Answer to Question #105579 in Differential Equations for Gayatri Yadav

Question

Solve the following equations:
(i) (D² + DD′ + D + D′ + 1)z = 0
(ii) (D² − 2DD′+ D′²)z = tan(y + x).

Answer 1

Equation (i)

$$\left(D^2+DD'+D+D'+1\right)z=0$$

Let us give some theory to understand how to solve this equation:

Put $D=h$ and $D'=k$ in $f(D, D')=0$ , then we get an

auxillary equation is

$$f(h, k)=0 \quad\quad\quad\quad(1)$$

Solve $(1)$ and find the value of $h$ interm of k or $k$ interm of $h$ . Let the $n$ values of $h$ be denoted by $f_1(k)$ , $f_2(k)$ , $f_3(k)$ , . . . , $f_n(k)$ .

Then

$$C.F.=\sum C_1e^{f_1(k)x+ky}+\ldots+\sum C_ne^{f_n(k)x+ky}$$

In our case,

$$\left(D^2+DD'+D+D'+1\right)\equiv h^2+hk+h+k+1=0\\[0.3cm] h^2+(1+k)h+(k+1)=0-\text{quadratic equation}\\[0.3cm] \sqrt{D}=\sqrt{(k+1)^2-4(k+1)}=\sqrt{(k+1)(k-3)}\\[0.3cm] \left[\begin{array}{l} h_1=\displaystyle\frac{-(k+1)+\sqrt{(k+1)(k-3)}}{2}\\[0.3cm] h_2=\displaystyle\frac{-(k+1)-\sqrt{(k+1)(k-3)}}{2} \end{array}\right.$$

Then,

$$C.F.=\sum C_1e^{h_1(k)x+ky}+\sum C_2e^{h_2(k)x+ky}=\\[0.3cm] =\sum C_1e^{x\cdot\left(\frac{-k-1-\sqrt{(k+1)(k-3)}}{2}\right)+ky}+\sum C_2e^{x\cdot\left(\frac{-k-1+\sqrt{(k+1)(k-3)}}{2}\right)+ky}=\\[0.3cm] =e^{-x/2}\cdot\sum C_1e^{x\cdot\left(\frac{-k-\sqrt{(k+1)(k-3)}}{2}\right)+ky}+\\[0.3cm] +e^{-x/2}\cdot\sum C_2e^{x\cdot\left(\frac{-k-1+\sqrt{(k+1)(k-3)}}{2}\right)+ky}=\\[0.3cm] =e^{-x/2}\cdot\left(f_1\left(L_1x+ky\right)+f_2\left(L_2x+ky\right)\right)$$

Conclusion,

$$\left(D^2+DD'+D+D'+1\right)z=0\longrightarrow\\[0.3cm] C.F.=e^{-x/2}\cdot\left(f_1(L_1x+ky)+f_2(L_2x+ky)\right),\text{where}\\[0.3cm] \left[\begin{array}{l} L_1=\displaystyle\frac{-k+\sqrt{(k+1)(k-3)}}{2}\\[0.3cm] L_2=\displaystyle\frac{-k-\sqrt{(k+1)(k-3)}}{2} \end{array}\right.$$

Question (ii)

$$(D^2-2DD'+D'^2)z=\tan(x+y)\longrightarrow\\[0.3cm] z=C.F.+P.I.$$

1 STEP: We try to find C.F.

$$(D^2-2DD'+D'^2)z=0$$

The auxillary equation is

$$m^2-2m+1=0\longrightarrow (m-1)^2=0\longrightarrow\\[0.3cm] m_{1,2}=1$$

Then,

$$C.F.=f_1(y+1\cdot x)+x\cdot f_2(y+1\cdot x)\\[0.3cm] \boxed{C.F.=f_1(y+x)+x\cdot f_2(y+x)}$$

2 STEP: We try to find P.I.

Let us give some theory to understand how to solve this equation:

If $F(x,y)$ is any function, resolve $F(D, D')$ into linear factors say

$$(D−m_1D'),(D−m_2D'), . . . ,(D−m_nD')$$

then

$$P.I.=\frac{1}{(D−m_1D')(D−m_2D'). . .(D−m_nD')} F(x, y)$$

Note:

$$(1)\,\,\,\frac{1}{(D−mD')}F(x, y)=\int F(x, c−mx)dx,\,\,\text{where}\,\,y=c−mx\\[0.3cm] (2)\,\,\,\frac{1}{(D+mD')}F(x, y)=\int F(x, c+mx)dx,\,\,\text{where}\,\,y=c+mx$$

In our case,

$$P.I.=\frac{1}{(D-1\cdot D')^2}\tan(x+y)=\\[0.3cm] =\frac{1}{(D-1\cdot D')}\cdot\frac{1}{(D-1\cdot D')}\tan(x+y)=\\[0.3cm] =\frac{1}{(D-1\cdot D')}\cdot\int\tan(x+c-x)dx=[x+y=c]=\\[0.3cm] =\frac{1}{(D-1\cdot D')}\cdot\int\tan(c)dx=\frac{1}{(D-1\cdot D')}\cdot x\tan(c)=\\[0.3cm] =\int x\tan(c)dx=\frac{x^2}{2}\tan{c}+C_1\equiv\frac{x^2}{2}\tan(x+y)+C_1$$

Conclusion,

$$\boxed{P.I.=\frac{x^2}{2}\cdot\tan(x+y)+C_1}$$

General conclusion,

$$z(x,y)=C.F.+P.I.\longrightarrow\\[0.3cm] \boxed{z(x,y)=f_1(y+x)+x\cdot f_2(y+x)+\frac{x^2}{2}\cdot\tan(x+y)+C_1}$$

ANSWER

Question(i)

$$\left(D^2+DD'+D+D'+1\right)z=0\longrightarrow\\[0.3cm] C.F.=e^{-x/2}\cdot\left(f_1(L_1x+ky)+f_2(L_2x+ky)\right),\text{where}\\[0.3cm] \left[\begin{array}{l} L_1=\displaystyle\frac{-k+\sqrt{(k+1)(k-3)}}{2}\\[0.3cm] L_2=\displaystyle\frac{-k-\sqrt{(k+1)(k-3)}}{2} \end{array}\right.$$

Question(ii)

$$z(x,y)=f_1(y+x)+x\cdot f_2(y+x)+\frac{x^2}{2}\cdot\tan(x+y)+C_1$$