Answer to Question #143211 in Differential Equations for Nikhil Singh

Question

Solve
(D^2+5DD'+5D'^2)z= x.sin(3x-2y)

Answer 1

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

The auxiliary equation is

$m^2+5m+5=0 \implies \frac{-b \pm \sqrt{b^2-4ac}}{2a} =\frac{-5 \pm \sqrt{5^2-4(1)(5)}}{2(1)} \\ i.e.\ m=-5,0\\$

whence complementary function (C.F.) =

$$\phi_1(y-5x)+\phi_2(y)\\$$

$P.I. = \frac{1}{D^2+5DD'+5D'^2}x \cdot sin(3x-2y)= \begin{cases} D^2=-9 \\ DD'=6 \\ D'=-4 \end{cases}$

(i.e., on replacing $D^2$ by $a^2,\ DD'$ by $-ab, D'^2=-b^2$ provided $φ(–a2, –ab, –b2) ≠ 0$ )

Clearly it is a case of failure as $φ(–a2, –ab, –b2) = 0.$

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

$P.I.=\frac{1}{D(D+5D')}x\ sin (3x-2y)\\ =\frac{1}{D} \intop_{D+5D'}x\ sin (3x-10x-2c_1)\ \therefore\ y-5x=c_1\\ =\frac{1}{D} \intop x\ sin (-7x-2c_1)\delta x =\frac{1}{D} \intop \frac{- x\ cos (-7x-2c_1)}{-7}+\frac{1}{-7}\frac{sin (-7x-2c_1)}{-7}\\ =\frac{1}{D}\frac{x}{7}cos(3x-2y)+\frac{1}{49}sin (3x-2y)\\ = \intop_D(\frac{x}{7}cos(3x-2y)+\frac{1}{49}sin(3x-2y))\delta x\\ = \intop_D(\frac{x}{7}cos(3x-2c_2)+\frac{1}{49}sin(3x-2c_2))\delta x\ \therefore y=c_2\\$

$= \intop(\frac{x}{7}cos(3x-2c_2)\delta x+\intop\frac{1}{49}sin(3x-2c_2))\delta x\$

Letting $a= \intop(\frac{x}{7}cos(3x-2c_2)\delta x$ and $b=\intop\frac{1}{49}sin(3x-2c_2))\delta x\$

$a= \intop(\frac{x}{7}cos(3x-2c_2)\delta x =\frac{x\ sin(3x-2c_2)}{21}+\frac13 \frac{cos(3x-2c_2)}{3}\\ a\implies \frac{x}{21}sin(3x-2c_2) +\frac{1}{9}cos(3x-2c_2)\\ b= \intop(\frac{1}{49}sin(3x-2c_2)\delta x =\frac{-cos(3x-2c_2)}{49}+\frac13 \frac{sin(3x-2c_2)}{49}\\ b\implies -\frac{1}{49}cos(3x-2c_2) +\frac{1}{147}sin(3x-2c_2)\\ \therefore a+b \implies \frac{x}{21}sin(3x-2c_2) +\frac{1}{9}cos(3x-2c_2)-\frac{1}{49}cos(3x-2c_2) +\frac{1}{147}sin(3x-2c_2)\\ \implies \frac{7x}{147}sin(3x-2c_2)+\frac{1}{147}sin(3x-2c_2) +\frac{40}{441}cos(3x-2c_2)\\ P.I. =\frac{7x}{147}sin(3x-2y)+\frac{1}{147}sin(3x-2y) +\frac{40}{441}cos(3x-2y)\\$

$\therefore$ The complete Integral $z=C.I.+P.I.$

Answer

$=\phi_1(y-5x)+\phi_2(y)+\frac{7x}{147}sin(3x-2y)+\frac{1}{147}sin(3x-2y) +\frac{40}{441}cos(3x-2y)\\$