\frac{1}{D^2}\left[(x+y)-\frac{3}{D}\left(\frac{\partial}{\partial y}(x+y)\right)+\frac{2}{D^2}\left(\frac{\partial^2}{\partial^2 y}(x+y)\right)+\ldots\right]=$$

$$=\frac{1}{D^2}\left[(x+y)-\frac{3}{D}\left(1\right)+\frac{2}{D^2}\left(0\right)+\ldots\right]=$$

$$=\frac{1}{D^2}\left[(x+y)-\frac{3}{D}\left(1\right)\right]=\left(\frac{1}{D}\equiv\int dx\right)=$$

$$=\frac{1}{D^2}\left[(x+y)-3\int1dx\right]=\frac{1}{D^2}\left[(x+y)-3x\right]=$$

$$=\frac{1}{D}\left[\int(y-2x)dx\right]=\int\left(xy-x^2\right)dx=\frac{x^2y}{2}-\frac{x^3}{3}$$

$$\boxed{P.I.=\frac{x^2y}{2}-\frac{x^3}{3}}$$

Conclusion,

$$z(x,y)=C.F.+P.I.=\varphi_1(y-x)+\varphi_2(y-2x)+\frac{x^2y}{2}-\frac{x^3}{3}$$

ANSWER

$$\left\{\begin{array}{c} z(x,y)=\varphi_1(y-x)+\varphi_2(y-2x)+\displaystyle\frac{x^2y}{2}-\frac{x^3}{3}\\[0.4cm] where\,\,\,\varphi_1\,\,\,and\,\,\,\varphi_2\,\,\,are\,\,\,arbitrary\,\,\,functions \end{array}\right.$$