Answer to Question #289449 in Differential Equations for Pankaj

Question

Find the integral curves of the differential equation:

(D³-D'³) z=x³y³

Answer 1

The given equation can be written as

$(D^3 – D'^3)z = x^3y^3$ --------(1)

Corresponding auxiliary equation is

$m^3 – 1 = 0$ , i.e $(m – 1)(m^2 + m + 1) = 0$

or

$(m – 1)(m – w)(m – w^2) = 0,$ for w to be the cube root of unity----(2)

$\displaystyle C.F. = φ_{1}(y + x) + φ_2(y + wx) + φ_3(y + w^2x)$ -----(3)

Now

$\displaystyle p.I=\frac{1}{D^3-D^{\prime3}}x^3y^3=\frac{1}{D^3\left[1-(\frac{D\prime}{D})^3\right]}x^3y^3$

$=\frac{1}{D^3}\left[1-(\frac{D\prime}{D})^3\right]^{-1}x^3y^3=\frac{1}{D^3}\left[1+\frac{D^\prime3}{D}+...\right]x^3y^3$

$=\frac{1}{D^3}x^3y^3+\frac{D^\prime3}{D^6}x^3y^3$

(on taking $\frac{D^\prime}{D}$  to the powers to which y is appearing in

$F(x, y) = x^ny^m$ i.e neglecting terms containing powers more than 3 in $D^\prime$

$\displaystyle =\frac{x^6y^3}{4\cdot5\cdot6}+\frac{6x^9}{4\cdot5\cdot6\cdot7\cdot8\cdot9}$

$P.I\displaystyle =\frac{x^6y^3}{ 120}+\frac{x^9}{ 10080}$

Hence complete solution 

$\displaystyle z=φ_{1}(y+x)+φ_{2}(y+wx)+φ_{3}(y+w^2x)+\frac{x^6y^3}{ 120}+\frac{x^9}{ 10080}$