Answer to Question #190820 in Differential Equations for Sourav Mondal

Question

Solve : 

(D² + 5DD' + 5D'²) z = x sin (3x — 2y)

Answer 1

We have given the differential equation,

$(D^2+5DD'+5D'^2)z = xsin(3x-2y)$

The auxiliary equation is

$m^2+5m+5 = 0$

$m = -5,0$

Hence, $CF = \phi_1(y-5x) + \phi_2(y)$

Particular integral PI can be calculated as :

$PI = \dfrac{1}{(D^2+5DD'+5D'^2)}x.sin(3x-2y)$

Factorize $\phi (D,D')$ and apply these factors turn by turn

$PI = \dfrac{1}{D(D+5D')}xsin(3x-2y)$

$= \dfrac{1}{D} \int_{D+5D'}^{} xsin(3x-10x-2c_1)$ because, $y-5x = c_1$

$= \dfrac{1}{D} \int xsin(-7x-2c_1)\delta x = \dfrac{1}{D} \int \dfrac{-xcos(-7x-2c_1)}{-7}+ \dfrac{sin(-7x-2c_1)}{49}$

$= \int_{D}^{}(\dfrac{x}{7}cos(3x-2y)+ \dfrac{1}{49} sin(3x-2y))\delta x$

$= \int_{D}^{}(\dfrac{x}{7}cos(3x-2c_2)+ \dfrac{1}{49} sin(3x-2y))\delta x$ because $y= c_2$

$= \int_{D}^{}(\dfrac{x}{7}cos(3x-2c_2)\delta x+ \dfrac{1}{49} sin(3x-2y))\delta x$

Letting $a = \int_{D}^{}(\dfrac{x}{7}cos(3x-2c_2)\delta x$

$b = \dfrac{1}{49} sin(3x-2y))\delta x$

$a = \int_{D}^{}(\dfrac{x}{7}cos(3x-2c_2)\delta x = \dfrac{xsin(3x-2c_2)}{21} + \dfrac{1}{3}\dfrac{cos(3x-c_2)}{3}$

$a = \int_{D}^{}(\dfrac{x}{7}cos(3x-2c_2)\delta x = \dfrac{xsin(3x-2c_2)}{21} + \dfrac{1}{9}{cos(3x-c_2)}$

$b= \dfrac{1}{49} sin(3x-2y))\delta x$

$b = \dfrac{-cos(3x-c_2)}{49}+\dfrac{sin(3x-2c_2)}{147}$

Therefore,

$a+b = \dfrac{xsin(3x-2c_2)}{21} + \dfrac{1}{9}{cos(3x-c_2)}+ \dfrac{-cos(3x-c_2)}{49}+\dfrac{sin(3x-2c_2)}{147}$

The complete Integral $z=C.I.+P.I.$

$z = \phi_1(y-5x) + \phi_2(y)+\dfrac{xsin(3x-2c_2)}{21} + \dfrac{1}{9}{cos(3x-c_2)}+ \dfrac{-cos(3x-c_2)}{49}+\dfrac{sin(3x-2c_2)}{147}$