Answer to Question #105381 in Differential Equations for Vaishali

Question

Solve the following equation :
(1) (D^2 - 2DD' + [D']^2 )z = tan(x - y)
(2) z(p - q) = z^2 + (x + y^2)

Answer 1

Question (i)

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

1 STEP: We try to find C.F.

$$\left(D^2-2DD'+[D']^2\right)z=0$$

The auxillary equation is

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

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

$$(D-m_1D'),(D-m_2D'),\ldots,(D-m_nD')$$

then,

$$P.I.=\displaystyle\frac{1}{(D-m_1D')\cdot\ldots(D-m_nD')}F(x,y)$$

Note:

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

In our case,

$$P.I.=\frac{1}{(D-1\cdot D')(D-1\cdot D')}\tan(x-y)=\\[0.3cm] =\frac{1}{(D-1\cdot D')}\int\tan(x-(c-1x))dx=[1x+y=c]=\\[0.3cm] =\frac{1}{(D-1\cdot D')}\int\tan(2x-c)dx=\\[0.3cm] =\frac{1}{(D-1\cdot D')}\left(\frac{-1}{2}\cdot\ln|\cos(2x-c)|\right)=\\[0.3cm] =\int\left(\frac{-1}{2}\cdot\ln|\cos(2x-c)|\right)dx$$

(More information about $\int\tan(x)dx$: https://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions)

Conclusion,

$$\boxed{P.I.=-\frac{1}{2}\cdot\int\ln|\cos(2x-c)|dx}$$

General conclusion,

$$z=C.F.+P.I.\longrightarrow\\[0.3cm] \boxed{z=f_1(y+x)+x\cdot f_2(y+x)-\frac{1}{2}\cdot\int\ln|\cos(2x-c)|dx}$$

Question (ii)

$$z(p-q)=z^2+(x+y)^2\longrightarrow\\[0.3cm] zp+(-z)q=(z^2+(x+y)^2)$$

The auxillary equation is

$$\frac{dx}{z}=\frac{dy}{-z}=\frac{dz}{z^2+(x+y)^2}\\[0.3cm]$$

Then,

$$\frac{dx}{z}=\frac{dy}{-z}\longrightarrow\frac{dx}{1}=\frac{dy}{-1}\longrightarrow\\[0.3cm] dx+dy=0\longrightarrow d(x+y)=0\longrightarrow\\[0.3cm] \boxed{x+y=C_1}\\[0.3cm] \frac{dx}{z}=\frac{dz}{z^2+(x+y)^2}\longrightarrow dx=\frac{zdz}{z^2+C_1^2}\longrightarrow\\[0.3cm] x=\frac{1}{2}\cdot\int\frac{d(z^2)}{z^2+C_1^2}\longrightarrow x=\frac{1}{2}\cdot\ln|z^2+C_1^2|-\frac{1}{2}\ln|C_2|\longrightarrow\\[0.3cm] 2x=\ln\left|\frac{z^2+C_1^2}{C_2}\right|\longrightarrow e^{2x}=\frac{z^2+C_1^2}{C_2}\\[0.3cm] \boxed{C_2=\frac{z^2+C_1^2}{e^{2x}}}\\[0.3cm]$$

Conclusion,

$$\boxed{\text{Solution of equation is}\,\,\varphi\left((x+y),e^{-2x}(z^2+(x+y)^2)\right)=0}$$

ANSWER

Question (i)

$$z=f_1(y+x)+x\cdot f_2(y+x)-\frac{1}{2}\cdot\int\ln|\cos(2x-c)|dx$$

Question (ii)

$$\varphi\left((x+y),e^{-2x}(z^2+(x+y)^2)\right)=0$$