Answer to Question #142181 in Differential Equations for Nikhil Singh

Question

Solve
(D^2+2DD'+D'2+2D+2D'+1)z=x

Answer 1

$$Solution$$

Let the given differential equation be

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

Factorize $F(D,D')=f(x,y)$ into linear factors. Then use the following results.

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

$$e(\frac{cx}{b})\phi(by+ax), if b \ne 0$$

We now have three particular cases of Rule I:

Rule IA. Take $c=0$ in Rule I. Hence corresponding to each linear factor $(bD+aD')$ , the part of C.F. is

Rule IB. Take $a=0$ in Rule I. Hence corresponding to each linear factor $(bD+c)$ , the part of C.F. is

$$e(\frac{cx}{b})\phi(by), if\ b \ne 0$$

Rule IC. Take $a=c=0\ and\ b=1$ in Rule I. Hence corresponding to each linear factor $(1\ \cdot D)$ , the part of C.F. is

$$\phi (y)$$

In our case,

$$(D^2+2DD'+D'^2+2D+2D'+1)z=x \implies z(x,y)=C.F+P.I$$

0 STEP: We factor the expression.

$D^2+2DD'+D'^2+2D+2D'+1=(D^2+2DD'+D'^2)+(2D+2D')+1=(D+D')^2+2\ \cdot\ (D+D')+1=(D+D')(D+D'+2)+1$

$\therefore (D^2+2DD'+D'^2+2D+2D'+1)z=((D+D')(D+D'+2)+1)z$

1 STEP: Let's find C.F.

$C.F=(C.F.)_1+(C.F.)_2+(C.F.)_3\\ \begin{cases} (D+D')z \\ (bD+aD'+c)z & -\begin{cases} b=1 \\a=1& \implies (C.F.)_1 = e(\frac{0 \cdot x}{1}) \cdot\ \phi(1\cdot y+1\cdot x)\\ c=0 \end{cases}\\ \end{cases}\\$

$(C.F.)_1 = \phi(y+x)$, where $\phi_1$ is an arbitrary function

$\begin{cases} (D+D'+2)z \\ (bD+aD'+c)z & -\begin{cases} b=1 \\a=1& \implies (C.F.)_2 = e(\frac{2 \cdot x}{1}) \cdot\ \phi(1\cdot y+1\cdot x)\\ c=2 \end{cases}\\ \end{cases}\\$$(C.F.)_2 = e^{2x}\cdot\phi_2(y+x)$, where $\phi_2$ is an arbitrary function

$\begin{cases} (1)z \\ (bD+aD'+c)z & -\begin{cases} b=0 \\a=0& \implies (C.F.)_3 = e(\frac{1 \cdot x}{1}) \cdot\ \phi(0\cdot y+0\cdot x)\\ c=1 \end{cases}\\ \end{cases}\\$$(C.F.)_3 = e^x$, where $\phi_3$ is an arbitrary function

Then

$C.F=(C.F.)_1+(C.F.)_2+(C.F.)_3\\ C.F=\phi_1(y+x)+e^{2x}\phi_2(y+x)+e^x\\$

2 STEP: Let's find P.I.

$P.I. = \frac{1}{D^2+2DD'+D'^2+2D+2D'+1} \cdot x =\frac{1}{D^2+2DD'+D'^2+2D+2D'+1}\cdot 1\cdot x= \frac{1}{(1 \cdot i)^2+2 \cdot(1 \cdot i)\cdot (2 \cdot i)+ (2 \cdot i)^2+ (2 \cdot i)^2+2D+2D'+1}\cdot 1\cdot x=\frac{1}{i^2+2 \cdot2 i^2+ 4 i^2+2D+2D'+1}\cdot x= \frac{1}{9i^2+2D+2D'+1} \cdot x =\frac{1}{-9+2D+2D'+1} \cdot x =\frac{1}{2D+2D'-8} \cdot x =\frac{1}{(2D+2D')-8} \cdot x =\frac{1\cdot (2D+2D)+8}{((2D+2D')-8)\cdot ((2D+2D')+8)} \cdot x =\frac{8+(2D+2D')}{(2D+2D')^2-64}\cdot x=\frac{8+(2D+2D')}{4D^2+8DD'+4D'^2-64}\cdot x=\frac{8+(2D+2D')}{4\cdot(1 \cdot i)^2+8(1 \cdot i)\cdot (2 \cdot i)+4(2 \cdot i)^2-64}\cdot x=\frac{8+(2D+2D')}{4 i^2+16 i^2+16 i^2-64}\cdot x=\frac{8+(2D+2D')}{36 i^2-64}\cdot x=\frac{8+(2D+2D')}{-36-64}\cdot x=\frac{8+(2D+2D')}{-100}\cdot x\\=-\frac{1}{100}(8+(2D+2D))\cdot x=-\frac{1}{100}(8+(2\cdot\frac{\delta}{\delta x}+2\cdot\frac{\delta}{\delta y}))\cdot x\\ \implies -\frac{1}{100}(2\cdot\frac{\delta}{\delta x}\cdot x+2\cdot\frac{\delta}{\delta y}\cdot x)+8 \cdot x)=-\frac{1}{100}(2+8x)=\frac{-2-8x}{100}=\frac{-2(1+4x)}{100}=\frac{-(1+4x)}{50}=\frac{4}{50}x-\frac{1}{50}$

General Solution,

$\begin{cases} z(x,y)=C.F.+P.I.=\phi_1(y+x)+e^{2x}\phi_2(y+x)+e^x+\frac{4}{50}x-\frac{1}{50} \end{cases}$

Where $\phi_1,\ \phi_2,\ and\ \phi_3$ are arbitrary functions